hypothesis testing year 1 questions

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Statistical Sampling

Years 1 & 2, other resources, data presentation and interpretation, probability, statistical distributions, hypothesis testing.

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Unit 12: Significance tests (hypothesis testing)

About this unit.

Significance tests give us a formal process for using sample data to evaluate the likelihood of some claim about a population value. Learn how to conduct significance tests and calculate p-values to see how likely a sample result is to occur by random chance. You'll also see how we use p-values to make conclusions about hypotheses.

The idea of significance tests

• Simple hypothesis testing (Opens a modal)
• Idea behind hypothesis testing (Opens a modal)
• Examples of null and alternative hypotheses (Opens a modal)
• P-values and significance tests (Opens a modal)
• Comparing P-values to different significance levels (Opens a modal)
• Estimating a P-value from a simulation (Opens a modal)
• Using P-values to make conclusions (Opens a modal)
• Simple hypothesis testing Get 3 of 4 questions to level up!
• Writing null and alternative hypotheses Get 3 of 4 questions to level up!
• Estimating P-values from simulations Get 3 of 4 questions to level up!

Error probabilities and power

• Introduction to Type I and Type II errors (Opens a modal)
• Type 1 errors (Opens a modal)
• Examples identifying Type I and Type II errors (Opens a modal)
• Introduction to power in significance tests (Opens a modal)
• Examples thinking about power in significance tests (Opens a modal)
• Consequences of errors and significance (Opens a modal)
• Type I vs Type II error Get 3 of 4 questions to level up!
• Error probabilities and power Get 3 of 4 questions to level up!

Tests about a population proportion

• Constructing hypotheses for a significance test about a proportion (Opens a modal)
• Conditions for a z test about a proportion (Opens a modal)
• Reference: Conditions for inference on a proportion (Opens a modal)
• Calculating a z statistic in a test about a proportion (Opens a modal)
• Calculating a P-value given a z statistic (Opens a modal)
• Making conclusions in a test about a proportion (Opens a modal)
• Writing hypotheses for a test about a proportion Get 3 of 4 questions to level up!
• Conditions for a z test about a proportion Get 3 of 4 questions to level up!
• Calculating the test statistic in a z test for a proportion Get 3 of 4 questions to level up!
• Calculating the P-value in a z test for a proportion Get 3 of 4 questions to level up!
• Making conclusions in a z test for a proportion Get 3 of 4 questions to level up!

Tests about a population mean

• Writing hypotheses for a significance test about a mean (Opens a modal)
• Conditions for a t test about a mean (Opens a modal)
• Reference: Conditions for inference on a mean (Opens a modal)
• When to use z or t statistics in significance tests (Opens a modal)
• Example calculating t statistic for a test about a mean (Opens a modal)
• Using TI calculator for P-value from t statistic (Opens a modal)
• Using a table to estimate P-value from t statistic (Opens a modal)
• Comparing P-value from t statistic to significance level (Opens a modal)
• Free response example: Significance test for a mean (Opens a modal)
• Writing hypotheses for a test about a mean Get 3 of 4 questions to level up!
• Conditions for a t test about a mean Get 3 of 4 questions to level up!
• Calculating the test statistic in a t test for a mean Get 3 of 4 questions to level up!
• Calculating the P-value in a t test for a mean Get 3 of 4 questions to level up!
• Making conclusions in a t test for a mean Get 3 of 4 questions to level up!

More significance testing videos

• Hypothesis testing and p-values (Opens a modal)
• One-tailed and two-tailed tests (Opens a modal)
• Z-statistics vs. T-statistics (Opens a modal)
• Small sample hypothesis test (Opens a modal)
• Large sample proportion hypothesis testing (Opens a modal)

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9.2: Hypothesis Testing

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All hypotheses tests have the same basic steps:

• Determine the hypothesis : What are we trying to figure out? This is formally written as the null and alternative hypotheses.
• Calculate the evidence : This will be a test statistics and either a critical value or a p-value.
• Make a decision : The options will be Reject the Null Hypothesis or Do not Reject the Null Hypothesis.
• Determine the conclusion : What does the decision mean in terms of the problem given?

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.

\(H_0\): The null hypothesis: It is a statement of no difference between the variables—they are not related. This can often be considered the status quo and as a result if you cannot accept the null it requires some action.

\(H_a\): The alternative hypothesis: It is a claim about the population that is contradictory to \(H_0\) and what we conclude when we reject \(H_0\). This is usually what the researcher is trying to prove.

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.

After you have determined which hypothesis the sample supports, you make a decision. There are two options for a decision. They are "reject \(H_0\)" if the sample information favors the alternative hypothesis or "do not reject \(H_0\)" or "decline to reject \(H_0\)" if the sample information is insufficient to reject the null hypothesis.

\(H_{0}\) always has a symbol with an equal in it. \(H_{a}\) never has a symbol with an equal in it. The choice of symbol depends on the wording of the hypothesis test. However, be aware that many researchers (including one of the co-authors in research work) use = in the null hypothesis, even with > or < as the symbol in the alternative hypothesis. This practice is acceptable because we only make the decision to reject or not reject the null hypothesis.

Example \(\PageIndex{1}\)

• \(H_{0}\): No more than 30% of the registered voters in Santa Clara County voted in the primary election. \(p \leq 30\)
• \(H_{a}\): More than 30% of the registered voters in Santa Clara County voted in the primary election. \(p > 30\)

Exercise \(\PageIndex{1}\)

A medical trial is conducted to test whether or not a new medicine reduces cholesterol by 25%. State the null and alternative hypotheses.

• \(H_{0}\): The drug reduces cholesterol by 25%. \(p = 0.25\)
• \(H_{a}\): The drug does not reduce cholesterol by 25%. \(p \neq 0.25\)

Example \(\PageIndex{2}\)

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:

• \(H_{0}: \mu = 2.0\)
• \(H_{a}: \mu \neq 2.0\)

Exercise \(\PageIndex{2}\)

We want to test whether the mean height of eighth graders is 66 inches. State the null and alternative hypotheses. Fill in the correct symbol \((=, \neq, \geq, <, \leq, >)\) for the null and alternative hypotheses.

• \(H_{0}: \mu_ 66\)
• \(H_{a}: \mu_ 66\)
• \(H_{0}: \mu = 66\)
• \(H_{a}: \mu \neq 66\)

Example \(\PageIndex{3}\)

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:

• \(H_{0}: \mu \geq 66\)
• \(H_{a}: \mu < 66\)

Exercise \(\PageIndex{3}\)

We want to test if it takes fewer than 45 minutes to teach a lesson plan. State the null and alternative hypotheses. Fill in the correct symbol ( =, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.

