hypothesis testing one sample

Statistics Made Easy

One Sample t-test: Definition, Formula, and Example

A  one sample t-test  is used to test whether or not the mean of a population is equal to some value.

This tutorial explains the following:

• The motivation for performing a one sample t-test.
• The formula to perform a one sample t-test.
• The assumptions that should be met to perform a one sample t-test.
• An example of how to perform a one sample t-test.

One Sample t-test: Motivation

Suppose we want to know whether or not the mean weight of a certain species of turtle in Florida is equal to 310 pounds. Since there are thousands of turtles in Florida, it would be extremely time-consuming and costly to go around and weigh each individual turtle.

Instead, we might take a simple random sample of 40 turtles and use the mean weight of the turtles in this sample to estimate the true population mean:

However, it’s virtually guaranteed that the mean weight of turtles in our sample will differ from 310 pounds. The question is whether or not this difference is statistically significant . Fortunately, a one sample t-test allows us to answer this question.

One Sample t-test: Formula

A one-sample t-test always uses the following null hypothesis:

• H 0 :  μ = μ 0 (population mean is equal to some hypothesized value μ 0 )

The alternative hypothesis can be either two-tailed, left-tailed, or right-tailed:

• H 1 (two-tailed):  μ ≠ μ 0 (population mean is not equal to some hypothesized value μ 0 )
• H 1 (left-tailed):  μ < μ 0 (population mean is less than some hypothesized value μ 0 )
• H 1 (right-tailed):  μ > μ 0 (population mean is greater than some hypothesized value μ 0 )

We use the following formula to calculate the test statistic t:

t = ( x  – μ) / (s/√ n )

• x : sample mean
• μ 0 : hypothesized population mean
• s:  sample standard deviation
• n:  sample size

If the p-value that corresponds to the test statistic t with (n-1) degrees of freedom is less than your chosen significance level (common choices are 0.10, 0.05, and 0.01) then you can reject the null hypothesis.

One Sample t-test: Assumptions

For the results of a one sample t-test to be valid, the following assumptions should be met:

• The variable under study should be either an interval or ratio variable .
• The observations in the sample should be independent .
• The variable under study should be approximately normally distributed.  You can check this assumption by creating a histogram and visually checking if the distribution has roughly a “bell shape.”
• The variable under study should have no outliers. You can check this assumption by creating a boxplot and visually checking for outliers.

One Sample t-test : Example

Suppose we want to know whether or not the mean weight of a certain species of turtle is equal to 310 pounds. To test this, will perform a one-sample t-test at significance level α = 0.05 using the following steps:

Step 1: Gather the sample data.

Suppose  we collect a random sample of turtles with the following information:

• Sample size n = 40
• Sample mean weight  x  = 300
• Sample standard deviation s = 18.5

Step 2: Define the hypotheses.

We will perform the one sample t-test with the following hypotheses:

• H 0 :  μ = 310 (population mean is equal to 310 pounds)
• H 1 :  μ ≠ 310 (population mean is not equal to 310 pounds)

Step 3: Calculate the test statistic  t .

t = ( x  – μ) / (s/√ n ) = (300-310) / (18.5/√ 40 ) =  -3.4187

Step 4: Calculate the p-value of the test statistic  t .

According to the T Score to P Value Calculator , the p-value associated with t = -3.4817 and degrees of freedom = n-1 = 40-1 = 39 is  0.00149 .

Step 5: Draw a conclusion.

Since this p-value is less than our significance level α = 0.05, we reject the null hypothesis. We have sufficient evidence to say that the mean weight of this species of turtle is not equal to 310 pounds.

Note:  You can also perform this entire one sample t-test by simply using the One Sample t-test calculator .

Additional Resources

The following tutorials explain how to perform a one-sample t-test using different statistical programs:

How to Perform a One Sample t-test in Excel How to Perform a One Sample t-test in SPSS How to Perform a One Sample t-test in Stata How to Perform a One Sample t-test in R How to Conduct a One Sample t-test in Python How to Perform a One Sample t-test on a TI-84 Calculator

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Hey there. My name is Zach Bobbitt. I have a Masters of Science degree in Applied Statistics and I’ve worked on machine learning algorithms for professional businesses in both healthcare and retail. I’m passionate about statistics, machine learning, and data visualization and I created Statology to be a resource for both students and teachers alike.  My goal with this site is to help you learn statistics through using simple terms, plenty of real-world examples, and helpful illustrations.

3 Replies to “One Sample t-test: Definition, Formula, and Example”

This is very helpful. Thanks a lot.

am impressed with the note statology links give us. thanks

The sample size for t test cannot be more than 30.

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11 Hypothesis Testing with One Sample

Student learning outcomes.

By the end of this chapter, the student should be able to:

• Be able to identify and develop the null and alternative hypothesis
• Identify the consequences of Type I and Type II error.
• Be able to perform an one-tailed and two-tailed hypothesis test using the critical value method
• Be able to perform a hypothesis test using the p-value method
• Be able to write conclusions based on hypothesis tests.

Introduction

Now we are down to the bread and butter work of the statistician: developing and testing hypotheses. It is important to   put this material in a broader context so that the method by which a hypothesis is formed is understood completely. Using textbook examples often clouds the real source of statistical hypotheses.

Statistical testing is part of a much larger process known as the scientific method. This method was developed more than two centuries ago as the accepted way that new knowledge could be created. Until then, and unfortunately even today, among some, “knowledge” could be created simply by some authority saying something was so, ipso dicta . Superstition and conspiracy theories were (are?) accepted uncritically.

The scientific method, briefly, states that only by following a careful and specific process can some assertion be included in the accepted body of knowledge. This process begins with a set of assumptions upon which a theory, sometimes called a model, is built. This theory, if it has any validity, will lead to predictions; what we call hypotheses.

As an example, in Microeconomics the theory of consumer choice begins with certain assumption concerning human behavior. From these assumptions a theory of how consumers make choices using indifference curves and the budget line. This theory gave rise to a very important prediction, namely, that there was an inverse relationship between price and quantity demanded. This relationship was known as the demand curve. The negative slope of the demand curve is really just a prediction, or a hypothesis, that can be tested with statistical tools.

Unless hundreds and hundreds of statistical tests of this hypothesis had not confirmed this relationship, the so-called Law of Demand would have been discarded years ago. This is the role of statistics, to test the hypotheses of various theories to determine if they should be admitted into the accepted body of knowledge; how we understand our world. Once admitted, however, they may be later discarded if new theories come along that make better predictions.

Not long ago two scientists claimed that they could get more energy out of a process than was put in. This caused a tremendous stir for obvious reasons. They were on the cover of Time and were offered extravagant sums to bring their research work to private industry and any number of universities. It was not long until their work was subjected to the rigorous tests of the scientific method and found to be a failure. No other lab could replicate their findings. Consequently they have sunk into obscurity and their theory discarded. It may surface again when someone can pass the tests of the hypotheses required by the scientific method, but until then it is just a curiosity. Many pure frauds have been attempted over time, but most have been found out by applying the process of the scientific method.

This discussion is meant to show just where in this process statistics falls. Statistics and statisticians are not necessarily in the business of developing theories, but in the business of testing others’ theories. Hypotheses come from these theories based upon an explicit set of assumptions and sound logic. The hypothesis comes first, before any data are gathered. Data do not create hypotheses; they are used to test them. If we bear this in mind as we study this section the process of forming and testing hypotheses will make more sense.

One job of a statistician is to make statistical inferences about populations based on samples taken from the population. Confidence intervals are one way to estimate a population parameter. Another way to make a statistical inference is to make a decision about the value of a specific parameter. For instance, a car dealer advertises that its new small truck gets 35 miles per gallon, on average. A tutoring service claims that its method of tutoring helps 90% of its students get an A or a B. A company says that women managers in their company earn an average of $60,000 per year.

A statistician will make a decision about these claims. This process is called ” hypothesis testing .” A hypothesis test involves collecting data from a sample and evaluating the data. Then, the statistician makes a decision as to whether or not there is sufficient evidence, based upon analyses of the data, to reject the null hypothesis.

In this chapter, you will conduct hypothesis tests on single means and single proportions. You will also learn about the errors associated with these tests.

Null and Alternative Hypotheses

The actual test begins by considering two hypotheses . They are called the null hypothesis and the alternative hypothesis . These hypotheses contain opposing viewpoints.

Since the null and alternative hypotheses are contradictory, you must examine evidence to decide if you have enough evidence to reject the null hypothesis or not. The evidence is in the form of sample data.

Table 1 presents the various hypotheses in the relevant pairs. For example, if the null hypothesis is equal to some value, the alternative has to be not equal to that value.

NOTE                                                                             

We want to test whether the mean GPA of students in American colleges is different from 2.0 (out of 4.0). The null and alternative hypotheses are:

We want to test if college students take less than five years to graduate from college, on the average. The null and alternative hypotheses are:

Outcomes and the Type I and Type II Errors

The four possible outcomes in the table are:

Each of the errors occurs with a particular probability. The Greek letters α and β represent the probabilities.

By way of example, the American judicial system begins with the concept that a defendant is “presumed innocent”. This is the status quo and is the null hypothesis. The judge will tell the jury that they can not find the defendant guilty unless the evidence indicates guilt beyond a “reasonable doubt” which is usually defined in criminal cases as 95% certainty of guilt. If the jury cannot accept the null, innocent, then action will be taken, jail time. The burden of proof always lies with the alternative hypothesis. (In civil cases, the jury needs only to be more than 50% certain of wrongdoing to find culpability, called “a preponderance of the evidence”).

The example above was for a test of a mean, but the same logic applies to tests of hypotheses for all statistical parameters one may wish to test.

The following are examples of Type I and Type II errors.

Type I error : Frank thinks that his rock climbing equipment may not be safe when, in fact, it really is safe.

Type II error : Frank thinks that his rock climbing equipment may be safe when, in fact, it is not safe.

Notice that, in this case, the error with the greater consequence is the Type II error. (If Frank thinks his rock climbing equipment is safe, he will go ahead and use it.)

This is a situation described as “accepting a false null”.

Type I error : The emergency crew thinks that the victim is dead when, in fact, the victim is alive. Type II error : The emergency crew does not know if the victim is alive when, in fact, the victim is dead.

The error with the greater consequence is the Type I error. (If the emergency crew thinks the victim is dead, they will not treat him.)

Distribution Needed for Hypothesis Testing

Particular distributions are associated with hypothesis testing.We will perform hypotheses tests of a population mean using a normal distribution or a Student’s t -distribution. (Remember, use a Student’s t -distribution when the population standard deviation is unknown and the sample size is small, where small is considered to be less than 30 observations.) We perform tests of a population proportion using a normal distribution when we can assume that the distribution is normally distributed. We consider this to be true if the sample proportion, p ‘ , times the sample size is greater than 5 and 1- p ‘ times the sample size is also greater then 5. This is the same rule of thumb we used when developing the formula for the confidence interval for a population proportion.

Hypothesis Test for the Mean

Going back to the standardizing formula we can derive the test statistic for testing hypotheses concerning means.

This gives us the decision rule for testing a hypothesis for a two-tailed test:

P-Value Approach

Both decision rules will result in the same decision and it is a matter of preference which one is used.

One and Two-tailed Tests

The claim would be in the alternative hypothesis. The burden of proof in hypothesis testing is carried in the alternative. This is because failing to reject the null, the status quo, must be accomplished with 90 or 95 percent significance that it cannot be maintained. Said another way, we want to have only a 5 or 10 percent probability of making a Type I error, rejecting a good null; overthrowing the status quo.

Figure 5 shows the two possible cases and the form of the null and alternative hypothesis that give rise to them.

Effects of Sample Size on Test Statistic

Table 3 summarizes test statistics for varying sample sizes and population standard deviation known and unknown.

A Systematic Approach for Testing A Hypothesis

A systematic approach to hypothesis testing follows the following steps and in this order. This template will work for all hypotheses that you will ever test.

• Set up the null and alternative hypothesis. This is typically the hardest part of the process. Here the question being asked is reviewed. What parameter is being tested, a mean, a proportion, differences in means, etc. Is this a one-tailed test or two-tailed test? Remember, if someone is making a claim it will always be a one-tailed test.
• Decide the level of significance required for this particular case and determine the critical value. These can be found in the appropriate statistical table. The levels of confidence typical for the social sciences are 90, 95 and 99. However, the level of significance is a policy decision and should be based upon the risk of making a Type I error, rejecting a good null. Consider the consequences of making a Type I error.
• Take a sample(s) and calculate the relevant parameters: sample mean, standard deviation, or proportion. Using the formula for the test statistic from above in step 2, now calculate the test statistic for this particular case using the parameters you have just calculated.
• Compare the calculated test statistic and the critical value. Marking these on the graph will give a good visual picture of the situation. There are now only two situations:

a.     The test statistic is in the tail: Cannot Accept the null, the probability that this sample mean (proportion) came from the hypothesized distribution is too small to believe that it is the real home of these sample data.

b.   The test statistic is not in the tail: Cannot Reject the null, the sample data are compatible with the hypothesized population parameter.

• Reach a conclusion. It is best to articulate the conclusion two different ways. First a formal statistical conclusion such as “With a 95 % level of significance we cannot accept the null hypotheses that the population mean is equal to XX (units of measurement)”. The second statement of the conclusion is less formal and states the action, or lack of action, required. If the formal conclusion was that above, then the informal one might be, “The machine is broken and we need to shut it down and call for repairs”.

All hypotheses tested will go through this same process. The only changes are the relevant formulas and those are determined by the hypothesis required to answer the original question.

Full Hypothesis Test Examples

Tests on means.

Jeffrey, as an eight-year old, established a mean time of 16.43 seconds for swimming the 25-yard freestyle, with a standard deviation of 0.8 seconds . His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for 15 25-yard freestyle swims . For the 15 swims, Jeffrey’s mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds. Conduct a hypothesis test using a preset α = 0.05.

Solution – Example 6

Set up the Hypothesis Test:

Since the problem is about a mean, this is a test of a single population mean . Set the null and alternative hypothesis:

In this case there is an implied challenge or claim. This is that the goggles will reduce the swimming time. The effect of this is to set the hypothesis as a one-tailed test. The claim will always be in the alternative hypothesis because the burden of proof always lies with the alternative. Remember that the status quo must be defeated with a high degree of confidence, in this case 95 % confidence. The null and alternative hypotheses are thus:

For Jeffrey to swim faster, his time will be less than 16.43 seconds. The “<” tells you this is left-tailed. Determine the distribution needed:

Distribution for the test statistic:

The sample size is less than 30 and we do not know the population standard deviation so this is a t-test and the proper formula is:

Our step 2, setting the level of significance, has already been determined by the problem, .05 for a 95 % significance level. It is worth thinking about the meaning of this choice. The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.) For this case the only concern with a Type I error would seem to be that Jeffery’s dad may fail to bet on his son’s victory because he does not have appropriate confidence in the effect of the goggles.