• \(H_{0}: \mu_ 45\)
• \(H_{a}: \mu_ 45\)
• \(H_{0}: \mu \geq 45\)
• \(H_{a}: \mu < 45\)

Example \(\PageIndex{4}\)

In an issue of U. S. News and World Report , an article on school standards stated that about half of all students in France, Germany, and Israel take advanced placement exams and a third pass. The same article stated that 6.6% of U.S. students take advanced placement exams and 4.4% pass. Test if the percentage of U.S. students who take advanced placement exams is more than 6.6%. State the null and alternative hypotheses.

• \(H_{0}: p \leq 0.066\)
• \(H_{a}: p > 0.066\)

Exercise \(\PageIndex{4}\)

On a state driver’s test, about 40% pass the test on the first try. We want to test if more than 40% pass on the first try. Fill in the correct symbol (\(=, \neq, \geq, <, \leq, >\)) for the null and alternative hypotheses.

• \(H_{0}: p_ 0.40\)
• \(H_{a}: p_ 0.40\)
• \(H_{0}: p = 0.40\)
• \(H_{a}: p > 0.40\)

COLLABORATIVE EXERCISE

Bring to class a newspaper, some news magazines, and some Internet articles . In groups, find articles from which your group can write null and alternative hypotheses. Discuss your hypotheses with the rest of the class.

Outcomes and the Type I and Type II Errors

When you perform a hypothesis test, there are four possible outcomes depending on the actual truth (or falseness) of the null hypothesis \(H_{0}\) and the decision to reject or not. The outcomes are summarized in the following table:

The four possible outcomes in the table are:

• The decision is not to reject \(H_{0}\) when \(H_{0}\) is true (correct decision).
• The decision is to reject \(H_{0}\) when \(H_{0}\) is true (incorrect decision known as aType I error).
• The decision is not to reject \(H_{0}\) when, in fact, \(H_{0}\) is false (incorrect decision known as a Type II error).
• The decision is to reject \(H_{0}\) when \(H_{0}\) is false ( correct decision whose probability is called the Power of the Test ).

Each of the errors occurs with a particular probability. The Greek letters \(\alpha\) and \(\beta\) represent the probabilities.

• \(\alpha =\) probability of a Type I error \(= P(\text{Type I error}) =\) probability of rejecting the null hypothesis when the null hypothesis is true.
• \(\beta =\) probability of a Type II error \(= P(\text{Type II error}) =\) probability of not rejecting the null hypothesis when the null hypothesis is false.

\(\alpha\) and \(\beta\) should be as small as possible because they are probabilities of errors. They are rarely zero.

The Power of the Test is \(1 - \beta\). Ideally, we want a high power that is as close to one as possible. Increasing the sample size can increase the Power of the Test. The following are examples of Type I and Type II errors.

Example \(\PageIndex{5}\): Type I vs. Type II errors

Suppose the null hypothesis, \(H_{0}\), is: Frank's rock climbing equipment is safe.

• 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.

\(\alpha =\) probability that Frank thinks his rock climbing equipment may not be safe when, in fact, it really is safe.

\(\beta =\) probability that Frank thinks 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.)

Exercise \(\PageIndex{5}\)

Suppose the null hypothesis, \(H_{0}\), is: the blood cultures contain no traces of pathogen \(X\). State the Type I and Type II errors.

• Type I error : The researcher thinks the blood cultures do contain traces of pathogen \(X\), when in fact, they do not.
• Type II error : The researcher thinks the blood cultures do not contain traces of pathogen \(X\), when in fact, they do.

Example \(\PageIndex{6}\)

Suppose the null hypothesis, \(H_{0}\), is: The victim of an automobile accident is alive when he arrives at the emergency room of a hospital.

• 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.

\(\alpha =\) probability that the emergency crew thinks the victim is dead when, in fact, he is really alive \(= P(\text{Type I error})\).

\(\beta =\) probability that the emergency crew does not know if the victim is alive when, in fact, the victim is dead \(= P(\text{Type II error})\).

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.)

Exercise \(\PageIndex{6}\)

Suppose the null hypothesis, \(H_{0}\), is: a patient is not sick. Which type of error has the greater consequence, Type I or Type II?

The error with the greater consequence is the Type II error: the patient will be thought well when, in fact, he is sick, so he will not get treatment.

Example \(\PageIndex{7}\)

It’s a Boy Genetic Labs claim to be able to increase the likelihood that a pregnancy will result in a boy being born. Statisticians want to test the claim. Suppose that the null hypothesis, \(H_{0}\), is: It’s a Boy Genetic Labs has no effect on gender outcome.

• Type I error : This results when a true null hypothesis is rejected. In the context of this scenario, we would state that we believe that It’s a Boy Genetic Labs influences the gender outcome, when in fact it has no effect. The probability of this error occurring is denoted by the Greek letter alpha, \(\alpha\).
• Type II error : This results when we fail to reject a false null hypothesis. In context, we would state that It’s a Boy Genetic Labs does not influence the gender outcome of a pregnancy when, in fact, it does. The probability of this error occurring is denoted by the Greek letter beta, \(\beta\).

The error of greater consequence would be the Type I error since couples would use the It’s a Boy Genetic Labs product in hopes of increasing the chances of having a boy.

Exercise \(\PageIndex{7}\)

“Red tide” is a bloom of poison-producing algae–a few different species of a class of plankton called dinoflagellates. When the weather and water conditions cause these blooms, shellfish such as clams living in the area develop dangerous levels of a paralysis-inducing toxin. In Massachusetts, the Division of Marine Fisheries (DMF) monitors levels of the toxin in shellfish by regular sampling of shellfish along the coastline. If the mean level of toxin in clams exceeds 800 μg (micrograms) of toxin per kg of clam meat in any area, clam harvesting is banned there until the bloom is over and levels of toxin in clams subside. Describe both a Type I and a Type II error in this context, and state which error has the greater consequence.

In this scenario, an appropriate null hypothesis would be \(H_{0}\): the mean level of toxins is at most \(800 \mu\text{g}\), \(H_{0}: \mu_{0} \leq 800 \mu\text{g}\).

Example \(\PageIndex{8}\)

A certain experimental drug claims a cure rate of at least 75% for males with prostate cancer. Describe both the Type I and Type II errors in context. Which error is the more serious?

• Type I : A cancer patient believes the cure rate for the drug is less than 75% when it actually is at least 75%.
• Type II : A cancer patient believes the experimental drug has at least a 75% cure rate when it has a cure rate that is less than 75%.

In this scenario, the Type II error contains the more severe consequence. If a patient believes the drug works at least 75% of the time, this most likely will influence the patient’s (and doctor’s) choice about whether to use the drug as a treatment option.