To find the critical value we need to select the appropriate test statistic. We have concluded that this is a t-test on the basis of the sample size and that we are interested in a population mean. We can now draw the graph of the t-distribution and mark the critical value (Figure 6). For this problem the degrees of freedom are n-1, or 14. Looking up 14 degrees of freedom at the 0.05 column of the t-table we find 1.761. This is the critical value and we can put this on our graph.

Step 3 is the calculation of the test statistic using the formula we have selected.

We find that the calculated test statistic is 2.08, meaning that the sample mean is 2.08 standard deviations away from the hypothesized mean of 16.43.

Step 4 has us compare the test statistic and the critical value and mark these on the graph. We see that the test statistic is in the tail and thus we move to step 4 and reach a conclusion. The probability that an average time of 16 minutes could come from a distribution with a population mean of 16.43 minutes is too unlikely for us to accept the null hypothesis. We cannot accept the null.

Step 5 has us state our conclusions first formally and then less formally. A formal conclusion would be stated as: “With a 95% level of significance we cannot accept the null hypothesis that the swimming time with goggles comes from a distribution with a population mean time of 16.43 minutes.” Less formally, “With 95% significance we believe that the goggles improves swimming speed”

If we wished to use the p-value system of reaching a conclusion we would calculate the statistic and take the additional step to find the probability of being 2.08 standard deviations from the mean on a t-distribution. This value is .0187. Comparing this to the α-level of .05 we see that we cannot accept the null. The p-value has been put on the graph as the shaded area beyond -2.08 and it shows that it is smaller than the hatched area which is the alpha level of 0.05. Both methods reach the same conclusion that we cannot accept the null hypothesis.

Jane has just begun her new job as on the sales force of a very competitive company. In a sample of 16 sales calls it was found that she closed the contract for an average value of $108 with a standard deviation of 12 dollars. Test at 5% significance that the population mean is at least $100 against the alternative that it is less than 100 dollars. Company policy requires that new members of the sales force must exceed an average of $100 per contract during the trial employment period. Can we conclude that Jane has met this requirement at the significance level of 95%?

Solution – Example 7

STEP 1 : Set the Null and Alternative Hypothesis.

STEP 2 : Decide the level of significance and draw the graph (Figure 7) showing the critical value.

STEP 3 : Calculate sample parameters and the test statistic.

STEP 4 : Compare test statistic and the critical values

STEP 5 : Reach a Conclusion

The test statistic is a Student’s t because the sample size is below 30; therefore, we cannot use the normal distribution. Comparing the calculated value of the test statistic and the critical value of t ( t a ) at a 5% significance level, we see that the calculated value is in the tail of the distribution. Thus, we conclude that 108 dollars per contract is significantly larger than the hypothesized value of 100 and thus we cannot accept the null hypothesis. There is evidence that supports Jane’s performance meets company standards.

Again we will follow the steps in our analysis of this problem.

Solution – Example 8

STEP 1 : Set the Null and Alternative Hypothesis. The random variable is the quantity of fluid placed in the bottles. This is a continuous random variable and the parameter we are interested in is the mean. Our hypothesis therefore is about the mean. In this case we are concerned that the machine is not filling properly. From what we are told it does not matter if the machine is over-filling or under-filling, both seem to be an equally bad error. This tells us that this is a two-tailed test: if the machine is malfunctioning it will be shutdown regardless if it is from over-filling or under-filling. The null and alternative hypotheses are thus:

STEP 2 : Decide the level of significance and draw the graph showing the critical value.

This problem has already set the level of significance at 99%. The decision seems an appropriate one and shows the thought process when setting the significance level. Management wants to be very certain, as certain as probability will allow, that they are not shutting down a machine that is not in need of repair. To draw the distribution and the critical value, we need to know which distribution to use. Because this is a continuous random variable and we are interested in the mean, and the sample size is greater than 30, the appropriate distribution is the normal distribution and the relevant critical value is 2.575 from the normal table or the t-table at 0.005 column and infinite degrees of freedom. We draw the graph and mark these points (Figure 8).

STEP 3 : Calculate sample parameters and the test statistic. The sample parameters are provided, the sample mean is 7.91 and the sample variance is .03 and the sample size is 35. We need to note that the sample variance was provided not the sample standard deviation, which is what we need for the formula. Remembering that the standard deviation is simply the square root of the variance, we therefore know the sample standard deviation, s, is 0.173. With this information we calculate the test statistic as -3.07, and mark it on the graph.

STEP 4 : Compare test statistic and the critical values Now we compare the test statistic and the critical value by placing the test statistic on the graph. We see that the test statistic is in the tail, decidedly greater than the critical value of 2.575. We note that even the very small difference between the hypothesized value and the sample value is still a large number of standard deviations. The sample mean is only 0.08 ounces different from the required level of 8 ounces, but it is 3 plus standard deviations away and thus we cannot accept the null hypothesis.

Three standard deviations of a test statistic will guarantee that the test will fail. The probability that anything is within three standard deviations is almost zero. Actually it is 0.0026 on the normal distribution, which is certainly almost zero in a practical sense. Our formal conclusion would be “ At a 99% level of significance we cannot accept the hypothesis that the sample mean came from a distribution with a mean of 8 ounces” Or less formally, and getting to the point, “At a 99% level of significance we conclude that the machine is under filling the bottles and is in need of repair”.

Media Attributions

• Type1Type2Error
• HypTestFig2
• HypTestFig3
• HypTestPValue
• OneTailTestFig5
• HypTestExam7
• HypTestExam8

Quantitative Analysis for Business Copyright © by Margo Bergman is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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The One-Sample t -Test

What is the one-sample t -test.

The one-sample t-test is a statistical hypothesis test used to determine whether an unknown population mean is different from a specific value.

When can I use the test?

You can use the test for continuous data. Your data should be a random sample from a normal population.

What if my data isn’t nearly normally distributed?

If your sample sizes are very small, you might not be able to test for normality. You might need to rely on your understanding of the data. When you cannot safely assume normality, you can perform a nonparametric test that doesn’t assume normality.

Using the one-sample t -test

See how to perform a one-sample t -test using statistical software.

• Download JMP to follow along using the sample data included with the software.
• To see more JMP tutorials, visit the JMP Learning Library .

The sections below discuss what we need for the test, checking our data, performing the test, understanding test results and statistical details.

What do we need?

For the one-sample t -test, we need one variable.

We also have an idea, or hypothesis, that the mean of the population has some value. Here are two examples:

• A hospital has a random sample of cholesterol measurements for men. These patients were seen for issues other than cholesterol. They were not taking any medications for high cholesterol. The hospital wants to know if the unknown mean cholesterol for patients is different from a goal level of 200 mg.
• We measure the grams of protein for a sample of energy bars. The label claims that the bars have 20 grams of protein. We want to know if the labels are correct or not.

One-sample t -test assumptions

For a valid test, we need data values that are:

• Independent (values are not related to one another).
• Continuous.
• Obtained via a simple random sample from the population.

Also, the population is assumed to be normally distributed .

One-sample t -test example

Imagine we have collected a random sample of 31 energy bars from a number of different stores to represent the population of energy bars available to the general consumer. The labels on the bars claim that each bar contains 20 grams of protein.

Table 1: Grams of protein in random sample of energy bars

If you look at the table above, you see that some bars have less than 20 grams of protein. Other bars have more. You might think that the data support the idea that the labels are correct. Others might disagree. The statistical test provides a sound method to make a decision, so that everyone makes the same decision on the same set of data values. 

Checking the data

Let’s start by answering: Is the t -test an appropriate method to test that the energy bars have 20 grams of protein ? The list below checks the requirements for the test.

• The data values are independent. The grams of protein in one energy bar do not depend on the grams in any other energy bar. An example of dependent values would be if you collected energy bars from a single production lot. A sample from a single lot is representative of that lot, not energy bars in general.
• The data values are grams of protein. The measurements are continuous.
• We assume the energy bars are a simple random sample from the population of energy bars available to the general consumer (i.e., a mix of lots of bars).
• We assume the population from which we are collecting our sample is normally distributed, and for large samples, we can check this assumption.

We decide that the t -test is an appropriate method.

Before jumping into analysis, we should take a quick look at the data. The figure below shows a histogram and summary statistics for the energy bars.

From a quick look at the histogram, we see that there are no unusual points, or outliers . The data look roughly bell-shaped, so our assumption of a normal distribution seems reasonable.

From a quick look at the statistics, we see that the average is 21.40, above 20. Does this  average from our sample of 31 bars invalidate the label's claim of 20 grams of protein for the unknown entire population mean? Or not?

How to perform the one-sample t -test

For the t -test calculations we need the mean, standard deviation and sample size. These are shown in the summary statistics section of Figure 1 above.

We round the statistics to two decimal places. Software will show more decimal places, and use them in calculations. (Note that Table 1 shows only two decimal places; the actual data used to calculate the summary statistics has more.)

We start by finding the difference between the sample mean and 20:

$ 21.40-20\ =\ 1.40$

Next, we calculate the standard error for the mean. The calculation is:

Standard Error for the mean = $ \frac{s}{\sqrt{n}}= \frac{2.54}{\sqrt{31}}=0.456 $

This matches the value in Figure 1 above.

We now have the pieces for our test statistic. We calculate our test statistic as:

$ t =  \frac{\text{Difference}}{\text{Standard Error}}= \frac{1.40}{0.456}=3.07 $

To make our decision, we compare the test statistic to a value from the t- distribution. This activity involves four steps.

• We calculate a test statistic. Our test statistic is 3.07.
• We decide on the risk we are willing to take for declaring a difference when there is not a difference. For the energy bar data, we decide that we are willing to take a 5% risk of saying that the unknown population mean is different from 20 when in fact it is not. In statistics-speak, we set α = 0.05. In practice, setting your risk level (α) should be made before collecting the data.

We find the value from the t- distribution based on our decision. For a t -test, we need the degrees of freedom to find this value. The degrees of freedom are based on the sample size. For the energy bar data:

degrees of freedom = $ n - 1 = 31 - 1 = 30 $

The critical value of t with α = 0.05 and 30 degrees of freedom is +/- 2.043. Most statistics books have look-up tables for the distribution. You can also find tables online. The most likely situation is that you will use software and will not use printed tables.

We compare the value of our statistic (3.07) to the t value. Since 3.07 > 2.043, we reject the null hypothesis that the mean grams of protein is equal to 20. We make a practical conclusion that the labels are incorrect, and the population mean grams of protein is greater than 20.

Statistical details

Let’s look at the energy bar data and the 1-sample t -test using statistical terms.

Our null hypothesis is that the underlying population mean is equal to 20. The null hypothesis is written as:

$ H_o:  \mathrm{\mu} = 20 $

The alternative hypothesis is that the underlying population mean is not equal to 20. The labels claiming 20 grams of protein would be incorrect. This is written as:

$ H_a:  \mathrm{\mu} ≠ 20 $

This is a two-sided test. We are testing if the population mean is different from 20 grams in either direction. If we can reject the null hypothesis that the mean is equal to 20 grams, then we make a practical conclusion that the labels for the bars are incorrect. If we cannot reject the null hypothesis, then we make a practical conclusion that the labels for the bars may be correct.

We calculate the average for the sample and then calculate the difference with the population mean, mu:

$  \overline{x} - \mathrm{\mu} $

We calculate the standard error as:

$ \frac{s}{ \sqrt{n}} $

The formula shows the sample standard deviation as s and the sample size as n .  

The test statistic uses the formula shown below:

$  \dfrac{\overline{x} - \mathrm{\mu}} {s / \sqrt{n}} $

We compare the test statistic to a t value with our chosen alpha value and the degrees of freedom for our data. Using the energy bar data as an example, we set α = 0.05. The degrees of freedom ( df ) are based on the sample size and are calculated as:

$ df = n - 1 = 31 - 1 = 30 $

Statisticians write the t value with α = 0.05 and 30 degrees of freedom as:

$ t_{0.05,30} $

The t value for a two-sided test with α = 0.05 and 30 degrees of freedom is +/- 2.042. There are two possible results from our comparison:

• The test statistic is less extreme than the critical  t  values; in other words, the test statistic is not less than -2.042, or is not greater than +2.042. You fail to reject the null hypothesis that the mean is equal to the specified value. In our example, you would be unable to conclude that the label for the protein bars should be changed.
• The test statistic is more extreme than the critical  t  values; in other words, the test statistic is less than -2.042, or is greater than +2.042. You reject the null hypothesis that the mean is equal to the specified value. In our example, you conclude that either the label should be updated or the production process should be improved to produce, on average, bars with 20 grams of protein.

Testing for normality

The normality assumption is more important for small sample sizes than for larger sample sizes.

Normal distributions are symmetric, which means they are “even” on both sides of the center. Normal distributions do not have extreme values, or outliers. You can check these two features of a normal distribution with graphs. Earlier, we decided that the energy bar data was “close enough” to normal to go ahead with the assumption of normality. The figure below shows a normal quantile plot for the data, and supports our decision.

You can also perform a formal test for normality using software. The figure below shows results of testing for normality with JMP software. We cannot reject the hypothesis of a normal distribution. 

We can go ahead with the assumption that the energy bar data is normally distributed.

What if my data are not from a Normal distribution?

If your sample size is very small, it is hard to test for normality. In this situation, you might need to use your understanding of the measurements. For example, for the energy bar data, the company knows that the underlying distribution of grams of protein is normally distributed. Even for a very small sample, the company would likely go ahead with the t -test and assume normality.

What if you know the underlying measurements are not normally distributed? Or what if your sample size is large and the test for normality is rejected? In this situation, you can use a nonparametric test. Nonparametric  analyses do not depend on an assumption that the data values are from a specific distribution. For the one-sample t ­-test, the one possible nonparametric test is the Wilcoxon Signed Rank test. 

Understanding p-values

Using a visual, you can check to see if your test statistic is more extreme than a specified value in the distribution. The figure below shows a t- distribution with 30 degrees of freedom.

Since our test is two-sided and we set α = 0.05, the figure shows that the value of 2.042 “cuts off” 5% of the data in the tails combined.

The next figure shows our results. You can see the test statistic falls above the specified critical value. It is far enough “out in the tail” to reject the hypothesis that the mean is equal to 20.

Putting it all together with Software

You are likely to use software to perform a t -test. The figure below shows results for the 1-sample t -test for the energy bar data from JMP software.  

The software shows the null hypothesis value of 20 and the average and standard deviation from the data. The test statistic is 3.07. This matches the calculations above.

The software shows results for a two-sided test and for one-sided tests. We want the two-sided test. Our null hypothesis is that the mean grams of protein is equal to 20. Our alternative hypothesis is that the mean grams of protein is not equal to 20.  The software shows a p- value of 0.0046 for the two-sided test. This p- value describes the likelihood of seeing a sample average as extreme as 21.4, or more extreme, when the underlying population mean is actually 20; in other words, the probability of observing a sample mean as different, or even more different from 20, than the mean we observed in our sample. A p -value of 0.0046 means there is about 46 chances out of 10,000. We feel confident in rejecting the null hypothesis that the population mean is equal to 20.

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• One sample t test

A one sample t test compares the mean with a hypothetical value. In most cases, the hypothetical value comes from theory. For example, if you express your data as 'percent of control', you can test whether the average differs significantly from 100. The hypothetical value can also come from previous data. For example, compare whether the mean systolic blood pressure differs from 135, a value determined in a previous study.

1. Choose data entry format

Caution: Changing format will erase your data.

2. Specify the hypothetical mean value

3. enter data, 4. view the results, learn more about the one sample t test.

In this article you will learn the requirements and assumptions of a one sample t test, how to format and interpret the results of a one sample t test, and when to use different types of t tests.

One sample t test: Overview

The one sample t test, also referred to as a single sample t test, is a statistical hypothesis test used to determine whether the mean calculated from sample data collected from a single group is different from a designated value specified by the researcher. This designated value does not come from the data itself, but is an external value chosen for scientific reasons. Often, this designated value is a mean previously established in a population, a standard value of interest, or a mean concluded from other studies. Like all hypothesis testing, the one sample t test determines if there is enough evidence reject the null hypothesis (H0) in favor of an alternative hypothesis (H1). The null hypothesis for a one sample t test can be stated as: "The population mean equals the specified mean value." The alternative hypothesis for a one sample t test can be stated as: "The population mean is different from the specified mean value."

The one sample t test differs from most statistical hypothesis tests because it does not compare two separate groups or look at a relationship between two variables. It is a straightforward comparison between data gathered on a single variable from one population and a specified value defined by the researcher. The one sample t test can be used to look for a difference in only one direction from the standard value (a one-tailed t test ) or can be used to look for a difference in either direction from the standard value (a two-tailed t test ).

Requirements and Assumptions for a one sample t test

A one sample t test should be used only when data has been collected on one variable for a single population and there is no comparison being made between groups. For a valid one sample t test analysis, data values must be all of the following:

The one sample t test assumes that all "errors" in the data are independent. The term "error" refers to the difference between each value and the group mean. The results of a t test only make sense when the scatter is random - that whatever factor caused a value to be too high or too low affects only that one value. Prism cannot test this assumption, but there are graphical ways to explore data to verify this assumption is met.

A t test is only appropriate to apply in situations where data represent variables that are continuous measurements. As they rely on the calculation of a mean value, variables that are categorical should not be analyzed using a t test.

The results of a t test should be based on a random sample and only be generalized to the larger population from which samples were drawn.

As with all parametric hypothesis testing, the one sample t test assumes that you have sampled your data from a population that follows a normal (or Gaussian) distribution. While this assumption is not as important with large samples, it is important with small sample sizes, especially less than 10. If your data do not come from a Gaussian distribution , there are three options to accommodate this. One option is to transform the values to make the distribution more Gaussian, perhaps by transforming all values to their reciprocals or logarithms. Another choice is to use the Wilcoxon signed rank nonparametric test instead of the t test. A final option is to use the t test anyway, knowing that the t test is fairly robust to departures from a Gaussian distribution with large samples.

How to format a one sample t test

Ideally, data for a one sample t test should be collected and entered as a single column from which a mean value can be easily calculated. If data is entered on a table with multiple subcolumns, Prism requires one of the following choices to be selected to perform the analysis:

• Each subcolumn of data can be analyzed separately
• An average of the values in the columns across each row can be calculated, and the analysis conducted on this new stack of means, or
• All values in all columns can be treated as one sample of data (paying no attention to which row or column any values are in).

How the one sample t test calculator works

Prism calculates the t ratio by dividing the difference between the actual and hypothetical means by the standard error of the actual mean. The equation is written as follows, where x is the calculated mean, μ is the hypothetical mean (specified value), S is the standard deviation of the sample, and n is the sample size:

A p value is computed based on the calculated t ratio and the numbers of degrees of freedom present (which equals sample size minus 1). The one sample t test calculator assumes it is a two-tailed one sample t test, meaning you are testing for a difference in either direction from the specified value.

How to interpret results of a one sample t test

As discussed, a one sample t test compares the mean of a single column of numbers against a hypothetical mean. This hypothetical mean can be based upon a specific standard or other external prediction. The test produces a P value which requires careful interpretation.

The p value answers this question: If the data were sampled from a Gaussian population with a mean equal to the hypothetical value you entered, what is the chance of randomly selecting N data points and finding a mean as far (or further) from the hypothetical value as observed here?

If the p value is large (usually defined to mean greater than 0.05), the data do not give you any reason to conclude that the population mean differs from the designated value to which it has been compared. This is not the same as saying that the true mean equals the hypothetical value, but rather states that there is no evidence of a difference. Thus, we cannot reject the null hypothesis (H0).

If the p value is small (usually defined to mean less than or equal to 0.05), then it is unlikely that the discrepancy observed between the sample mean and hypothetical mean is due to a coincidence arising from random sampling. There is evidence to reject the idea that the difference is coincidental and conclude instead that the population has a mean that is different from the hypothetical value to which it has been compared. The difference is statistically significant, and the null hypothesis is therefore rejected.

If the null hypothesis is rejected, the question of whether the difference is scientifically important still remains. The confidence interval can be a useful tool in answering this question. Prism reports the 95% confidence interval for the difference between the actual and hypothetical mean. In interpreting these results, one can be 95% sure that this range includes the true difference. It requires scientific judgment to determine if this difference is truly meaningful.

Performing t tests? We can help.

Sign up for more information on how to perform t tests and other common statistical analyses.

When to use different types of t tests

There are three types of t tests which can be used for hypothesis testing:

• Independent two-sample (or unpaired) t test
• Paired sample t test

As described, a one sample t test should be used only when data has been collected on one variable for a single population and there is no comparison being made between groups. It only applies when the mean value for data is intended to be compared to a fixed and defined number.

In most cases involving data analysis, however, there are multiple groups of data either representing different populations being compared, or the same population being compared at different times or conditions. For these situations, it is not appropriate to use a one sample t test. Other types of t tests are appropriate for these specific circumstances:

Independent Two-Sample t test (Unpaired t test)

The independent sample t test, also referred to as the unpaired t test, is used to compare the means of two different samples. The independent two-sample t test comes in two different forms:

• the standard Student's t test, which assumes that the variance of the two groups are equal.
• the Welch's t test , which is less restrictive compared to the original Student's test. This is the test where you do not assume that the variance is the same in the two groups, which results in fractional degrees of freedom.

The two methods give very similar results when the sample sizes are equal and the variances are similar.

Paired Sample t test

The paired sample t test is used to compare the means of two related groups of samples. Put into other words, it is used in a situation where you have two values (i.e., a pair of values) for the same group of samples. Often these two values are measured from the same samples either at two different times, under two different conditions, or after a specific intervention.

You can perform multiple independent two-sample comparison tests simultaneously in Prism. Select from parametric and nonparametric tests and specify if the data are unpaired or paired. Try performing a t test with a 30-day free trial of Prism .

Watch this video to learn how to choose between a paired and unpaired t test.

Example of how to apply the appropriate t test

"Alkaline" labeled bottled drinking water has become fashionable over the past several years. Imagine we have collected a random sample of 30 bottles of "alkaline" drinking water from a number of different stores to represent the population of "alkaline" bottled water for a particular brand available to the general consumer. The labels on each of the bottles claim that the pH of the "alkaline" water is 8.5. A laboratory then proceeds to measure the exact pH of the water in each bottle.

Table 1: pH of water in random sample of "alkaline bottled water"

If you look at the table above, you see that some bottles have a pH measured to be lower than 8.5, while other bottles have a pH measured to be higher. What can the data tell us about the actual pH levels found in this brand of "alkaline" water bottles marketed to the public as having a pH of 8.5? Statistical hypothesis testing provides a sound method to evaluate this question. Which specific test to use, however, depends on the specific question being asked.

Is a t test appropriate to apply to this data?

Let's start by asking: Is a t test an appropriate method to analyze this set of pH data? The following list reviews the requirements and assumptions for using a t test:

• Independent sampling : In an independent sample t test, the data values are independent. The pH of one bottle of water does not depend on the pH of any other water bottle. (An example of dependent values would be if you collected water bottles from a single production lot. A sample from a single lot is representative only of that lot, not of alkaline bottled water in general).
• Continuous variable : The data values are pH levels, which are numerical measurements that are continuous.
• Random sample : We assume the water bottles are a simple random sample from the population of "alkaline" water bottles produced by this brand as they are a mix of many production lots.
• Normal distribution : We assume the population from which we collected our samples has pH levels that are normally distributed. To verify this, we should visualize the data graphically. The figure below shows a histogram for the pH measurements of the water bottles. From a quick look at the histogram, we see that there are no unusual points, or outliers. The data look roughly bell-shaped, so our assumption of a normal distribution seems reasonable. The QQ plot can also be used to graphically assess normality and is the preferred choice when the sample size is small.

Based upon these features and assumptions being met, we can conclude that a t test is an appropriate method to be applied to this set of data.

Which t test is appropriate to use?

The next decision is which t test to apply, and this depends on the exact question we would like our analysis to answer. This example illustrates how each type of t test could be chosen for a specific analysis, and why the one sample t test is the correct choice to determine if the measured pH of the bottled water samples match the advertised pH of 8.5.

We could be interested in determining whether a certain characteristic of a water bottle is associated with having a higher or lower pH, such as whether bottles are glass or plastic. For this questions, we would effectively be dividing the bottles into 2 separate groups and comparing the means of the pH between the 2 groups. For this analysis, we would elect to use a two sample t test because we are comparing the means of two independent groups.

We could also be interested in learning if pH is affected by a water bottle being opened and exposed to the air for a week. In this case, each original sample would be tested for pH level after a week had elapsed and the water had been exposed to the air, creating a second set of sample data. To evaluate whether this exposure affected pH, we would again be comparing two different groups of data, but this time the data are in paired samples each having an original pH measurement and a second measurement from after the week of exposure to the open air. For this analysis, it is appropriate to use a paired t test so that data for each bottle is assembled in rows, and the change in pH is considered bottle by bottle.

Returning to the original question we set out to answer-whether bottled water that is advertised to have a pH of 8.5 actually meets this claim-it is now clear that neither an independent two sample t test or a paired t test would be appropriate. In this case, all 30 pH measurements are sampled from one group representing bottled drinking water labeled "alkaline" available to the general consumer. We wish to compare this measured mean with an expected advertised value of 8.5. This is the exact situation for which one should employ a one sample t test!

From a quick look at the descriptive statistics, we see that the mean of the sample measurements is 8.513, slightly above 8.5. Does this average from our sample of 30 bottles validate the advertised claim of pH 8.5? By applying Prism's one sample t test analysis to this data set, we will get results by which we can evaluate whether the null hypothesis (that there is no difference between the mean pH level in the water bottles and the pH level advertised on the bottles) should be accepted or rejected.

How to Perform a One Sample T Test in Prism

In prior versions of Prism, the one sample t test and the Wilcoxon rank sum tests were computed as part of Prism's Column Statistics analysis. Now, starting with Prism 8, performing one sample t tests is even easier with a separate analysis in Prism.

Steps to perform a one sample t test in Prism

• Create a Column data table.
• Enter each data set in a single Y column so all values from each group are stacked into a column. Prism will perform a one sample t test (or Wilcoxon rank sum test) on each column you enter.
• Click Analyze, look in the list of Column analyses, and choose one sample t test and Wilcoxon test.

It's that simple! Prism streamlines your t test analysis so you can make more accurate and more informed data interpretations. Start your 30-day free trial of Prism and try performing your first one sample t test in Prism.

Watch this video for a step-by-step tutorial on how to perform a t test in Prism.

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Statistics and probability

Course: statistics and probability   >   unit 12, hypothesis testing and p-values.

• One-tailed and two-tailed tests
• Z-statistics vs. T-statistics
• Small sample hypothesis test
• Large sample proportion hypothesis testing

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8.6: Hypothesis Test of a Single Population Mean with Examples

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Steps for performing Hypothesis Test of a Single Population Mean

Step 1: State your hypotheses about the population mean. Step 2: Summarize the data. State a significance level. State and check conditions required for the procedure

• Find or identify the sample size, n, the sample mean, \(\bar{x}\) and the sample standard deviation, s .

The sampling distribution for the one-mean test statistic is, approximately, T- distribution if the following conditions are met

• Sample is random with independent observations .
• Sample is large. The population must be Normal or the sample size must be at least 30.

Step 3: Perform the procedure based on the assumption that \(H_{0}\) is true

• Find the Estimated Standard Error: \(SE=\frac{s}{\sqrt{n}}\).
• Compute the observed value of the test statistic: \(T_{obs}=\frac{\bar{x}-\mu_{0}}{SE}\).
• Check the type of the test (right-, left-, or two-tailed)
• Find the p-value in order to measure your level of surprise.

Step 4: Make a decision about \(H_{0}\) and \(H_{a}\)

• Do you reject or not reject your null hypothesis?

Step 5: Make a conclusion

• What does this mean in the context of the data?

The following examples illustrate a left-, right-, and two-tailed test.

Example \(\pageindex{1}\).

\(H_{0}: \mu = 5, H_{a}: \mu < 5\)

Test of a single population mean. \(H_{a}\) tells you the test is left-tailed. The picture of the \(p\)-value is as follows:

Exercise \(\PageIndex{1}\)

\(H_{0}: \mu = 10, H_{a}: \mu < 10\)

Assume the \(p\)-value is 0.0935. What type of test is this? Draw the picture of the \(p\)-value.

left-tailed test

Example \(\PageIndex{2}\)

\(H_{0}: \mu \leq 0.2, H_{a}: \mu > 0.2\)

This is a test of a single population proportion. \(H_{a}\) tells you the test is right-tailed . The picture of the p -value is as follows:

Exercise \(\PageIndex{2}\)

\(H_{0}: \mu \leq 1, H_{a}: \mu > 1\)

Assume the \(p\)-value is 0.1243. What type of test is this? Draw the picture of the \(p\)-value.

right-tailed test

Example \(\PageIndex{3}\)

\(H_{0}: \mu = 50, H_{a}: \mu \neq 50\)

This is a test of a single population mean. \(H_{a}\) tells you the test is two-tailed . The picture of the \(p\)-value is as follows.

Exercise \(\PageIndex{3}\)

\(H_{0}: \mu = 0.5, H_{a}: \mu \neq 0.5\)

Assume the p -value is 0.2564. What type of test is this? Draw the picture of the \(p\)-value.

two-tailed test

Full Hypothesis Test Examples

Example \(\pageindex{4}\).

Statistics students believe that the mean score on the first statistics test is 65. A statistics instructor thinks the mean score is higher than 65. He samples ten statistics students and obtains the scores 65 65 70 67 66 63 63 68 72 71. He performs a hypothesis test using a 5% level of significance. The data are assumed to be from a normal distribution.