Exercise \(\PageIndex{8}\)

Determine both Type I and Type II errors for the following scenario:

Assume a null hypothesis, \(H_{0}\), that states the percentage of adults with jobs is at least 88%. Identify the Type I and Type II errors from these four statements.

• Not to reject the null hypothesis that the percentage of adults who have jobs is at least 88% when that percentage is actually less than 88%
• Not to reject the null hypothesis that the percentage of adults who have jobs is at least 88% when the percentage is actually at least 88%.
• Reject the null hypothesis that the percentage of adults who have jobs is at least 88% when the percentage is actually at least 88%.
• Reject the null hypothesis that the percentage of adults who have jobs is at least 88% when that percentage is actually less than 88%.

Type I error: c

Type I error: b

Distribution Needed for Hypothesis Testing

Earlier in the course, we discussed sampling distributions. Particular distributions are associated with hypothesis testing. Perform 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 distribution of the sample mean is approximately normal.) We perform tests of a population proportion using a normal distribution (usually \(n\) is large or the sample size is large).

If you are testing a single population mean, the distribution for the test is for means :

\[\bar{X} - N\left(\mu_{x}, \frac{\sigma_{x}}{\sqrt{n}}\right)\]

The population parameter is \(\mu\). The estimated value (point estimate) for \(\mu\) is \(\bar{x}\), the sample mean.

If you are testing a single population proportion, the distribution for the test is for proportions or percentages:

\[P' - N\left(p, \sqrt{\frac{p-q}{n}}\right)\]

The population parameter is \(p\). The estimated value (point estimate) for \(p\) is \(p′\). \(p' = \frac{x}{n}\) where \(x\) is the number of successes and n is the sample size.

Assumptions

When you perform a hypothesis test of a single population mean \(\mu\) using a Student's \(t\)-distribution (often called a \(t\)-test), there are fundamental assumptions that need to be met in order for the test to work properly. Your data should be a simple random sample that comes from a population that is approximately normally distributed. You use the sample standard deviation to approximate the population standard deviation. (Note that if the sample size is sufficiently large, a \(t\)-test will work even if the population is not approximately normally distributed).

When you perform a hypothesis test of a single population mean \(\mu\) using a normal distribution (often called a \(z\)-test), you take a simple random sample from the population. The population you are testing is normally distributed or your sample size is sufficiently large. You know the value of the population standard deviation which, in reality, is rarely known.

When you perform a hypothesis test of a single population proportion \(p\), you take a simple random sample from the population. You must meet the conditions for a binomial distribution which are: there are a certain number \(n\) of independent trials, the outcomes of any trial are success or failure, and each trial has the same probability of a success \(p\). The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities \(np\) and \(nq\) must both be greater than five \((np > 5\) and \(nq > 5)\). Then the binomial distribution of a sample (estimated) proportion can be approximated by the normal distribution with \(\mu = p\) and \(\sigma = \sqrt{\frac{pq}{n}}\). Remember that \(q = 1 – p\).

Rare Events, the Sample, Decision and Conclusion

Establishing the type of distribution, sample size, and known or unknown standard deviation can help you figure out how to go about a hypothesis test. However, there are several other factors you should consider when working out a hypothesis test.

Rare Events

Suppose you make an assumption about a property of the population (this assumption is the null hypothesis). Then you gather sample data randomly. If the sample has properties that would be very unlikely to occur if the assumption is true, then you would conclude that your assumption about the population is probably incorrect. (Remember that your assumption is just an assumption—it is not a fact and it may or may not be true. But your sample data are real and the data are showing you a fact that seems to contradict your assumption.)

For example, Didi and Ali are at a birthday party of a very wealthy friend. They hurry to be first in line to grab a prize from a tall basket that they cannot see inside because they will be blindfolded. There are 200 plastic bubbles in the basket and Didi and Ali have been told that there is only one with a $100 bill. Didi is the first person to reach into the basket and pull out a bubble. Her bubble contains a $100 bill. The probability of this happening is \(\frac{1}{200} = 0.005\). Because this is so unlikely, Ali is hoping that what the two of them were told is wrong and there are more $100 bills in the basket. A "rare event" has occurred (Didi getting the $100 bill) so Ali doubts the assumption about only one $100 bill being in the basket.

Using the Sample to Test the Null Hypothesis

Use the sample data to calculate the actual probability of getting the test result, called the \(p\)-value. The \(p\)-value is the probability that, if the null hypothesis is true, the results from another randomly selected sample will be as extreme or more extreme as the results obtained from the given sample.

A large \(p\)-value calculated from the data indicates that we should not reject the null hypothesis. The smaller the \(p\)-value, the more unlikely the outcome, and the stronger the evidence is against the null hypothesis. We would reject the null hypothesis if the evidence is strongly against it.

Draw a graph that shows the \(p\)-value. The hypothesis test is easier to perform if you use a graph because you see the problem more clearly.

Example \(\PageIndex{9}\)

Suppose a baker claims that his bread height is more than 15 cm, on average. Several of his customers do not believe him. To persuade his customers that he is right, the baker decides to do a hypothesis test. He bakes 10 loaves of bread. The mean height of the sample loaves is 17 cm. The baker knows from baking hundreds of loaves of bread that the standard deviation for the height is 0.5 cm. and the distribution of heights is normal.

• The null hypothesis could be \(H_{0}: \mu \leq 15\)
• The alternate hypothesis is \(H_{a}: \mu > 15\)

The words "is more than" translates as a "\(>\)" so "\(\mu > 15\)" goes into the alternate hypothesis. The null hypothesis must contradict the alternate hypothesis.

Since \(\sigma\) is known (\(\sigma = 0.5 cm.\)), the distribution for the population is known to be normal with mean \(μ = 15\) and standard deviation

\[\dfrac{\sigma}{\sqrt{n}} = \frac{0.5}{\sqrt{10}} = 0.16. \nonumber\]

Suppose the null hypothesis is true (the mean height of the loaves is no more than 15 cm). Then is the mean height (17 cm) calculated from the sample unexpectedly large? The hypothesis test works by asking the question how unlikely the sample mean would be if the null hypothesis were true. The graph shows how far out the sample mean is on the normal curve. The p -value is the probability that, if we were to take other samples, any other sample mean would fall at least as far out as 17 cm.

The \(p\) -value, then, is the probability that a sample mean is the same or greater than 17 cm. when the population mean is, in fact, 15 cm. We can calculate this probability using the normal distribution for means.

\(p\text{-value} = P(\bar{x} > 17)\) which is approximately zero.