Set up the hypothesis test:

A 5% level of significance means that \(\alpha = 0.05\). This is a test of a single population mean .

\(H_{0}: \mu = 65  H_{a}: \mu > 65\)

Since the instructor thinks the average score is higher, use a "\(>\)". The "\(>\)" means the test is right-tailed.

Determine the distribution needed:

Random variable: \(\bar{X} =\) average score on the first statistics test.

Distribution for the test: If you read the problem carefully, you will notice that there is no population standard deviation given . You are only given \(n = 10\) sample data values. Notice also that the data come from a normal distribution. This means that the distribution for the test is a student's \(t\).

Use \(t_{df}\). Therefore, the distribution for the test is \(t_{9}\) where \(n = 10\) and \(df = 10 - 1 = 9\).

The sample mean and sample standard deviation are calculated as 67 and 3.1972 from the data.

Calculate the \(p\)-value using the Student's \(t\)-distribution:

\[t_{obs} = \dfrac{\bar{x}-\mu_{\bar{x}}}{\left(\dfrac{s}{\sqrt{n}}\right)}=\dfrac{67-65}{\left(\dfrac{3.1972}{\sqrt{10}}\right)}\]

Use the T-table or Excel's t_dist() function to find p-value:

\(p\text{-value} = P(\bar{x} > 67) =P(T >1.9782 )= 1-0.9604=0.0396\)

Interpretation of the p -value: If the null hypothesis is true, then there is a 0.0396 probability (3.96%) that the sample mean is 65 or more.

Compare \(\alpha\) and the \(p-\text{value}\):

Since \(α = 0.05\) and \(p\text{-value} = 0.0396\). \(\alpha > p\text{-value}\).

Make a decision: Since \(\alpha > p\text{-value}\), reject \(H_{0}\).

This means you reject \(\mu = 65\). In other words, you believe the average test score is more than 65.

Conclusion: At a 5% level of significance, the sample data show sufficient evidence that the mean (average) test score is more than 65, just as the math instructor thinks.

The \(p\text{-value}\) can easily be calculated.

Put the data into a list. Press STAT and arrow over to TESTS . Press 2:T-Test . Arrow over to Data and press ENTER . Arrow down and enter 65 for \(\mu_{0}\), the name of the list where you put the data, and 1 for Freq: . Arrow down to \(\mu\): and arrow over to \(> \mu_{0}\). Press ENTER . Arrow down to Calculate and press ENTER . The calculator not only calculates the \(p\text{-value}\) (p = 0.0396) but it also calculates the test statistic ( t -score) for the sample mean, the sample mean, and the sample standard deviation. \(\mu > 65\) is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with \(t = 1.9781\) (test statistic) and \(p = 0.0396\) (\(p\text{-value}\)). Make sure when you use Draw that no other equations are highlighted in \(Y =\) and the plots are turned off.

Exercise \(\PageIndex{4}\)

It is believed that a stock price for a particular company will grow at a rate of $5 per week with a standard deviation of $1. An investor believes the stock won’t grow as quickly. The changes in stock price is recorded for ten weeks and are as follows: $4, $3, $2, $3, $1, $7, $2, $1, $1, $2. Perform a hypothesis test using a 5% level of significance. State the null and alternative hypotheses, find the p -value, state your conclusion, and identify the Type I and Type II errors.

• \(H_{0}: \mu = 5\)
• \(H_{a}: \mu < 5\)
• \(p = 0.0082\)

Because \(p < \alpha\), we reject the null hypothesis. There is sufficient evidence to suggest that the stock price of the company grows at a rate less than $5 a week.

• Type I Error: To conclude that the stock price is growing slower than $5 a week when, in fact, the stock price is growing at $5 a week (reject the null hypothesis when the null hypothesis is true).
• Type II Error: To conclude that the stock price is growing at a rate of $5 a week when, in fact, the stock price is growing slower than $5 a week (do not reject the null hypothesis when the null hypothesis is false).

Example \(\PageIndex{5}\)

The National Institute of Standards and Technology provides exact data on conductivity properties of materials. Following are conductivity measurements for 11 randomly selected pieces of a particular type of glass.

1.11; 1.07; 1.11; 1.07; 1.12; 1.08; .98; .98 1.02; .95; .95

Is there convincing evidence that the average conductivity of this type of glass is greater than one? Use a significance level of 0.05. Assume the population is normal.

Let’s follow a four-step process to answer this statistical question.

• \(H_{0}: \mu \leq 1\)
• \(H_{a}: \mu > 1\)
• Plan : We are testing a sample mean without a known population standard deviation. Therefore, we need to use a Student's-t distribution. Assume the underlying population is normal.
• Do the calculations : \(p\text{-value} ( = 0.036)\)

4. State the Conclusions : Since the \(p\text{-value} (= 0.036)\) is less than our alpha value, we will reject the null hypothesis. It is reasonable to state that the data supports the claim that the average conductivity level is greater than one.

The hypothesis test itself has an established process. This can be summarized as follows:

• Determine \(H_{0}\) and \(H_{a}\). Remember, they are contradictory.
• Determine the random variable.
• Determine the distribution for the test.
• Draw a graph, calculate the test statistic, and use the test statistic to calculate the \(p\text{-value}\). (A t -score is an example of test statistics.)
• Compare the preconceived α with the p -value, make a decision (reject or do not reject H 0 ), and write a clear conclusion using English sentences.

Notice that in performing the hypothesis test, you use \(\alpha\) and not \(\beta\). \(\beta\) is needed to help determine the sample size of the data that is used in calculating the \(p\text{-value}\). Remember that the quantity \(1 – \beta\) is called the Power of the Test . A high power is desirable. If the power is too low, statisticians typically increase the sample size while keeping α the same.If the power is low, the null hypothesis might not be rejected when it should be.

• Data from Amit Schitai. Director of Instructional Technology and Distance Learning. LBCC.
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• Data from Gallup®. Available online at www.gallup.com (accessed June 27, 2013).
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• Data from the United States Census Bureau. Available online at www.census.gov/hhes/socdemo/language/.
• Data from Toastmasters International. Available online at http://toastmasters.org/artisan/deta...eID=429&Page=1 .
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• Federal Bureau of Investigations. “Uniform Crime Reports and Index of Crime in Daviess in the State of Kentucky enforced by Daviess County from 1985 to 2005.” Available online at http://www.disastercenter.com/kentucky/crime/3868.htm (accessed June 27, 2013).
• “Foothill-De Anza Community College District.” De Anza College, Winter 2006. Available online at research.fhda.edu/factbook/DA...t_da_2006w.pdf.
• Johansen, C., J. Boice, Jr., J. McLaughlin, J. Olsen. “Cellular Telephones and Cancer—a Nationwide Cohort Study in Denmark.” Institute of Cancer Epidemiology and the Danish Cancer Society, 93(3):203-7. Available online at http://www.ncbi.nlm.nih.gov/pubmed/11158188 (accessed June 27, 2013).
• Rape, Abuse & Incest National Network. “How often does sexual assault occur?” RAINN, 2009. Available online at www.rainn.org/get-information...sexual-assault (accessed June 27, 2013).

Introduction to Statistics and Data Science

Chapter 14 hypothesis testing: one sample, 14.1 introduction and warning.

We now turn to the art of testing specific hypotheses using data. This is called Hypothesis testing . Unfortunately, hypothesis testing is probably the most abused concept in statistics. It can be very subtle and should only be used when the question being considered fits snugly into the hypothesis testing framework. We will see that very often a confidence interval is a better choice and conveys more information than is contained in a statistical hypothesis test.

14.2 A Starting Example

Lets say that we are investigating whether a large chemical company has been leaking toxic chemicals into the water supply of a town aptly called Chemical City. Let’s say that the toxic chemical is known to stunt the growth of children. Therefore, we collect a random sample of heights of 3rd Grade children in Chemical City by measuring the heights of 42 children. Lets load it into R for analysis.

We can estimate the population mean \(\mu\) for all the 3rd grade children in Chemical City using the sample mean:

Suppose that we also know from a large study that the mean height for healthy 3rd grade children is \(\mu=3.85\) feet. The question we can now ask is whether we have collected sufficient evidence that the growth of children in Chemical City has been stunted? Could the average height of the children been less than the national average in our sample by random chance?

We can now form what is called the null hypothesis \(H_0\) we wish to test.

• \(H_0\) is the hypothesis that the mean height of children in Chemical city is equal to or greater than the national average \(E[\bar{X}]=\mu \geq 3.85\)

The null hypothesis is often the boring hypothesis which doesn’t change the status quo. The opposite of the null hypothesis is called the alternative hypothesis . In this example our alternative hypothesis would be:

• \(H_a\) is the hypothesis that the mean height of children in Chemical city is less than the national average \(E[\bar{X}]< \mu\) .

As usual, our first approach will be based on simulations. We can re-sample of height data to get an idea of how variable our sample mean \(\bar{X}\) will be for children in Chemical City.

Therefore, we can say that their is a small (about 1%) chance that our sample produced a smaller result than the national average by random chance.

Figure 14.1: The red area shows those times we found a sample mean greater than or equal to the national average. Notice this distribution is mound shaped (normally distributed) this is the CLT at work again.

14.3 The t.test command: Hypothesis Tests for the Population Mean \(\mu\)

For large enough \(n \geq 30\) samples sizes (the same conditions for using the t.test command for confidence intervals) we can also do hypothesis testing using the t.test command for a hypothesis concerning the population mean . For example, here is the command for doing the Chemical City problem.

The first parameter is the data set, the second is the null hypothesis value for the mean and the third option tells R that our alternative hypothesis is that the mean is less than 3.85. We will see examples of the other options for this. Notice that R will give you a sentence in the output of the t.test spelling out what the alternative hypothesis is. Also, notice the line which gives us the p.value .

The p-value may be used to determine how well the data align with results expected when the null hypothesis is true. It is a measurement of how compatible the data are with the null hypothesis. How likely is the effect observed in the sample data if the null hypothesis is true. The p value gives the probability of observing our data as extreme (or more extreme) assuming the null hypothesis is true.

• High p-values The data are likely when the null is true
• Low p-values The data are unlikely under the null hypothesis.

A low p-value implies we can reject the null hypothesis . In this case we can see that the p.value is pretty small \(p\approx0.01\) so we have strong evidence against the null hypothesis in our data. Therefore, we might reject the hypothesis that the children of Chemical City have average or greater heights. Notice that we haven’t “proved” the alternative hypothesis conclusively, rather we have found it is unlikely that the null hypothesis holds in light of the evidence present in the sample.

The one sample t test is making use of the central limit theorem under the hood. Therefore it comes with the same limitations as the CLT.

• Only works for the hypothesis tests concerning the mean
• Need sample sizes of at least 30 to be safe
• If we apply this with smaller samples, we are assuming that the underlying population distribution is roughly mound shaped. Generally, we have no way to verify this.

The statements above might remind you of a conditional probability statement. In fact we can think of the p value as the conditional probability \[P(D|H_0),\] that is the probability of observing the data given that the null hypothesis is true.

14.4 Theory of Hypothesis Testing

We have already seen the main ingredients of statistical hypothesis testing. However, here are the basic steps for single sample hypothesis tests for the population mean \(\mu\) :

Select the alternative hypothesis as that which the sampling experiment is intended to establish. The alternative hypothesis will take one of the following forms:

• One-tailed, upper tailed (e.g \(H_a: \mu > 2400\) )
• One tailed, lower tailed (e.g \(H_a: \mu < 2400\) )
• Two-tailed (e.g \(H_a \neq 2400\) )

Select the null hypothesis \(H_0\) as the status quo. It will be the opposite of the alternative hypothesis, although if the alternative is one-tailed we can fix the value at the closest value to the alternative hypothesis.

Run the test in R using the t.test command.

• Interpret the results using the p.value. If p is small we may reject the null hypothesis , if it is large we say that we retain the null hypothesis , or fail to reject the null hypothesis . More on the reasons for this jargon later, although I will say that using the correct jargon is very important to avoiding misinterpretations of hypothesis testing results.

For each of the following situations form the null and alternative hypothesis and give the options for the command in R .

14.5 Under the Hood (t tests)

Suppose we have designed a questionnaire system (SMARTY PANTS) to select people with higher than average IQs. The IQ test by definition gives the average person a score of 100. To evaluate this system we take a random sample of 50 people which SMARTY PANTS tells us have a higher than average IQ and have them complete an actual IQ test.

We find that the mean IQ score of our SMARTY PANTS selected people is \(m=103\) with a sample standard deviation of \(s=15\) .

This sample mean (m) is certainly greater than 100, but we know this could have occurred by chance. Perhaps we were just lucky in our samples? To evaluate this we want to conduct a statistical hypothesis test with:

• \(H_0\) : \(\mu \leq 100\) The mean of those selected people by SMARTY pants in less than or equal to 100
• \(H_a\) : \(\mu > 100\) The mean of those selected people by SMARTY is greater than 100.

To evaluate this we assume that the null hypothesis is true . As we specified above the null hypothesis here owns all values less than or equal to 100 ( \(H_0:\mu \leq 100\) ). The closest value in the null hypothesis range to our sample mean (m=103) is 100. So let’s assume that the population (true) mean of our SMARTY PANTS sample is 100.

Given this assumption we now want to calculate the probability that we would collect a sample of 50 people with a sample mean greater than or equal to what we actually observed \(m \geq 103\) ?

Well, we know from our study of sampling that the sampling distribution for the sample mean \(\bar{x}\) will be approximately normally distributed (for samples with at least 30) with a mean= \(\mu\) and a standard deviation given by the standard error \(\sigma_{\bar{x}}=s/\sqrt{N}\) . Notice this statement is using the central limit theorem.

Figure 14.2 shows the sampling distribution assuming the null hypothesis is true.

Figure 14.2: Sampling distribution under the assumption of the null hypothesis. Red area shows the probability of observing a sample mean greater than or equal to 103.

Recall, we can compute the Z score for normally distributed values to get an idea of how extreme they are. \[Z=\frac{D-\mu}{\sigma}=\frac{m-\mu}{\frac{s}{\sqrt{N}}}\]

This tells us that the results we observed in our data are 1.4142136 standard deviations above the mean. Thus, the data we observed is not particularly extreme under the null hypothesis. We can find the probability we observe a Z score greater than or equal to 0.2 using pnorm :

This is in fact the p value for this statistical hypothesis test. The Z value is called the test statistic in this case.

To be more precise we should really use the student t distribution instead of the assuming a normal distribution for our samples. Recall, we use the student t distribution when we have used the data to estimate both the sample mean and the population standard deviation . This is almost always the case for real data sets.

Using the student t distribution in R is pretty easy we just switch pnorm for pt and specify the second parameter for a student t distribution (called the degree of freedom). The degrees of freedom for our t distribution is just the number of samples (100) minus 1.

Notice, this is essentially the same thing as what we found with the normal approximation above. This is because the student t distribution becomes closer and closer to a normal distribution for large sample sizes.

In summary when we use a t.test command in R . This is what R computes for us:

The test statistic (t): \[t=\frac{m-\mu}{\frac{s}{\sqrt{N}}}\]

• \(m\) is the sample mean
• \(\mu\) is the mean under the null hypothesis
• \(s\) is the sample standard deviation
• \(N\) is the sample size.

The p value using the alternative keyword:

• Greater: 1-pt(t, df=N-1)
• Less: pt(t, df=N-1)
• Two-tailed: 2*pt(abs(t), df=N-1) where abs is the absolute value of our t statistic.

14.6 Errors in Hypothesis Testing

In general two things can go wrong when we use statistical hypothesis testing (Fig. 14.3 ).

Figure 14.3: Errors in Hypothesis Testing

Type I Error (False Positive): occurs when the researcher rejects the null hypothesis in favor of the alternative hypothesis, while in fact the null hypothesis is true. The probability of committing a type I error is denoted by \(\alpha\) .

Type II Error (False Negative): occurs when the researcher retains the null hypothesis when in fact the null hypothesis is false. The probability of commuting a type II error in denoted by \(\beta\) .

An analogy that some people find helpful in understanding the two types of error is to consider a defendant in a trial.

Court Analogy for Hypothesis Tests:

• \(H_0:\) The defendant is innocent
• \(H_a:\) The defendant is guilty
• Type I Error: Convict an innocent person
• Type II Error: Let a guilty person go free

Notice that we could always set one of these error rates to zero if we wanted. For example, if I just convict everyone that comes to trial then my Type II error would be zero! I cannot let a guilty person go free if I convict everyone. On the other hand I can set my Type I error rate to zero by acquitting every case.

14.6.1 Statistical Significance ( \(\alpha\) )

We can control the \(\alpha\) or false positive rate of our statistical test by setting a significance level before we test the hypothesis (really this should be set before we even collect the data). Typically we use \(\alpha=0.05\) or \(\alpha=0.01\) which mean that we have a 5% chance or 1% chance respectively of rejecting the null hypothesis incorrectly. When the statistical test is performed then we reject the null hypothesis only if the p-value produced is less than the significance level used.

To understand this lets perform some simulations. The below code runs a t test for the null hypothesis \(\mu=1.0\) for a random sample from a normal distribution which has a mean of 1.0.

In this artificial case we know the null hypothesis is true , but due to sampling error we may occasionally get an p value which indicates that the null hypothesis should be rejected. Notice that the p.value which comes out changes every time we collect a sample and run our test.

This is a key-point the p value depends on the data we collect, which the result of random sampling, therefore p values are random variables themselves!

By setting our significance level in advance to \(\alpha=0.05\) we say that we will reject the null hypothesis if and only if \(p < 0.05\) . Lets run our simulation thousands of times to see how many times we commit a Type I error at this significance level.

We can see that we get about what we expect from this simulation. We incorrectly reject the null hypothesis about 5% of the time as expected. If we were to compute the p value of the test, and then decide to set our significance level we can end up with a much higher false positive rate. This is because very often we are searching for evidence to reject the null hypothesis.

If we conduct enough experiments even when their is no effect whatsoever eventually we will find convincing statistical evidence against the null hypothesis.

For example, if we conduct 100 experiments and only report the smallest p value we find then we can find an effect:

Figure 14.4: Figure downloaded from fivethirtyeight.com showing studies on food types that increase/decrease the risk for cancer

14.6.2 Type II Error

The type II (False Negative) error rate for statistical hypothesis tests is much more difficult to control than the Type I error rate. However, we can get an idea of this rate by doing some simulations where we know the alternative hypothesis is true.

Notice how large this \(\beta\) rate is! In the general \(\beta\) will usually increase as we decrease \(\alpha\) .

That is if we risk more False Negatives if we make our criterion for rejecting the null hypothesis very stringent. In the court analogy this is equivalent to saying by making the burden of proof very high we can control for the rate of false convictions, but making this criteria overly stringent will lead to an increase in the false acquittals. The best way to decrease both \(\alpha\) and \(\beta\) is to shed more light on the true nature of the population by increasing the sample size . We can control the false positive rate by choosing \(\alpha\) small, and decrease \(\beta\) by increasing the sample size.

We can see that by increasing the sample size to large samples we can get a more reasonable Type II (False Negative) error rate.

14.6.3 Practical Significance versus Statistical Significance

It is also important to note that just because we find the differences are statistically significant it doesn’t mean that they are of any practical significance. For example, suppose we wanted to examine whether an online statistics tutor increased the mean grade point average of the students. If the mean GPA before the tutoring was 3.50 we could do a upper tailed significance test to see if the GPA increased.

\(H_0: \mu=3.5\)

\(H_a: \mu > 3.5\)

Lets simulate this process but we will set the mean GPA after the tutoring to \(3.51\) .

We can see that the difference in GPAs before and after the tutoring is highly, highly significant with a tiny p value. However, if we payed 500 dollars for the tutoring a difference in 0.01 in the final GPA is of no practical significance. This means we need to pay attention to the actual differences rather than just whether the result in statistically significant. If we have large enough sample sizes we can find statistically significant differences between groups which have no practical significance. This is like the twix bar from the left and right factories.

As a matter of point if we have small sample sizes we can fail to find differences between groups as well. For example, if we had some nefarious wish to show that Turtles sizes on golf courses were no different than normal turtles, we might collect the data for only a few turtles (say 10) and then do a two tailed hypothesis test. Here is a simulation of this scenario:

We can see that the mean of the golf course turtles mass is actually different than the population mean ( \(\mu=19.0\) ), however since our sample size is so small we will in general get large p values. Therefore, we fail to reject the null hypothesis. This is why we don’t say that we “accept the null hypothesis” , because failing to reject the null hypothesis might occur because there truly is no difference OR it may occur because we didn’t collect enough data. We can never prove the null hypothesis as being true, only collect and measure (using a p value) the evidence against it.

The golf course owner might misinterpret the results of the statistical test and report that the results of the turtle survey concluded that no difference existed between the golf course turtles and normal turtles.

We are careful to say we retain the null hypothesis or if you are a fan of double negative we fail to reject the null hypothesis because when we retain the null hypothesis this could happen for two reasons:

• The null hypothesis is in fact true
• We have insufficient data to reject the null hypothesis

14.7 Hypothesis Testing for Population Fraction

We have seen how to perform hypothesis testing for the population mean \(\mu\) . We can also easily do hypothesis testing for the population proportion in R . The same principles apply to this case, except now we are testing for evidence that our samples proportion \(p\) differs from a larger population.

14.7.1 Example:

Suppose we flip a coin 20 times and obtain 15 heads and 5 tails. Can we conclude that the coin is biased? Here are alternative hypothesis is that \(H_a: p \neq 0.5\) and therefore the null hypothesis is \(H_0=0.5\) . To run this test in R we use the prop.test command:

You will notice that the syntax for this command is similar to the t.test hypothesis test. Here the first entry is the number of successes in our sample, the second entry is the total number of trials in the sample, p sets the null hypothesis value (like mu in t.test ), and the alternative option is the same as the t.test .

The conditions for performing a hypothesis test for a population proportion are the same as forming a confidence interval. We need….

• At least five sucesses in our data set
• At least five failures in our data set

14.8 Hypothesis Testing in Linear Regression

We have seen some t tests being performed already in this class when we learned about linear regression. Let’s consider the simple example using the alligator data set. Recall this data set has two columns lnLength and lnWeight which gives the snout length of alligators against their weight. When plotting these two variable together we get a nice linear relationship in the scatter plot.

Therefore, we are led to doing a linear regression analysis of the relationship between these variables. From this analysis we get the familiar summary.

However, now that we know a bit about hypothesis testing take a look at the last two columns in the Coefficients section. One column has t values and the next column gives us p values (3.08e-10, 1.49e-12). These p values are the result of performing a two tailed t test on the coefficients. These tell us if we have sufficient evidence to reject the null hypothesis that the coefficient is equal to zero.

The other p value that you can see in this summary is on the very last line. This p value is produced by a new type of hypothesis test (rather inventively called an F-test). An F-test for a regression model allows us to compare the fit of our model against a model with all the slopes set to zero. The null hypothesis of the F-test is that a model with no slope \(y=\alpha\) can explain our data just as well as our linear model \(y=\beta x + \alpha\) .

In this simple case notice the F-test is redundant with the t test for the slope. In fact they give the exact same p value. In multiple regression models the F-test won’t be redundant with one of the slope t tests and gives us a way to assess the overall fit of our model.

14.9 Power of a Statistical Test

The final note for this section is to define the power of a statistical hypothesis test. The power of a test range from 0 to 1, with 1 being the most powerful test. The statistical power of a hypothesis test is given by \[\text{Power}=1-\beta\] where \(\beta\) is the Type II (False Negative) error rate. The power of a test a measurement of the ability of the test to distinguish truth from illusions. We saw that if we collect only a small amount of data, then we have very little evidence from which to draw our conclusions. Therefore, our statistical test has little power in this case. In addition, since collecting more data will in general decrease our Type II error rate we can increase the power of any test by collecting more data.

In Statistics much effort is devoted to finding the most powerful test available for a given scenario. Generally, to increase the power of a test without collecting more data we will need to add some assumptions about the data. If these are justified then all is well and we will get reduced error rates.

14.10 Homework

14.10.1 concept questions:.

• If I test 100 hypotheses using a significance level of \(\alpha=0.05\) , what is the expected number of Type I (False Positive) errors?
• Collecting larger samples lowers both the Type I and Type II error rates for a Hypothesis test (True or False?) Why?
• By testing a hypothesis many times against data, eventually I will collect a sample which allows me to reject the null hypothesis (True/False)
• The p value measures the weight of evidence against the null hypothesis present in the data. (True/False)
• The p value of a test is a random variable. (True/False)
• By collecting enough data we can prove the null hypothesis (True/False)
• Decreasing the significance parameter \(\alpha\) will decrease the risk of False Negatives (Type II errors) (True/False)
• A p value is the probability that the null hypothesis is true. (True/False)
• A p value can be thought of as a conditional probability \(P(D|H_0)\) where D is the data. (True/False)

14.10.2 Practice Problems:

For each of the scenarios below form the null and alternative hypothesis.

• We want to measure if people in Kerrville consume on average more fried food in a week than the national average of 1 pound of fried food a week.
• We would like to evaluate if a smaller fraction of people in Kerrville support banning those on the no-fly list from purchasing a gun than the national average. A national poll found that around 80% of the public supported this measure.
• You have had have collected the number of cyber attacks each week on your website during the last year. You would like to test whether there has been a change in the long-time average of \(425\) attacks per year.

14.10.3 Advanced Problems

• Within the Turtle data set conduct a hypothesis test to determine if we have sufficient evidence to conclude that Turtles which were collected on the Property “BGAD” have a mean mass greater than 2100 grams. The following R command filters the mass data to give only those Turtles collected on the BGAD property and drops any turtles whose weights were not measured.
• Using Lebron James Data Set (from the Confidence Interval homework) conduct a hypothesis test to see if we have sufficient evidence that Lebron’s true scoring average PPG exceeded 20 PPG in 2016-2017.

You suspect that an online poker game is not giving you random cards. The odds a receiving two pair in a five card hand determined randomly is 42.25%. You have played 100 hands and only started with a single pair 32 times. Do you have sufficient evidence to conclude that the website is cheating?

Explain why we are careful to say we “fail to reject the null hypothesis” or “retain the null hypothesis” versus we “accept the null hypothesis”.

Load the wine_tasting data set. Conduct a hypothesis test to determine if the mean price of a bottle of wine from France is truly greater than the average price of a bottle of wine from all countries in the data set (the mean of the price column). Report and interpret the p value obtained from your test.

Load the NBA_Salaries_2017 data set. Determine whether we have sufficient evidence to conclude that the average guaranteed salary of NBA players ( NBA_Salaries_2017$Guaranteed) exceeded 16 million in 2017. Report and interpret the p value obtained as well as justifying the use of the statistical hypothesis test chosen.

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S.3.3 hypothesis testing examples.

• Example: Right-Tailed Test
• Example: Left-Tailed Test
• Example: Two-Tailed Test

Brinell Hardness Scores

An engineer measured the Brinell hardness of 25 pieces of ductile iron that were subcritically annealed. The resulting data were:

The engineer hypothesized that the mean Brinell hardness of all such ductile iron pieces is greater than 170. Therefore, he was interested in testing the hypotheses:

H 0 : μ = 170 H A : μ > 170

The engineer entered his data into Minitab and requested that the "one-sample t -test" be conducted for the above hypotheses. He obtained the following output:

Descriptive Statistics

$\mu$: mean of Brinelli

Null hypothesis    H₀: $\mu$ = 170 Alternative hypothesis    H₁: $\mu$ > 170

The output tells us that the average Brinell hardness of the n = 25 pieces of ductile iron was 172.52 with a standard deviation of 10.31. (The standard error of the mean "SE Mean", calculated by dividing the standard deviation 10.31 by the square root of n = 25, is 2.06). The test statistic t * is 1.22, and the P -value is 0.117.

If the engineer set his significance level α at 0.05 and used the critical value approach to conduct his hypothesis test, he would reject the null hypothesis if his test statistic t * were greater than 1.7109 (determined using statistical software or a t -table):

Since the engineer's test statistic, t * = 1.22, is not greater than 1.7109, the engineer fails to reject the null hypothesis. That is, the test statistic does not fall in the "critical region." There is insufficient evidence, at the \(\alpha\) = 0.05 level, to conclude that the mean Brinell hardness of all such ductile iron pieces is greater than 170.

If the engineer used the P -value approach to conduct his hypothesis test, he would determine the area under a t n - 1 = t 24 curve and to the right of the test statistic t * = 1.22:

In the output above, Minitab reports that the P -value is 0.117. Since the P -value, 0.117, is greater than \(\alpha\) = 0.05, the engineer fails to reject the null hypothesis. There is insufficient evidence, at the \(\alpha\) = 0.05 level, to conclude that the mean Brinell hardness of all such ductile iron pieces is greater than 170.

Note that the engineer obtains the same scientific conclusion regardless of the approach used. This will always be the case.

Height of Sunflowers

A biologist was interested in determining whether sunflower seedlings treated with an extract from Vinca minor roots resulted in a lower average height of sunflower seedlings than the standard height of 15.7 cm. The biologist treated a random sample of n = 33 seedlings with the extract and subsequently obtained the following heights:

The biologist's hypotheses are:

H 0 : μ = 15.7 H A : μ < 15.7

The biologist entered her data into Minitab and requested that the "one-sample t -test" be conducted for the above hypotheses. She obtained the following output:

$\mu$: mean of Height

Null hypothesis    H₀: $\mu$ = 15.7 Alternative hypothesis    H₁: $\mu$ < 15.7

The output tells us that the average height of the n = 33 sunflower seedlings was 13.664 with a standard deviation of 2.544. (The standard error of the mean "SE Mean", calculated by dividing the standard deviation 13.664 by the square root of n = 33, is 0.443). The test statistic t * is -4.60, and the P -value, 0.000, is to three decimal places.