A \(p\)-value of approximately zero tells us that it is highly unlikely that a loaf of bread rises no more than 15 cm, on average. That is, almost 0% of all loaves of bread would be at least as high as 17 cm. purely by CHANCE had the population mean height really been 15 cm. Because the outcome of 17 cm. is so unlikely (meaning it is happening NOT by chance alone) , we conclude that the evidence is strongly against the null hypothesis (the mean height is at most 15 cm.). There is sufficient evidence that the true mean height for the population of the baker's loaves of bread is greater than 15 cm.

Exercise \(\PageIndex{9}\)

A normal distribution has a standard deviation of 1. We want to verify a claim that the mean is greater than 12. A sample of 36 is taken with a sample mean of 12.5.

• \(H_{0}: \mu leq 12\)
• \(H_{a}: \mu > 12\)

The \(p\)-value is 0.0013

Draw a graph that shows the \(p\)-value.

\(p\text{-value} = 0.0013\)

Decision and Conclusion

A systematic way to make a decision of whether to reject or not reject the null hypothesis is to compare the \(p\)-value and a preset or preconceived \(\alpha\) (also called a " significance level "). A preset \(\alpha\) is the probability of a Type I error (rejecting the null hypothesis when the null hypothesis is true). It may or may not be given to you at the beginning of the problem.

When you make a decision to reject or not reject \(H_{0}\), do as follows:

• If \(\alpha > p\text{-value}\), reject \(H_{0}\). The results of the sample data are significant. There is sufficient evidence to conclude that \(H_{0}\) is an incorrect belief and that the alternative hypothesis, \(H_{a}\), may be correct.
• If \(\alpha \leq p\text{-value}\), do not reject \(H_{0}\). The results of the sample data are not significant.There is not sufficient evidence to conclude that the alternative hypothesis,\(H_{a}\), may be correct.

When you "do not reject \(H_{0}\)", it does not mean that you should believe that H 0 is true. It simply means that the sample data have failed to provide sufficient evidence to cast serious doubt about the truthfulness of \(H_{0}\).

Conclusion: After you make your decision, write a thoughtful conclusion about the hypotheses in terms of the given problem.

Example \(\PageIndex{10}\)

When using the \(p\)-value to evaluate a hypothesis test, it is sometimes useful to use the following memory device

• If the \(p\)-value is low, the null must go.
• If the \(p\)-value is high, the null must fly.

This memory aid relates a \(p\)-value less than the established alpha (the \(p\) is low) as rejecting the null hypothesis and, likewise, relates a \(p\)-value higher than the established alpha (the \(p\) is high) as not rejecting the null hypothesis.

Fill in the blanks.

Reject the null hypothesis when ______________________________________.

The results of the sample data _____________________________________.

Do not reject the null when hypothesis when __________________________________________.

The results of the sample data ____________________________________________.

Reject the null hypothesis when the \(p\) -value is less than the established alpha value . The results of the sample data support the alternative hypothesis .

Do not reject the null hypothesis when the \(p\) -value is greater than the established alpha value . The results of the sample data do not support the alternative hypothesis .

Exercise \(\PageIndex{10}\)

It’s a Boy Genetics Labs claim their procedures improve the chances of a boy being born. The results for a test of a single population proportion are as follows:

• \(H_{0}: p = 0.50, H_{a}: p > 0.50\)
• \(\alpha = 0.01\)
• \(p\text{-value} = 0.025\)

Interpret the results and state a conclusion in simple, non-technical terms.

Since the \(p\)-value is greater than the established alpha value (the \(p\)-value is high), we do not reject the null hypothesis. There is not enough evidence to support It’s a Boy Genetics Labs' stated claim that their procedures improve the chances of a boy being born.

In a hypothesis test , sample data is evaluated in order to arrive at a decision about some type of claim. If certain conditions about the sample are satisfied, then the claim can be evaluated for a population. In a hypothesis test, we:

• Evaluate the null hypothesis , typically denoted with \(H_{0}\). The null is not rejected unless the hypothesis test shows otherwise. The null statement must always contain some form of equality \((=, \leq \text{or} \geq)\)
• Always write the alternative hypothesis , typically denoted with \(H_{a}\) or \(H_{1}\), using less than, greater than, or not equals symbols, i.e., \((\neq, >, \text{or} <)\).
• If we reject the null hypothesis, then we can assume there is enough evidence to support the alternative hypothesis.
• Never state that a claim is proven true or false. Keep in mind the underlying fact that hypothesis testing is based on probability laws; therefore, we can talk only in terms of non-absolute certainties.

In every hypothesis test, the outcomes are dependent on a correct interpretation of the data. Incorrect calculations or misunderstood summary statistics can yield errors that affect the results. A Type I error occurs when a true null hypothesis is rejected. A Type II error occurs when a false null hypothesis is not rejected. The probabilities of these errors are denoted by the Greek letters \(\alpha\) and \(\beta\), for a Type I and a Type II error respectively. The power of the test, \(1 - \beta\), quantifies the likelihood that a test will yield the correct result of a true alternative hypothesis being accepted. A high power is desirable.

In order for a hypothesis test’s results to be generalized to a population, certain requirements must be satisfied.

When testing for a single population mean:

• A Student's \(t\)-test should be used if the data come from a simple, random sample and the population is approximately normally distributed, or the sample size is large, with an unknown standard deviation.
• The normal test will work if the data come from a simple, random sample and the population is approximately normally distributed, or the sample size is large, with a known standard deviation.

When testing a single population proportion use a normal test for a single population proportion if the data comes from a simple, random sample, fill the requirements for a binomial distribution, and the mean number of successes and the mean number of failures satisfy the conditions: \(np > 5\) and \(nq > 5\) where \(n\) is the sample size, \(p\) is the probability of a success, and \(q\) is the probability of a failure.

When the probability of an event occurring is low, and it happens, it is called a rare event. Rare events are important to consider in hypothesis testing because they can inform your willingness not to reject or to reject a null hypothesis. To test a null hypothesis, find the p -value for the sample data and graph the results. When deciding whether or not to reject the null the hypothesis, keep these two parameters in mind:

• \(\alpha > p-value\), reject the null hypothesis
• \(\alpha \leq p-value\), do not reject the null hypothesis

Formula Review

\(H_{0}\) and \(H_{a}\) are contradictory.

• If \(\alpha \leq p\)-value, then do not reject \(H_{0}\).
• If\(\alpha > p\)-value, then reject \(H_{0}\).

\(\alpha\) is preconceived. Its value is set before the hypothesis test starts. The \(p\)-value is calculated from the data.

If there is no given preconceived \(\alpha\), then use \(\alpha = 0.05\).