Minitab Note. Minitab will always report P -values to only 3 decimal places. If Minitab reports the P -value as 0.000, it really means that the P -value is 0.000....something. Throughout this course (and your future research!), when you see that Minitab reports the P -value as 0.000, you should report the P -value as being "< 0.001."

If the biologist set her significance level \(\alpha\) at 0.05 and used the critical value approach to conduct her hypothesis test, she would reject the null hypothesis if her test statistic t * were less than -1.6939 (determined using statistical software or a t -table):s-3-3

Since the biologist's test statistic, t * = -4.60, is less than -1.6939, the biologist rejects the null hypothesis. That is, the test statistic falls in the "critical region." There is sufficient evidence, at the α = 0.05 level, to conclude that the mean height of all such sunflower seedlings is less than 15.7 cm.

If the biologist used the P -value approach to conduct her hypothesis test, she would determine the area under a t n - 1 = t 32 curve and to the left of the test statistic t * = -4.60:

In the output above, Minitab reports that the P -value is 0.000, which we take to mean < 0.001. Since the P -value is less than 0.001, it is clearly less than \(\alpha\) = 0.05, and the biologist rejects the null hypothesis. There is sufficient evidence, at the \(\alpha\) = 0.05 level, to conclude that the mean height of all such sunflower seedlings is less than 15.7 cm.

Note again that the biologist obtains the same scientific conclusion regardless of the approach used. This will always be the case.

Gum Thickness

A manufacturer claims that the thickness of the spearmint gum it produces is 7.5 one-hundredths of an inch. A quality control specialist regularly checks this claim. On one production run, he took a random sample of n = 10 pieces of gum and measured their thickness. He obtained:

The quality control specialist's hypotheses are:

H 0 : μ = 7.5 H A : μ ≠ 7.5

The quality control specialist entered his data into Minitab and requested that the "one-sample t -test" be conducted for the above hypotheses. He obtained the following output:

$\mu$: mean of Thickness

Null hypothesis    H₀: $\mu$ = 7.5 Alternative hypothesis    H₁: $\mu \ne$ 7.5

The output tells us that the average thickness of the n = 10 pieces of gums was 7.55 one-hundredths of an inch with a standard deviation of 0.1027. (The standard error of the mean "SE Mean", calculated by dividing the standard deviation 0.1027 by the square root of n = 10, is 0.0325). The test statistic t * is 1.54, and the P -value is 0.158.

If the quality control specialist sets his significance level \(\alpha\) at 0.05 and used the critical value approach to conduct his hypothesis test, he would reject the null hypothesis if his test statistic t * were less than -2.2616 or greater than 2.2616 (determined using statistical software or a t -table):

Since the quality control specialist's test statistic, t * = 1.54, is not less than -2.2616 nor greater than 2.2616, the quality control specialist fails to reject the null hypothesis. That is, the test statistic does not fall in the "critical region." There is insufficient evidence, at the \(\alpha\) = 0.05 level, to conclude that the mean thickness of all of the manufacturer's spearmint gum differs from 7.5 one-hundredths of an inch.

If the quality control specialist used the P -value approach to conduct his hypothesis test, he would determine the area under a t n - 1 = t 9 curve, to the right of 1.54 and to the left of -1.54:

In the output above, Minitab reports that the P -value is 0.158. Since the P -value, 0.158, is greater than \(\alpha\) = 0.05, the quality control specialist fails to reject the null hypothesis. There is insufficient evidence, at the \(\alpha\) = 0.05 level, to conclude that the mean thickness of all pieces of spearmint gum differs from 7.5 one-hundredths of an inch.

Note that the quality control specialist obtains the same scientific conclusion regardless of the approach used. This will always be the case.

In our review of hypothesis tests, we have focused on just one particular hypothesis test, namely that concerning the population mean \(\mu\). The important thing to recognize is that the topics discussed here — the general idea of hypothesis tests, errors in hypothesis testing, the critical value approach, and the P -value approach — generally extend to all of the hypothesis tests you will encounter.

This calculator runs a one sample proportion test for a given sample data set and specified null and alternative hypotheses. In the fields below enter the sample size \(n\) and the number of scores with the trait of interest, \(f\).

Enter a value for the null hypothesis. This value should indicate the absence of an effect in your data. It must be between the values 0 and 1. Indicate whether your alternative hypothesis involves one-tail or two-tails. If it is a one-tailed test, then you need to indicate whether it is a positive (right tail) test or a negative (left tail) test.

Enter an \(\alpha\) value for the hypothesis test. This is the Type I error rate for your hypothesis test. It also determines the confidence level \(100 \times (1-\alpha)\) for a confidence interval. The confidence interval is based on the normal distribution, which is an approximation.

Press the Run Test button and a table summarizing the computations and conclusions will appear below.

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9.E: Hypothesis Testing with One Sample (Exercises)

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These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.

9.1: Introduction

9.2: null and alternative hypotheses.

Some of the following statements refer to the null hypothesis, some to the alternate hypothesis.

State the null hypothesis, \(H_{0}\), and the alternative hypothesis. \(H_{a}\), in terms of the appropriate parameter \((\mu \text{or} p)\).

• The mean number of years Americans work before retiring is 34.
• At most 60% of Americans vote in presidential elections.
• The mean starting salary for San Jose State University graduates is at least $100,000 per year.
• Twenty-nine percent of high school seniors get drunk each month.
• Fewer than 5% of adults ride the bus to work in Los Angeles.
• The mean number of cars a person owns in her lifetime is not more than ten.
• About half of Americans prefer to live away from cities, given the choice.
• Europeans have a mean paid vacation each year of six weeks.
• The chance of developing breast cancer is under 11% for women.
• Private universities' mean tuition cost is more than $20,000 per year.
• \(H_{0}: \mu = 34; H_{a}: \mu \neq 34\)
• \(H_{0}: p \leq 0.60; H_{a}: p > 0.60\)
• \(H_{0}: \mu \geq 100,000; H_{a}: \mu < 100,000\)
• \(H_{0}: p = 0.29; H_{a}: p \neq 0.29\)
• \(H_{0}: p = 0.05; H_{a}: p < 0.05\)
• \(H_{0}: \mu \leq 10; H_{a}: \mu > 10\)
• \(H_{0}: p = 0.50; H_{a}: p \neq 0.50\)
• \(H_{0}: \mu = 6; H_{a}: \mu \neq 6\)
• \(H_{0}: p ≥ 0.11; H_{a}: p < 0.11\)
• \(H_{0}: \mu \leq 20,000; H_{a}: \mu > 20,000\)

Over the past few decades, public health officials have examined the link between weight concerns and teen girls' smoking. Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls were surveyed again. Sixty-three said they smoked to stay thin. Is there good evidence that more than thirty percent of the teen girls smoke to stay thin? The alternative hypothesis is:

• \(p < 0.30\)
• \(p \leq 0.30\)
• \(p \geq 0.30\)
• \(p > 0.30\)

A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 attended the midnight showing. An appropriate alternative hypothesis is:

• \(p = 0.20\)
• \(p > 0.20\)
• \(p < 0.20\)
• \(p \leq 0.20\)

Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. The null and alternative hypotheses are:

• \(H_{0}: \bar{x} = 4.5, H_{a}: \bar{x} > 4.5\)
• \(H_{0}: \mu \geq 4.5, H_{a}: \mu < 4.5\)
• \(H_{0}: \mu = 4.75, H_{a}: \mu > 4.75\)
• \(H_{0}: \mu = 4.5, H_{a}: \mu > 4.5\)

9.3: Outcomes and the Type I and Type II Errors

State the Type I and Type II errors in complete sentences given the following statements.

• The mean number of cars a person owns in his or her lifetime is not more than ten.
• Private universities mean tuition cost is more than $20,000 per year.
• Type I error: We conclude that the mean is not 34 years, when it really is 34 years. Type II error: We conclude that the mean is 34 years, when in fact it really is not 34 years.
• Type I error: We conclude that more than 60% of Americans vote in presidential elections, when the actual percentage is at most 60%.Type II error: We conclude that at most 60% of Americans vote in presidential elections when, in fact, more than 60% do.
• Type I error: We conclude that the mean starting salary is less than $100,000, when it really is at least $100,000. Type II error: We conclude that the mean starting salary is at least $100,000 when, in fact, it is less than $100,000.
• Type I error: We conclude that the proportion of high school seniors who get drunk each month is not 29%, when it really is 29%. Type II error: We conclude that the proportion of high school seniors who get drunk each month is 29% when, in fact, it is not 29%.
• Type I error: We conclude that fewer than 5% of adults ride the bus to work in Los Angeles, when the percentage that do is really 5% or more. Type II error: We conclude that 5% or more adults ride the bus to work in Los Angeles when, in fact, fewer that 5% do.
• Type I error: We conclude that the mean number of cars a person owns in his or her lifetime is more than 10, when in reality it is not more than 10. Type II error: We conclude that the mean number of cars a person owns in his or her lifetime is not more than 10 when, in fact, it is more than 10.
• Type I error: We conclude that the proportion of Americans who prefer to live away from cities is not about half, though the actual proportion is about half. Type II error: We conclude that the proportion of Americans who prefer to live away from cities is half when, in fact, it is not half.
• Type I error: We conclude that the duration of paid vacations each year for Europeans is not six weeks, when in fact it is six weeks. Type II error: We conclude that the duration of paid vacations each year for Europeans is six weeks when, in fact, it is not.
• Type I error: We conclude that the proportion is less than 11%, when it is really at least 11%. Type II error: We conclude that the proportion of women who develop breast cancer is at least 11%, when in fact it is less than 11%.
• Type I error: We conclude that the average tuition cost at private universities is more than $20,000, though in reality it is at most $20,000. Type II error: We conclude that the average tuition cost at private universities is at most $20,000 when, in fact, it is more than $20,000.

For statements a-j in Exercise 9.109 , answer the following in complete sentences.

• State a consequence of committing a Type I error.
• State a consequence of committing a Type II error.

When a new drug is created, the pharmaceutical company must subject it to testing before receiving the necessary permission from the Food and Drug Administration (FDA) to market the drug. Suppose the null hypothesis is “the drug is unsafe.” What is the Type II Error?

• To conclude the drug is safe when in, fact, it is unsafe.
• Not to conclude the drug is safe when, in fact, it is safe.
• To conclude the drug is safe when, in fact, it is safe.
• Not to conclude the drug is unsafe when, in fact, it is unsafe.

A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing. The Type I error is to conclude that the percent of EVC students who attended is ________.

• at least 20%, when in fact, it is less than 20%.
• 20%, when in fact, it is 20%.
• less than 20%, when in fact, it is at least 20%.
• less than 20%, when in fact, it is less than 20%.

It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5%, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average?

The Type II error is not to reject that the mean number of hours of sleep LTCC students get per night is at least seven when, in fact, the mean number of hours

• is more than seven hours.
• is at most seven hours.
• is at least seven hours.
• is less than seven hours.

Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test, the Type I error is:

• to conclude that the current mean hours per week is higher than 4.5, when in fact, it is higher
• to conclude that the current mean hours per week is higher than 4.5, when in fact, it is the same
• to conclude that the mean hours per week currently is 4.5, when in fact, it is higher
• to conclude that the mean hours per week currently is no higher than 4.5, when in fact, it is not higher

9.4: Distribution Needed for Hypothesis Testing

It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5%, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average? The distribution to be used for this test is \(\bar{X} \sim\) ________________

• \(N\left(7.24, \frac{1.93}{\sqrt{22}}\right)\)
• \(N\left(7.24, 1.93\right)\)

9.5: Rare Events, the Sample, Decision and Conclusion

The National Institute of Mental Health published an article stating that in any one-year period, approximately 9.5 percent of American adults suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, seven of them suffered from depression or a depressive illness. Conduct a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult American population.

• Is this a test of one mean or proportion?
• State the null and alternative hypotheses. \(H_{0}\) : ____________________ \(H_{a}\) : ____________________
• Is this a right-tailed, left-tailed, or two-tailed test?
• What symbol represents the random variable for this test?
• In words, define the random variable for this test.
• \(x =\) ________________
• \(n =\) ________________
• \(p′ =\) _____________
• Calculate \(\sigma_{x} =\) __________. Show the formula set-up.
• State the distribution to use for the hypothesis test.
• Find the \(p\text{-value}\).
• Reason for the decision:
• Conclusion (write out in a complete sentence):

9.6: Additional Information and Full Hypothesis Test Examples

For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in [link] . Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.

If you are using a Student's \(t\) - distribution for one of the following homework problems, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, however.)

A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles. Using \(\alpha = 0.05\), is the data highly inconsistent with the claim?

• \(H_{0}: \mu \geq 50,000\)
• \(H_{a}: \mu < 50,000\)
• Let \(\bar{X} =\) the average lifespan of a brand of tires.
• normal distribution
• \(z = -2.315\)
• \(p\text{-value} = 0.0103\)
• Check student’s solution.
• alpha: 0.05
• Decision: Reject the null hypothesis.
• Reason for decision: The \(p\text{-value}\) is less than 0.05.
• Conclusion: There is sufficient evidence to conclude that the mean lifespan of the tires is less than 50,000 miles.
• \((43,537, 49,463)\)

From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant of around 2.1 years. A survey of 40 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.1 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level?

The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of 20¢. A study was done to test the claim that the mean cost of a daily newspaper is $1.00. Twelve costs yield a mean cost of 95¢ with a standard deviation of 18¢. Do the data support the claim at the 1% level?

• \(H_{0}: \mu = $1.00\)
• \(H_{a}: \mu \neq $1.00\)
• Let \(\bar{X} =\) the average cost of a daily newspaper.
• \(z = –0.866\)
• \(p\text{-value} = 0.3865\)
• \(\alpha: 0.01\)
• Decision: Do not reject the null hypothesis.
• Reason for decision: The \(p\text{-value}\) is greater than 0.01.
• Conclusion: There is sufficient evidence to support the claim that the mean cost of daily papers is $1. The mean cost could be $1.
• \(($0.84, $1.06)\)

An article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of 49 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1% level?

The mean number of sick days an employee takes per year is believed to be about ten. Members of a personnel department do not believe this figure. They randomly survey eight employees. The number of sick days they took for the past year are as follows: 12; 4; 15; 3; 11; 8; 6; 8. Let \(x =\) the number of sick days they took for the past year. Should the personnel team believe that the mean number is ten?

• \(H_{0}: \mu = 10\)
• \(H_{a}: \mu \neq 10\)
• Let \(\bar{X}\) the mean number of sick days an employee takes per year.
• Student’s t -distribution
• \(t = –1.12\)
• \(p\text{-value} = 0.300\)
• \(\alpha: 0.05\)
• Reason for decision: The \(p\text{-value}\) is greater than 0.05.
• Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean number of sick days is not ten.
• \((4.9443, 11.806)\)

In 1955, Life Magazine reported that the 25 year-old mother of three worked, on average, an 80 hour week. Recently, many groups have been studying whether or not the women's movement has, in fact, resulted in an increase in the average work week for women (combining employment and at-home work). Suppose a study was done to determine if the mean work week has increased. 81 women were surveyed with the following results. The sample mean was 83; the sample standard deviation was ten. Does it appear that the mean work week has increased for women at the 5% level?

Your statistics instructor claims that 60 percent of the students who take her Elementary Statistics class go through life feeling more enriched. For some reason that she can't quite figure out, most people don't believe her. You decide to check this out on your own. You randomly survey 64 of her past Elementary Statistics students and find that 34 feel more enriched as a result of her class. Now, what do you think?

• \(H_{0}: p \geq 0.6\)
• \(H_{a}: p < 0.6\)
• Let \(P′ =\) the proportion of students who feel more enriched as a result of taking Elementary Statistics.
• normal for a single proportion
• \(p\text{-value} = 0.1308\)
• Conclusion: There is insufficient evidence to conclude that less than 60 percent of her students feel more enriched.

The “plus-4s” confidence interval is \((0.411, 0.648)\)

A Nissan Motor Corporation advertisement read, “The average man’s I.Q. is 107. The average brown trout’s I.Q. is 4. So why can’t man catch brown trout?” Suppose you believe that the brown trout’s mean I.Q. is greater than four. You catch 12 brown trout. A fish psychologist determines the I.Q.s as follows: 5; 4; 7; 3; 6; 4; 5; 3; 6; 3; 8; 5. Conduct a hypothesis test of your belief.

Refer to Exercise 9.119 . Conduct a hypothesis test to see if your decision and conclusion would change if your belief were that the brown trout’s mean I.Q. is not four.

• \(H_{0}: \mu = 4\)
• \(H_{a}: \mu \neq 4\)
• Let \(\bar{X}\) the average I.Q. of a set of brown trout.
• two-tailed Student's t-test
• \(t = 1.95\)
• \(p\text{-value} = 0.076\)
• Reason for decision: The \(p\text{-value}\) is greater than 0.05
• Conclusion: There is insufficient evidence to conclude that the average IQ of brown trout is not four.
• \((3.8865,5.9468)\)

According to an article in Newsweek , the natural ratio of girls to boys is 100:105. In China, the birth ratio is 100: 114 (46.7% girls). Suppose you don’t believe the reported figures of the percent of girls born in China. You conduct a study. In this study, you count the number of girls and boys born in 150 randomly chosen recent births. There are 60 girls and 90 boys born of the 150. Based on your study, do you believe that the percent of girls born in China is 46.7?

A poll done for Newsweek found that 13% of Americans have seen or sensed the presence of an angel. A contingent doubts that the percent is really that high. It conducts its own survey. Out of 76 Americans surveyed, only two had seen or sensed the presence of an angel. As a result of the contingent’s survey, would you agree with the Newsweek poll? In complete sentences, also give three reasons why the two polls might give different results.

• \(H_{a}: p < 0.13\)
• Let \(P′ =\) the proportion of Americans who have seen or sensed angels
• –2.688
• \(p\text{-value} = 0.0036\)
• Reason for decision: The \(p\text{-value}\)e is less than 0.05.
• Conclusion: There is sufficient evidence to conclude that the percentage of Americans who have seen or sensed an angel is less than 13%.

The“plus-4s” confidence interval is (0.0022, 0.0978)

The mean work week for engineers in a start-up company is believed to be about 60 hours. A newly hired engineer hopes that it’s shorter. She asks ten engineering friends in start-ups for the lengths of their mean work weeks. Based on the results that follow, should she count on the mean work week to be shorter than 60 hours?

Data (length of mean work week): 70; 45; 55; 60; 65; 55; 55; 60; 50; 55.

Use the “Lap time” data for Lap 4 (see [link] ) to test the claim that Terri finishes Lap 4, on average, in less than 129 seconds. Use all twenty races given.

• \(H_{0}: \mu \geq 129\)
• \(H_{a}: \mu < 129\)
• Let \(\bar{X} =\) the average time in seconds that Terri finishes Lap 4.
• Student's t -distribution
• \(t = 1.209\)
• Conclusion: There is insufficient evidence to conclude that Terri’s mean lap time is less than 129 seconds.
• \((128.63, 130.37)\)

Use the “Initial Public Offering” data (see [link] ) to test the claim that the mean offer price was $18 per share. Do not use all the data. Use your random number generator to randomly survey 15 prices.

The following questions were written by past students. They are excellent problems!

"Asian Family Reunion," by Chau Nguyen

Every two years it comes around.

We all get together from different towns.

In my honest opinion,

It's not a typical family reunion.

Not forty, or fifty, or sixty,

But how about seventy companions!

The kids would play, scream, and shout

One minute they're happy, another they'll pout.

The teenagers would look, stare, and compare

From how they look to what they wear.

The men would chat about their business

That they make more, but never less.

Money is always their subject

And there's always talk of more new projects.

The women get tired from all of the chats

They head to the kitchen to set out the mats.

Some would sit and some would stand

Eating and talking with plates in their hands.

Then come the games and the songs

And suddenly, everyone gets along!

With all that laughter, it's sad to say

That it always ends in the same old way.

They hug and kiss and say "good-bye"

And then they all begin to cry!

I say that 60 percent shed their tears

But my mom counted 35 people this year.

She said that boys and men will always have their pride,

So we won't ever see them cry.

I myself don't think she's correct,

So could you please try this problem to see if you object?

• \(H_{0}: p = 0.60\)
• \(H_{a}: p < 0.60\)
• Let \(P′ =\) the proportion of family members who shed tears at a reunion.
• –1.71
• Reason for decision: \(p\text{-value} < \alpha\)
• Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of family members who shed tears at a reunion is less than 0.60. However, the test is weak because the \(p\text{-value}\) and alpha are quite close, so other tests should be done.
• We are 95% confident that between 38.29% and 61.71% of family members will shed tears at a family reunion. \((0.3829, 0.6171)\). The“plus-4s” confidence interval (see chapter 8) is \((0.3861, 0.6139)\)

Note that here the “large-sample” \(1 - \text{PropZTest}\) provides the approximate \(p\text{-value}\) of 0.0438. Whenever a \(p\text{-value}\) based on a normal approximation is close to the level of significance, the exact \(p\text{-value}\) based on binomial probabilities should be calculated whenever possible. This is beyond the scope of this course.

"The Problem with Angels," by Cyndy Dowling

Although this problem is wholly mine,

The catalyst came from the magazine, Time.

On the magazine cover I did find

The realm of angels tickling my mind.

Inside, 69% I found to be

In angels, Americans do believe.

Then, it was time to rise to the task,

Ninety-five high school and college students I did ask.

Viewing all as one group,

Random sampling to get the scoop.

So, I asked each to be true,

"Do you believe in angels?" Tell me, do!

Hypothesizing at the start,

Totally believing in my heart

That the proportion who said yes

Would be equal on this test.

Lo and behold, seventy-three did arrive,

Out of the sample of ninety-five.

Now your job has just begun,

Solve this problem and have some fun.

"Blowing Bubbles," by Sondra Prull

Studying stats just made me tense,

I had to find some sane defense.

Some light and lifting simple play

To float my math anxiety away.

Blowing bubbles lifts me high

Takes my troubles to the sky.

POIK! They're gone, with all my stress

Bubble therapy is the best.

The label said each time I blew

The average number of bubbles would be at least 22.

I blew and blew and this I found

From 64 blows, they all are round!

But the number of bubbles in 64 blows

Varied widely, this I know.

20 per blow became the mean

They deviated by 6, and not 16.

From counting bubbles, I sure did relax

But now I give to you your task.

Was 22 a reasonable guess?

Find the answer and pass this test!

• \(H_{0}: \mu \geq 22\)
• \(H_{a}: \mu < 22\)
• Let \(\bar{X} =\) the mean number of bubbles per blow.
• –2.667
• \(p\text{-value} = 0.00486\)
• Conclusion: There is sufficient evidence to conclude that the mean number of bubbles per blow is less than 22.
• \((18.501, 21.499)\)

"Dalmatian Darnation," by Kathy Sparling

A greedy dog breeder named Spreckles

Bred puppies with numerous freckles

The Dalmatians he sought

Possessed spot upon spot

The more spots, he thought, the more shekels.

His competitors did not agree

That freckles would increase the fee.

They said, “Spots are quite nice

But they don't affect price;

One should breed for improved pedigree.”

The breeders decided to prove

This strategy was a wrong move.

Breeding only for spots

Would wreak havoc, they thought.

His theory they want to disprove.

They proposed a contest to Spreckles

Comparing dog prices to freckles.

In records they looked up

One hundred one pups:

Dalmatians that fetched the most shekels.

They asked Mr. Spreckles to name

An average spot count he'd claim

To bring in big bucks.

Said Spreckles, “Well, shucks,

It's for one hundred one that I aim.”

Said an amateur statistician

Who wanted to help with this mission.

“Twenty-one for the sample

Standard deviation's ample:

They examined one hundred and one

Dalmatians that fetched a good sum.

They counted each spot,

Mark, freckle and dot

And tallied up every one.

Instead of one hundred one spots

They averaged ninety six dots

Can they muzzle Spreckles’

Obsession with freckles

Based on all the dog data they've got?

"Macaroni and Cheese, please!!" by Nedda Misherghi and Rachelle Hall

As a poor starving student I don't have much money to spend for even the bare necessities. So my favorite and main staple food is macaroni and cheese. It's high in taste and low in cost and nutritional value.

One day, as I sat down to determine the meaning of life, I got a serious craving for this, oh, so important, food of my life. So I went down the street to Greatway to get a box of macaroni and cheese, but it was SO expensive! $2.02 !!! Can you believe it? It made me stop and think. The world is changing fast. I had thought that the mean cost of a box (the normal size, not some super-gigantic-family-value-pack) was at most $1, but now I wasn't so sure. However, I was determined to find out. I went to 53 of the closest grocery stores and surveyed the prices of macaroni and cheese. Here are the data I wrote in my notebook:

Price per box of Mac and Cheese:

• 5 stores @ $2.02
• 15 stores @ $0.25
• 3 stores @ $1.29
• 6 stores @ $0.35
• 4 stores @ $2.27
• 7 stores @ $1.50
• 5 stores @ $1.89
• 8 stores @ 0.75.

I could see that the cost varied but I had to sit down to figure out whether or not I was right. If it does turn out that this mouth-watering dish is at most $1, then I'll throw a big cheesy party in our next statistics lab, with enough macaroni and cheese for just me. (After all, as a poor starving student I can't be expected to feed our class of animals!)

• \(H_{0}: \mu \leq 1\)
• \(H_{a}: \mu > 1\)
• Let \(\bar{X} =\) the mean cost in dollars of macaroni and cheese in a certain town.
• Student's \(t\)-distribution
• \(t = 0.340\)
• \(p\text{-value} = 0.36756\)
• Conclusion: The mean cost could be $1, or less. At the 5% significance level, there is insufficient evidence to conclude that the mean price of a box of macaroni and cheese is more than $1.
• \((0.8291, 1.241)\)

"William Shakespeare: The Tragedy of Hamlet, Prince of Denmark," by Jacqueline Ghodsi

THE CHARACTERS (in order of appearance):

• HAMLET, Prince of Denmark and student of Statistics
• POLONIUS, Hamlet’s tutor
• HOROTIO, friend to Hamlet and fellow student

Scene: The great library of the castle, in which Hamlet does his lessons

(The day is fair, but the face of Hamlet is clouded. He paces the large room. His tutor, Polonius, is reprimanding Hamlet regarding the latter’s recent experience. Horatio is seated at the large table at right stage.)

POLONIUS: My Lord, how cans’t thou admit that thou hast seen a ghost! It is but a figment of your imagination!

HAMLET: I beg to differ; I know of a certainty that five-and-seventy in one hundred of us, condemned to the whips and scorns of time as we are, have gazed upon a spirit of health, or goblin damn’d, be their intents wicked or charitable.

POLONIUS If thou doest insist upon thy wretched vision then let me invest your time; be true to thy work and speak to me through the reason of the null and alternate hypotheses. (He turns to Horatio.) Did not Hamlet himself say, “What piece of work is man, how noble in reason, how infinite in faculties? Then let not this foolishness persist. Go, Horatio, make a survey of three-and-sixty and discover what the true proportion be. For my part, I will never succumb to this fantasy, but deem man to be devoid of all reason should thy proposal of at least five-and-seventy in one hundred hold true.

HORATIO (to Hamlet): What should we do, my Lord?

HAMLET: Go to thy purpose, Horatio.

HORATIO: To what end, my Lord?

HAMLET: That you must teach me. But let me conjure you by the rights of our fellowship, by the consonance of our youth, but the obligation of our ever-preserved love, be even and direct with me, whether I am right or no.

(Horatio exits, followed by Polonius, leaving Hamlet to ponder alone.)

(The next day, Hamlet awaits anxiously the presence of his friend, Horatio. Polonius enters and places some books upon the table just a moment before Horatio enters.)

POLONIUS: So, Horatio, what is it thou didst reveal through thy deliberations?

HORATIO: In a random survey, for which purpose thou thyself sent me forth, I did discover that one-and-forty believe fervently that the spirits of the dead walk with us. Before my God, I might not this believe, without the sensible and true avouch of mine own eyes.

POLONIUS: Give thine own thoughts no tongue, Horatio. (Polonius turns to Hamlet.) But look to’t I charge you, my Lord. Come Horatio, let us go together, for this is not our test. (Horatio and Polonius leave together.)

HAMLET: To reject, or not reject, that is the question: whether ‘tis nobler in the mind to suffer the slings and arrows of outrageous statistics, or to take arms against a sea of data, and, by opposing, end them. (Hamlet resignedly attends to his task.)

(Curtain falls)

"Untitled," by Stephen Chen

I've often wondered how software is released and sold to the public. Ironically, I work for a company that sells products with known problems. Unfortunately, most of the problems are difficult to create, which makes them difficult to fix. I usually use the test program X, which tests the product, to try to create a specific problem. When the test program is run to make an error occur, the likelihood of generating an error is 1%.

So, armed with this knowledge, I wrote a new test program Y that will generate the same error that test program X creates, but more often. To find out if my test program is better than the original, so that I can convince the management that I'm right, I ran my test program to find out how often I can generate the same error. When I ran my test program 50 times, I generated the error twice. While this may not seem much better, I think that I can convince the management to use my test program instead of the original test program. Am I right?