Types of Hypothesis Tests

• Single population mean, known population variance (or standard deviation): Normal test .
• Single population mean, unknown population variance (or standard deviation): Student's \(t\)-test .
• Single population proportion: Normal test .
• For a single population mean , we may use a normal distribution with the following mean and standard deviation. Means: \(\mu = \mu_{\bar{x}}\) and \(\\sigma_{\bar{x}} = \frac{\sigma_{x}}{\sqrt{n}}\)
• A single population proportion , we may use a normal distribution with the following mean and standard deviation. Proportions: \(\mu = p\) and \(\sigma = \sqrt{\frac{pq}{n}}\).

Data from the National Institute of Mental Health. Available online at http://www.nimh.nih.gov/publicat/depression.cfm .

• It is continuous and assumes any real values.
• The pdf is symmetrical about its mean of zero. However, it is more spread out and flatter at the apex than the normal distribution.
• It approaches the standard normal distribution as \(n\) gets larger.
• There is a "family" of \(t\)-distributions: every representative of the family is completely defined by the number of degrees of freedom which is one less than the number of data items.

Contributors and Attributions

Barbara Illowsky and Susan Dean (De Anza College) with many other contributing authors. Content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] .

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Year 1: Hypothesis Testing

A4.1 Hypothesis testing with the binomial distribution

Introduction

Conducting a hypothesis test.

The first video on the topic of Hypothesis testing. This video introduces the idea of how to conduct a hypothesis test and also what is meant by significance levels.

The second video from the Year 1 Statistics course on the topic of hypothesis testing. This video builds on the first introductory, video and goes through the process of how to conduct a hypothesis test using a binomial distribution.

Step Three:

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S.3 hypothesis testing.

In reviewing hypothesis tests, we start first with the general idea. Then, we keep returning to the basic procedures of hypothesis testing, each time adding a little more detail.

The general idea of hypothesis testing involves:

• Making an initial assumption.
• Collecting evidence (data).
• Based on the available evidence (data), deciding whether to reject or not reject the initial assumption.

Every hypothesis test — regardless of the population parameter involved — requires the above three steps.

Example S.3.1

Is normal body temperature really 98.6 degrees f section  .

Consider the population of many, many adults. A researcher hypothesized that the average adult body temperature is lower than the often-advertised 98.6 degrees F. That is, the researcher wants an answer to the question: "Is the average adult body temperature 98.6 degrees? Or is it lower?" To answer his research question, the researcher starts by assuming that the average adult body temperature was 98.6 degrees F.

Then, the researcher went out and tried to find evidence that refutes his initial assumption. In doing so, he selects a random sample of 130 adults. The average body temperature of the 130 sampled adults is 98.25 degrees.

Then, the researcher uses the data he collected to make a decision about his initial assumption. It is either likely or unlikely that the researcher would collect the evidence he did given his initial assumption that the average adult body temperature is 98.6 degrees:

• If it is likely , then the researcher does not reject his initial assumption that the average adult body temperature is 98.6 degrees. There is not enough evidence to do otherwise.
• either the researcher's initial assumption is correct and he experienced a very unusual event;
• or the researcher's initial assumption is incorrect.

In statistics, we generally don't make claims that require us to believe that a very unusual event happened. That is, in the practice of statistics, if the evidence (data) we collected is unlikely in light of the initial assumption, then we reject our initial assumption.

Example S.3.2

Criminal trial analogy section  .

One place where you can consistently see the general idea of hypothesis testing in action is in criminal trials held in the United States. Our criminal justice system assumes "the defendant is innocent until proven guilty." That is, our initial assumption is that the defendant is innocent.

In the practice of statistics, we make our initial assumption when we state our two competing hypotheses -- the null hypothesis ( H 0 ) and the alternative hypothesis ( H A ). Here, our hypotheses are:

• H 0 : Defendant is not guilty (innocent)
• H A : Defendant is guilty

In statistics, we always assume the null hypothesis is true . That is, the null hypothesis is always our initial assumption.

The prosecution team then collects evidence — such as finger prints, blood spots, hair samples, carpet fibers, shoe prints, ransom notes, and handwriting samples — with the hopes of finding "sufficient evidence" to make the assumption of innocence refutable.

In statistics, the data are the evidence.

The jury then makes a decision based on the available evidence:

• If the jury finds sufficient evidence — beyond a reasonable doubt — to make the assumption of innocence refutable, the jury rejects the null hypothesis and deems the defendant guilty. We behave as if the defendant is guilty.
• If there is insufficient evidence, then the jury does not reject the null hypothesis . We behave as if the defendant is innocent.

In statistics, we always make one of two decisions. We either "reject the null hypothesis" or we "fail to reject the null hypothesis."

Errors in Hypothesis Testing Section  

Did you notice the use of the phrase "behave as if" in the previous discussion? We "behave as if" the defendant is guilty; we do not "prove" that the defendant is guilty. And, we "behave as if" the defendant is innocent; we do not "prove" that the defendant is innocent.

This is a very important distinction! We make our decision based on evidence not on 100% guaranteed proof. Again:

• If we reject the null hypothesis, we do not prove that the alternative hypothesis is true.
• If we do not reject the null hypothesis, we do not prove that the null hypothesis is true.

We merely state that there is enough evidence to behave one way or the other. This is always true in statistics! Because of this, whatever the decision, there is always a chance that we made an error .

Let's review the two types of errors that can be made in criminal trials:

Table S.3.2 shows how this corresponds to the two types of errors in hypothesis testing.

Note that, in statistics, we call the two types of errors by two different  names -- one is called a "Type I error," and the other is called  a "Type II error." Here are the formal definitions of the two types of errors:

There is always a chance of making one of these errors. But, a good scientific study will minimize the chance of doing so!

Making the Decision Section  

Recall that it is either likely or unlikely that we would observe the evidence we did given our initial assumption. If it is likely , we do not reject the null hypothesis. If it is unlikely , then we reject the null hypothesis in favor of the alternative hypothesis. Effectively, then, making the decision reduces to determining "likely" or "unlikely."

In statistics, there are two ways to determine whether the evidence is likely or unlikely given the initial assumption:

• We could take the " critical value approach " (favored in many of the older textbooks).
• Or, we could take the " P -value approach " (what is used most often in research, journal articles, and statistical software).

In the next two sections, we review the procedures behind each of these two approaches. To make our review concrete, let's imagine that μ is the average grade point average of all American students who major in mathematics. We first review the critical value approach for conducting each of the following three hypothesis tests about the population mean $\mu$:

In Practice

• We would want to conduct the first hypothesis test if we were interested in concluding that the average grade point average of the group is more than 3.
• We would want to conduct the second hypothesis test if we were interested in concluding that the average grade point average of the group is less than 3.
• And, we would want to conduct the third hypothesis test if we were only interested in concluding that the average grade point average of the group differs from 3 (without caring whether it is more or less than 3).

Upon completing the review of the critical value approach, we review the P -value approach for conducting each of the above three hypothesis tests about the population mean \(\mu\). The procedures that we review here for both approaches easily extend to hypothesis tests about any other population parameter.