• \(H_{0}: p = 0.01\)
• \(H_{a}: p > 0.01\)
• Let \(P′ =\) the proportion of errors generated
• Normal for a single proportion
• Decision: Reject the null hypothesis
• Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of errors generated is more than 0.01.

The“plus-4s” confidence interval is \((0.004, 0.144)\).

"Japanese Girls’ Names"

by Kumi Furuichi

It used to be very typical for Japanese girls’ names to end with “ko.” (The trend might have started around my grandmothers’ generation and its peak might have been around my mother’s generation.) “Ko” means “child” in Chinese characters. Parents would name their daughters with “ko” attaching to other Chinese characters which have meanings that they want their daughters to become, such as Sachiko—happy child, Yoshiko—a good child, Yasuko—a healthy child, and so on.

However, I noticed recently that only two out of nine of my Japanese girlfriends at this school have names which end with “ko.” More and more, parents seem to have become creative, modernized, and, sometimes, westernized in naming their children.

I have a feeling that, while 70 percent or more of my mother’s generation would have names with “ko” at the end, the proportion has dropped among my peers. I wrote down all my Japanese friends’, ex-classmates’, co-workers, and acquaintances’ names that I could remember. Following are the names. (Some are repeats.) Test to see if the proportion has dropped for this generation.

Ai, Akemi, Akiko, Ayumi, Chiaki, Chie, Eiko, Eri, Eriko, Fumiko, Harumi, Hitomi, Hiroko, Hiroko, Hidemi, Hisako, Hinako, Izumi, Izumi, Junko, Junko, Kana, Kanako, Kanayo, Kayo, Kayoko, Kazumi, Keiko, Keiko, Kei, Kumi, Kumiko, Kyoko, Kyoko, Madoka, Maho, Mai, Maiko, Maki, Miki, Miki, Mikiko, Mina, Minako, Miyako, Momoko, Nana, Naoko, Naoko, Naoko, Noriko, Rieko, Rika, Rika, Rumiko, Rei, Reiko, Reiko, Sachiko, Sachiko, Sachiyo, Saki, Sayaka, Sayoko, Sayuri, Seiko, Shiho, Shizuka, Sumiko, Takako, Takako, Tomoe, Tomoe, Tomoko, Touko, Yasuko, Yasuko, Yasuyo, Yoko, Yoko, Yoko, Yoshiko, Yoshiko, Yoshiko, Yuka, Yuki, Yuki, Yukiko, Yuko, Yuko.

"Phillip’s Wish," by Suzanne Osorio

My nephew likes to play

Chasing the girls makes his day.

He asked his mother

If it is okay

To get his ear pierced.

She said, “No way!”

To poke a hole through your ear,

Is not what I want for you, dear.

He argued his point quite well,

Says even my macho pal, Mel,

Has gotten this done.

It’s all just for fun.

C’mon please, mom, please, what the hell.

Again Phillip complained to his mother,

Saying half his friends (including their brothers)

Are piercing their ears

And they have no fears

He wants to be like the others.

She said, “I think it’s much less.

We must do a hypothesis test.

And if you are right,

I won’t put up a fight.

But, if not, then my case will rest.”

We proceeded to call fifty guys

To see whose prediction would fly.

Nineteen of the fifty

Said piercing was nifty

And earrings they’d occasionally buy.

Then there’s the other thirty-one,

Who said they’d never have this done.

So now this poem’s finished.

Will his hopes be diminished,

Or will my nephew have his fun?

• \(H_{0}: p = 0.50\)
• \(H_{a}: p < 0.50\)
• Let \(P′ =\) the proportion of friends that has a pierced ear.
• –1.70
• \(p\text{-value} = 0.0448\)
• Reason for decision: The \(p\text{-value}\) is less than 0.05. (However, they are very close.)
• Conclusion: There is sufficient evidence to support the claim that less than 50% of his friends have pierced ears.
• Confidence Interval: \((0.245, 0.515)\): The “plus-4s” confidence interval is \((0.259, 0.519)\).

"The Craven," by Mark Salangsang

Once upon a morning dreary

In stats class I was weak and weary.

Pondering over last night’s homework

Whose answers were now on the board

This I did and nothing more.

While I nodded nearly napping

Suddenly, there came a tapping.

As someone gently rapping,

Rapping my head as I snore.

Quoth the teacher, “Sleep no more.”

“In every class you fall asleep,”

The teacher said, his voice was deep.

“So a tally I’ve begun to keep

Of every class you nap and snore.

The percentage being forty-four.”

“My dear teacher I must confess,

While sleeping is what I do best.

The percentage, I think, must be less,

A percentage less than forty-four.”

This I said and nothing more.

“We’ll see,” he said and walked away,

And fifty classes from that day

He counted till the month of May

The classes in which I napped and snored.

The number he found was twenty-four.

At a significance level of 0.05,

Please tell me am I still alive?

Or did my grade just take a dive

Plunging down beneath the floor?

Upon thee I hereby implore.

Toastmasters International cites a report by Gallop Poll that 40% of Americans fear public speaking. A student believes that less than 40% of students at her school fear public speaking. She randomly surveys 361 schoolmates and finds that 135 report they fear public speaking. Conduct a hypothesis test to determine if the percent at her school is less than 40%.

• \(H_{0}: p = 0.40\)
• \(H_{a}: p < 0.40\)
• Let \(P′ =\) the proportion of schoolmates who fear public speaking.
• –1.01
• \(p\text{-value} = 0.1563\)
• Conclusion: There is insufficient evidence to support the claim that less than 40% of students at the school fear public speaking.
• Confidence Interval: \((0.3241, 0.4240)\): The “plus-4s” confidence interval is \((0.3257, 0.4250)\).

Sixty-eight percent of online courses taught at community colleges nationwide were taught by full-time faculty. To test if 68% also represents California’s percent for full-time faculty teaching the online classes, Long Beach City College (LBCC) in California, was randomly selected for comparison. In the same year, 34 of the 44 online courses LBCC offered were taught by full-time faculty. Conduct a hypothesis test to determine if 68% represents California. NOTE: For more accurate results, use more California community colleges and this past year's data.

According to an article in Bloomberg Businessweek , New York City's most recent adult smoking rate is 14%. Suppose that a survey is conducted to determine this year’s rate. Nine out of 70 randomly chosen N.Y. City residents reply that they smoke. Conduct a hypothesis test to determine if the rate is still 14% or if it has decreased.

• \(H_{0}: p = 0.14\)
• \(H_{a}: p < 0.14\)
• Let \(P′ =\) the proportion of NYC residents that smoke.
• –0.2756
• \(p\text{-value} = 0.3914\)
• At the 5% significance level, there is insufficient evidence to conclude that the proportion of NYC residents who smoke is less than 0.14.
• Confidence Interval: \((0.0502, 0.2070)\): The “plus-4s” confidence interval (see chapter 8) is \((0.0676, 0.2297)\).

The mean age of De Anza College students in a previous term was 26.6 years old. An instructor thinks the mean age for online students is older than 26.6. She randomly surveys 56 online students and finds that the sample mean is 29.4 with a standard deviation of 2.1. Conduct a hypothesis test.

Registered nurses earned an average annual salary of $69,110. For that same year, a survey was conducted of 41 California registered nurses to determine if the annual salary is higher than $69,110 for California nurses. The sample average was $71,121 with a sample standard deviation of $7,489. Conduct a hypothesis test.

• \(H_{0}: \mu = 69,110\)
• \(H_{0}: \mu > 69,110\)
• Let \(\bar{X} =\) the mean salary in dollars for California registered nurses.
• \(t = 1.719\)
• \(p\text{-value}: 0.0466\)
• Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean salary of California registered nurses exceeds $69,110.
• \(($68,757, $73,485)\)

La Leche League International reports that the mean age of weaning a child from breastfeeding is age four to five worldwide. In America, most nursing mothers wean their children much earlier. Suppose a random survey is conducted of 21 U.S. mothers who recently weaned their children. The mean weaning age was nine months (3/4 year) with a standard deviation of 4 months. Conduct a hypothesis test to determine if the mean weaning age in the U.S. is less than four years old.

Over the past few decades, public health officials have examined the link between weight concerns and teen girls' smoking. Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls were surveyed again. Sixty-three said they smoked to stay thin. Is there good evidence that more than thirty percent of the teen girls smoke to stay thin?

After conducting the test, your decision and conclusion are

• Reject \(H_{0}\): There is sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin.
• Do not reject \(H_{0}\): There is not sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin.
• Do not reject \(H_{0}\): There is not sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin.
• Reject \(H_{0}\): There is sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin.

A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing.

At a 1% level of significance, an appropriate conclusion is:

• There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20%.
• There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is more than 20%.
• There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20%.
• There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is at least 20%.

Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test.

At a significance level of \(a = 0.05\), what is the correct conclusion?

• There is enough evidence to conclude that the mean number of hours is more than 4.75
• There is enough evidence to conclude that the mean number of hours is more than 4.5
• There is not enough evidence to conclude that the mean number of hours is more than 4.5
• There is not enough evidence to conclude that the mean number of hours is more than 4.75

Instructions: For the following ten exercises,

Hypothesis testing: For the following ten exercises, answer each question.

State the null and alternate hypothesis.

State the \(p\text{-value}\).

State \(\alpha\).

What is your decision?

Write a conclusion.

Answer any other questions asked in the problem.

According to the Center for Disease Control website, in 2011 at least 18% of high school students have smoked a cigarette. An Introduction to Statistics class in Davies County, KY conducted a hypothesis test at the local high school (a medium sized–approximately 1,200 students–small city demographic) to determine if the local high school’s percentage was lower. One hundred fifty students were chosen at random and surveyed. Of the 150 students surveyed, 82 have smoked. Use a significance level of 0.05 and using appropriate statistical evidence, conduct a hypothesis test and state the conclusions.

A recent survey in the N.Y. Times Almanac indicated that 48.8% of families own stock. A broker wanted to determine if this survey could be valid. He surveyed a random sample of 250 families and found that 142 owned some type of stock. At the 0.05 significance level, can the survey be considered to be accurate?

• \(H_{0}: p = 0.488\) \(H_{a}: p \neq 0.488\)
• \(p\text{-value} = 0.0114\)
• \(\alpha = 0.05\)
• Reject the null hypothesis.
• At the 5% level of significance, there is enough evidence to conclude that 48.8% of families own stocks.
• The survey does not appear to be accurate.

Driver error can be listed as the cause of approximately 54% of all fatal auto accidents, according to the American Automobile Association. Thirty randomly selected fatal accidents are examined, and it is determined that 14 were caused by driver error. Using \(\alpha = 0.05\), is the AAA proportion accurate?

The US Department of Energy reported that 51.7% of homes were heated by natural gas. A random sample of 221 homes in Kentucky found that 115 were heated by natural gas. Does the evidence support the claim for Kentucky at the \(\alpha = 0.05\) level in Kentucky? Are the results applicable across the country? Why?

• \(H_{0}: p = 0.517\) \(H_{0}: p \neq 0.517\)
• \(p\text{-value} = 0.9203\).
• \(\alpha = 0.05\).
• Do not reject the null hypothesis.
• At the 5% significance level, there is not enough evidence to conclude that the proportion of homes in Kentucky that are heated by natural gas is 0.517.
• However, we cannot generalize this result to the entire nation. First, the sample’s population is only the state of Kentucky. Second, it is reasonable to assume that homes in the extreme north and south will have extreme high usage and low usage, respectively. We would need to expand our sample base to include these possibilities if we wanted to generalize this claim to the entire nation.

For Americans using library services, the American Library Association claims that at most 67% of patrons borrow books. The library director in Owensboro, Kentucky feels this is not true, so she asked a local college statistic class to conduct a survey. The class randomly selected 100 patrons and found that 82 borrowed books. Did the class demonstrate that the percentage was higher in Owensboro, KY? Use \(\alpha = 0.01\) level of significance. What is the possible proportion of patrons that do borrow books from the Owensboro Library?

The Weather Underground reported that the mean amount of summer rainfall for the northeastern US is at least 11.52 inches. Ten cities in the northeast are randomly selected and the mean rainfall amount is calculated to be 7.42 inches with a standard deviation of 1.3 inches. At the \(\alpha = 0.05 level\), can it be concluded that the mean rainfall was below the reported average? What if \(\alpha = 0.01\)? Assume the amount of summer rainfall follows a normal distribution.

• \(H_{0}: \mu \geq 11.52\) \(H_{a}: \mu < 11.52\)
• \(p\text{-value} = 0.000002\) which is almost 0.
• At the 5% significance level, there is enough evidence to conclude that the mean amount of summer rain in the northeaster US is less than 11.52 inches, on average.
• We would make the same conclusion if alpha was 1% because the \(p\text{-value}\) is almost 0.

A survey in the N.Y. Times Almanac finds the mean commute time (one way) is 25.4 minutes for the 15 largest US cities. The Austin, TX chamber of commerce feels that Austin’s commute time is less and wants to publicize this fact. The mean for 25 randomly selected commuters is 22.1 minutes with a standard deviation of 5.3 minutes. At the \(\alpha = 0.10\) level, is the Austin, TX commute significantly less than the mean commute time for the 15 largest US cities?

A report by the Gallup Poll found that a woman visits her doctor, on average, at most 5.8 times each year. A random sample of 20 women results in these yearly visit totals

3; 2; 1; 3; 7; 2; 9; 4; 6; 6; 8; 0; 5; 6; 4; 2; 1; 3; 4; 1

At the \(\alpha = 0.05\) level can it be concluded that the sample mean is higher than 5.8 visits per year?

• \(H_{0}: \mu \leq 5.8\) \(H_{a}: \mu > 5.8\)
• \(p\text{-value} = 0.9987\)
• At the 5% level of significance, there is not enough evidence to conclude that a woman visits her doctor, on average, more than 5.8 times a year.

According to the N.Y. Times Almanac the mean family size in the U.S. is 3.18. A sample of a college math class resulted in the following family sizes:

5; 4; 5; 4; 4; 3; 6; 4; 3; 3; 5; 5; 6; 3; 3; 2; 7; 4; 5; 2; 2; 2; 3; 2

At \(\alpha = 0.05\) level, is the class’ mean family size greater than the national average? Does the Almanac result remain valid? Why?

The student academic group on a college campus claims that freshman students study at least 2.5 hours per day, on average. One Introduction to Statistics class was skeptical. The class took a random sample of 30 freshman students and found a mean study time of 137 minutes with a standard deviation of 45 minutes. At α = 0.01 level, is the student academic group’s claim correct?

• \(H_{0}: \mu \geq 150\) \(H_{0}: \mu < 150\)
• \(p\text{-value} = 0.0622\)
• \(\alpha = 0.01\)
• At the 1% significance level, there is not enough evidence to conclude that freshmen students study less than 2.5 hours per day, on average.
• The student academic group’s claim appears to be correct.

9.7: Hypothesis Testing of a Single Mean and Single Proportion