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Hypothesis Testing Solved Examples(Questions and Solutions)

Here is a list hypothesis testing exercises and solutions. Try to solve a question by yourself first before you look at the solution.

Question 1 In the population, the average IQ is 100 with a standard deviation of 15. A team of scientists want to test a new medication to see if it has either a positive or negative effect on intelligence, or not effect at all. A sample of 30 participants who have taken the medication  has a mean of 140. Did the medication affect intelligence? View Solution to Question 1

A professor wants to know if her introductory statistics class has a good grasp of basic math. Six students are chosen at random from the class and given a math proficiency test. The professor wants the class to be able to score above 70 on the test. The six students get the following scores:62, 92, 75, 68, 83, 95. Can the professor have 90% confidence that the mean score for the class on the test would be above 70. Solution to Question 2

Question 3 In a packaging plant, a machine packs cartons with jars. It is supposed that a new machine would pack faster on the average than the machine currently used. To test the hypothesis, the time it takes each machine to pack ten cartons are recorded. The result in seconds is as follows.

Do the data provide sufficient evidence to conclude that, on the average, the new machine packs faster? Perform  the required hypothesis test at the 5% level of significance. Solution to Question 3 

Question 4 We want to compare the heights in inches of two groups of individuals. Here are the measurements: X: 175, 168, 168, 190, 156, 181, 182, 175, 174, 179 Y:  120, 180, 125, 188, 130, 190, 110, 185, 112, 188 Solution to Question 4 

Question 5 A clinic provides a program to help their clients lose weight and asks a consumer agency to investigate the effectiveness of the program. The agency takes a sample of 15 people, weighing each person in the sample before the program begins and 3 months later. The results a tabulated below

Determine is the program is effective. Solution to Question 5

Question 6 A sample of 20 students were selected and given a diagnostic module prior to studying for a test. And then they were given the test again after completing the module. . The result of the students scores in the test before and after the test is tabulated below.

We want to see if there is significant improvement in the student’s performance due to this teaching method Solution to Question 6 

Question 7 A study was performed to test wether cars get better mileage on premium gas than on regular gas. Each of 10 cars was first filled with regular or premium gas, decided by a coin toss, and the mileage for the tank was recorded. The mileage was recorded again for the same cars using other kind of gasoline. Determine wether cars get significantly better mileage with premium gas.

Mileage with regular gas: 16,20,21,22,23,22,27,25,27,28 Mileage with premium gas: 19, 22,24,24,25,25,26,26,28,32 Solution to Question 7 

Question 8  An automatic cutter machine must cut steel strips of 1200 mm length. From a preliminary data, we checked that the lengths of the pieces produced by the machine can be considered as normal random variables  with a 3mm standard deviation. We want to make sure that the machine is set correctly. Therefore 16 pieces of the products are randomly selected and weight. The figures were in mm: 1193,1196,1198,1195,1198,1199,1204,1193,1203,1201,1196,1200,1191,1196,1198,1191 Examine wether there is any significant deviation from the required size Solution to Question 8

Question 9 Blood pressure reading of ten patients before and after medication for reducing the blood pressure are as follows

Patient: 1,2,3,4,5,6,7,8,9,10 Before treatment: 86,84,78,90,92,77,89,90,90,86 After treatment:    80,80,92,79,92,82,88,89,92,83

Test the null hypothesis of no effect agains the alternate hypothesis that medication is effective. Execute it with Wilcoxon test Solution to Question 9

Question on ANOVA Sussan Sound predicts that students will learn most effectively with a constant background sound, as opposed to an unpredictable sound or no sound at all. She randomly divides 24 students into three groups of 8 each. All students study a passage of text for 30 minutes. Those in group 1 study with background sound at a constant volume in the background. Those in group 2 study with nose that changes volume periodically. Those in group 3 study with no sound at all. After studying, all students take a 10 point multiple choice test over the material. Their scores are tabulated below.

Group1: Constant sound: 7,4,6,8,6,6,2,9 Group 2: Random sound: 5,5,3,4,4,7,2,2 Group 3: No sound at all: 2,4,7,1,2,1,5,5 Solution to Question 10

Question 11 Using the following three groups of data, perform a one-way analysis of variance using α  = 0.05.

Solution to Question 11

Question 12 In a packaging plant, a machine packs cartons with jars. It is supposed that a new machine would pack faster on the average than the machine currently used. To test the hypothesis, the time it takes each machine to pack ten cartons are recorded. The result in seconds is as follows.

New Machine: 42,41,41.3,41.8,42.4,42.8,43.2,42.3,41.8,42.7 Old Machine:  42.7,43.6,43.8,43.3,42.5,43.5,43.1,41.7,44,44.1

Perform an F-test to determine if the null hypothesis should be accepted. Solution to Question 12

Question 13 A random sample 500 U.S adults are questioned about their political affiliation and opinion on a tax reform bill. We need to test if the political affiliation and their opinon on a tax reform bill are dependent, at 5% level of significance. The observed contingency table is given below.

Solution to Question 13

Question 14 Can a dice be considered regular which is showing the following frequency distribution during 1000 throws?

Solution to Question 14

Solution to Question 15

Question 16 A newly developed muesli contains five types of seeds (A, B, C, D and E). The percentage of which is 35%, 25%, 20%, 10% and 10% according to the product information. In a randomly selected muesli, the following volume distribution was found.

Lets us decide about the null hypothesis whether the composition of the sample corresponds to the distribution indicated on the packaging at alpha = 0.1 significance level. Solution to Question 16

Question 17 A research team investigated whether there was any significant correlation between the severity of a certain disease runoff and the age of the patients. During the study, data for n = 200 patients were collected and grouped according to the severity of the disease and the age of the patient. The table below shows the result

Let us decided about the correlation between the age of the patients and the severity of disease progression. Solution to Question 17

Question 18 A publisher is interested in determine which of three book cover is most attractive. He interviews 400 people in each of the three states (California, Illinois and New York), and asks each person which of the  cover he or she prefers. The number of preference for each cover is as follows:

Do these data indicate that there are regional differences in people’s preferences concerning these covers? Use the 0.05 level of significance. Solution to Question 18

Question 19 Trees planted along the road were checked for which ones are healthy(H) or diseased (D) and the following arrangement of the trees were obtained:

H H H H D D D H H H H H H H D D H H D D D

Test at the    = 0.05 significance wether this arrangement may be regarded as random

Solution to Question 19 

Question 20 Suppose we flip a coin n = 15 times and come up with the following arrangements

H T T T H H T T T T H H T H H

(H = head, T = tail)

Test at the alpha = 0.05 significance level whether this arrangement may be regarded as random.

Solution to Question 20

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9.4 Full Hypothesis Test Examples

Tests on means, example 9.8.

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. Assume that the swim times for the 25-yard freestyle are normal.

Set up the Hypothesis Test:

Since the problem is about a mean, this is a test of a single population mean .

H 0 : μ = 16.43   H a : μ < 16.43

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:

Random variable: X ¯ X ¯ = the mean time to swim the 25-yard freestyle.

Distribution for the test: X ¯ X ¯ is normal (population standard deviation is known: σ = 0.8)

X ¯ ~ N ( μ , σ X n ) X ¯ ~ N ( μ , σ X n ) Therefore, X ¯ ~ N ( 16.43 , 0.8 15 ) X ¯ ~ N ( 16.43 , 0.8 15 )

μ = 16.43 comes from H 0 and not the data. σ = 0.8, and n = 15.

Calculate the p -value using the normal distribution for a mean:

p -value = P ( x ¯ x ¯ < 16) = 0.0187 where the sample mean in the problem is given as 16.

p -value = 0.0187 (This is called the actual level of significance .) The p -value is the area to the left of the sample mean is given as 16.

μ = 16.43 comes from H 0 . Our assumption is μ = 16.43.

Interpretation of the p -value: If H 0 is true , there is a 0.0187 probability (1.87%)that Jeffrey's mean time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87% chance is small, the mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event.

Compare α and the p -value:

α = 0.05 p -value = 0.0187 α > p -value

Make a decision: Since α > α > p -value, reject H 0 .

This indicates that you reject the null hypothesis that the mean time to swim the 25-yard freestyle is at least 16.43 seconds.

Conclusion: At the 5% significance level, there is sufficient evidence that Jeffrey's mean time to swim the 25-yard freestyle is less than 16.43 seconds. Thus, based on the sample data, we conclude that Jeffrey swims faster using the new goggles.

The Type I and Type II errors for this problem are as follows: 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 at least 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.)

The Type II error is that there is not evidence to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually does swim the 25-yard free-style, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis is false.)

The mean throwing distance of a football for Marco, a high school quarterback, is 40 yards, with a standard deviation of two yards. The team coach tells Marco to adjust his grip to get more distance. The coach records the distances for 20 throws. For the 20 throws, Marco’s mean distance was 45 yards. The coach thought the different grip helped Marco throw farther than 40 yards. Conduct a hypothesis test using a preset α = 0.05. Assume the throw distances for footballs are normal.

First, determine what type of test this is, set up the hypothesis test, find the p -value, sketch the graph, and state your conclusion.

Example 9.9

Jasmine has just begun her new job 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 dollars with a standard deviation of 12 dollars. Test at 5% significance that the population mean is at least 100 dollars 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 Jasmine has met this requirement at the significance level of 95%?

• H 0 : µ ≤ 100 H a : µ > 100 The null and alternative hypothesis are for the parameter µ because the number of dollars of the contracts is a continuous random variable. Also, this is a one-tailed test because the company has only an interested if the number of dollars per contact is below a particular number not "too high" a number. This can be thought of as making a claim that the requirement is being met and thus the claim is in the alternative hypothesis.
• Test statistic: t c = x ¯ − µ 0 s n = 108 − 100 ( 12 16 ) = 2.67 t c = x ¯ − µ 0 s n = 108 − 100 ( 12 16 ) = 2.67
• Critical value: t a = 1.753 t a = 1.753 with n-1 degrees of freedom= 15

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 ( t a ) ( 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 Jasmine's performance meets company standards.

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, state your conclusion, and identify the Type I errors.

Example 9.10

A manufacturer of salad dressings uses machines to dispense liquid ingredients into bottles that move along a filling line. The machine that dispenses salad dressings is working properly when 8 ounces are dispensed. Suppose that the average amount dispensed in a particular sample of 35 bottles is 7.91 ounces with a variance of 0.03 ounces squared, s 2 s 2 . Is there evidence that the machine should be stopped and production wait for repairs? The lost production from a shutdown is potentially so great that management feels that the level of significance in the analysis should be 99%.

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

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.

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.

STEP 5 : Reach a Conclusion

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”.

Try It 9.10

A company records the mean time of employees working in a day. The mean comes out to be 475 minutes, with a standard deviation of 45 minutes. A manager recorded times of 20 employees. The times of working were (frequencies are in parentheses) 460(3); 465(2); 470(3); 475(1); 480(6); 485(3); 490(2).

Conduct a hypothesis test using a 2.5% level of significance to determine if the mean time is more than 475 .

Hypothesis Test for Proportions

Just as there were confidence intervals for proportions, or more formally, the population parameter p of the binomial distribution, there is the ability to test hypotheses concerning p .

The population parameter for the binomial is p . The estimated value (point estimate) for p is p′ where p′ = x/n , x is the number of successes in the sample and n is the sample size.

When you perform a hypothesis test of a population proportion p , you take a simple random sample from the population. The conditions for a binomial distribution must be met, which are: there are a certain number n of independent trials meaning random sampling, the outcomes of any trial are binary, success or failure, and each trial has the same probability of a success p . The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np′ and nq′ must both be greater than five ( np′ > 5 and nq′ > 5). In this case the binomial distribution of a sample (estimated) proportion can be approximated by the normal distribution with μ = np μ = np and σ = npq σ = npq . Remember that q = 1 – p q = 1 – p . There is no distribution that can correct for this small sample bias and thus if these conditions are not met we simply cannot test the hypothesis with the data available at that time. We met this condition when we first were estimating confidence intervals for p .

Again, we begin with the standardizing formula modified because this is the distribution of a binomial.

Substituting p 0 p 0 , the hypothesized value of p , we have:

This is the test statistic for testing hypothesized values of p , where the null and alternative hypotheses take one of the following forms:

The decision rule stated above applies here also: if the calculated value of Z c shows that the sample proportion is "too many" standard deviations from the hypothesized proportion, the null hypothesis cannot be accepted. The decision as to what is "too many" is pre-determined by the analyst depending on the level of significance required in the test.

Example 9.11

The mortgage department of a large bank is interested in the nature of loans of first-time borrowers. This information will be used to tailor their marketing strategy. They believe that 50% of first-time borrowers take out smaller loans than other borrowers. They perform a hypothesis test to determine if the percentage is the same or different from 50% . They sample 100 first-time borrowers and find 53 of these loans are smaller that the other borrowers. For the hypothesis test, they choose a 5% level of significance.

STEP 1 : Set the null and alternative hypothesis.

H 0 : p = 0.50   H a : p ≠ 0.50

The words "is the same or different from" tell you this is a two-tailed test. The Type I and Type II errors are as follows: The Type I error is to conclude that the proportion of borrowers is different from 50% when, in fact, the proportion is actually 50%. (Reject the null hypothesis when the null hypothesis is true). The Type II error is there is not enough evidence to conclude that the proportion of first time borrowers differs from 50% when, in fact, the proportion does differ from 50%. (You fail to reject the null hypothesis when the null hypothesis is false.)

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

The level of significance has been set by the problem at the 5% level. Because this is two-tailed test one-half of the alpha value will be in the upper tail and one-half in the lower tail as shown on the graph. The critical value for the normal distribution at the 95% level of confidence is 1.96. This can easily be found on the student’s t-table at the very bottom at infinite degrees of freedom remembering that at infinity the t-distribution is the normal distribution. Of course the value can also be found on the normal table but you have go looking for one-half of 95 (0.475) inside the body of the table and then read out to the sides and top for the number of standard deviations.

STEP 3 : Calculate the sample parameters and critical value of the test statistic.

The test statistic is a normal distribution, Z, for testing proportions and is:

For this case, the sample of 100 found 53 of these loans were smaller than those of other borrowers. The sample proportion, p′ = 53/100= 0.53 The test question, therefore, is : “Is 0.53 significantly different from .50?” Putting these values into the formula for the test statistic we find that 0.53 is only 0.60 standard deviations away from .50. This is barely off of the mean of the standard normal distribution of zero. There is virtually no difference from the sample proportion and the hypothesized proportion in terms of standard deviations.

STEP 4 : Compare the test statistic and the critical value.

The calculated value is well within the critical values of ± 1.96 standard deviations and thus we cannot reject the null hypothesis. To reject the null hypothesis we need significant evident of difference between the hypothesized value and the sample value. In this case the sample value is very nearly the same as the hypothesized value measured in terms of standard deviations.

STEP 5 : Reach a conclusion

The formal conclusion would be “At a 5% level of significance we cannot reject the null hypothesis that 50% of first-time borrowers take out smaller loans than other borrowers.” Notice the length to which the conclusion goes to include all of the conditions that are attached to the conclusion. Statisticians, for all the criticism they receive, are careful to be very specific even when this seems trivial. Statisticians cannot say more than they know, and the data constrain the conclusion to be within the metes and bounds of the data.

Try It 9.11

A teacher believes that 85% of students in the class will want to go on a field trip to the local zoo. The teacher performs a hypothesis test to determine if the percentage is the same or different from 85%. The teacher samples 50 students and 39 reply that they would want to go to the zoo. For the hypothesis test, use a 1% level of significance.

Example 9.12

Suppose a consumer group suspects that the proportion of households that have three or more cell phones is 30%. A cell phone company has reason to believe that the proportion is not 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 43 of the households have three or more cell phones.

Here is an abbreviate version of the system to solve hypothesis tests applied to a test on a proportions.

Try It 9.12

Marketers believe that 92% of adults in the United States own a cell phone. A cell phone manufacturer believes that number is actually lower. 200 American adults are surveyed, of which, 174 report having cell phones. Use a 5% level of significance. State the null and alternative hypothesis, find the p -value, state your conclusion, and identify the Type I and Type II errors.

Example 9.13

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.

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

• H 0 : μ ≤ 1
• H a : μ > 1
• Plan : We are testing a sample mean without a known population standard deviation with less than 30 observations. Therefore, we need to use a Student's-t distribution. Assume the underlying population is normal.
• Do the calculations and draw the graph .
• State the Conclusions : We cannot accept the null hypothesis. It is reasonable to state that the data supports the claim that the average conductivity level is greater than one.

Try It 9.13

The boiling point of a specific liquid is measured for 15 samples, and the boiling points are obtained as follows:

205; 206; 206; 202; 199; 194; 197; 198; 198; 201; 201; 202; 207; 211; 205

Is there convincing evidence that the average boiling point is greater than 200? Use a significance level of 0.1. Assume the population is normal.

Example 9.14

In a study of 420,019 cell phone users, 172 of the subjects developed brain cancer. Test the claim that cell phone users developed brain cancer at a greater rate than that for non-cell phone users (the rate of brain cancer for non-cell phone users is 0.0340%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error.

• H 0 : p ≤ 0.00034
• H a : p > 0.00034

If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes cancer-causing environments, we want to minimize the chances of incorrectly identifying causes of cancer.

• We will be testing a sample proportion with x = 172 and n = 420,019. The sample is sufficiently large because we have np' = 420,019(0.00034) = 142.8, nq' = 420,019(0.99966) = 419,876.2, two independent outcomes, and a fixed probability of success p' = 0.00034. Thus we will be able to generalize our results to the population.

Try It 9.14

In a study of 390,000 moisturizer users, 138 of the subjects developed skin diseases. Test the claim that moisturizer users developed skin diseases at a greater rate than that for non-moisturizer users (the rate of skin diseases for non-moisturizer users is 0.041%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error.

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Hypothesis Testing

Year 1 students, year 1 course page.

• Hypothesis testing with binomial distributions.
• Critical regions.

Hypothesis Testing

When you conduct a piece of quantitative research, you are inevitably attempting to answer a research question or hypothesis that you have set. One method of evaluating this research question is via a process called hypothesis testing , which is sometimes also referred to as significance testing . Since there are many facets to hypothesis testing, we start with the example we refer to throughout this guide.

An example of a lecturer's dilemma

Two statistics lecturers, Sarah and Mike, think that they use the best method to teach their students. Each lecturer has 50 statistics students who are studying a graduate degree in management. In Sarah's class, students have to attend one lecture and one seminar class every week, whilst in Mike's class students only have to attend one lecture. Sarah thinks that seminars, in addition to lectures, are an important teaching method in statistics, whilst Mike believes that lectures are sufficient by themselves and thinks that students are better off solving problems by themselves in their own time. This is the first year that Sarah has given seminars, but since they take up a lot of her time, she wants to make sure that she is not wasting her time and that seminars improve her students' performance.

The research hypothesis

The first step in hypothesis testing is to set a research hypothesis. In Sarah and Mike's study, the aim is to examine the effect that two different teaching methods – providing both lectures and seminar classes (Sarah), and providing lectures by themselves (Mike) – had on the performance of Sarah's 50 students and Mike's 50 students. More specifically, they want to determine whether performance is different between the two different teaching methods. Whilst Mike is skeptical about the effectiveness of seminars, Sarah clearly believes that giving seminars in addition to lectures helps her students do better than those in Mike's class. This leads to the following research hypothesis:

Before moving onto the second step of the hypothesis testing process, we need to take you on a brief detour to explain why you need to run hypothesis testing at all. This is explained next.

Sample to population

If you have measured individuals (or any other type of "object") in a study and want to understand differences (or any other type of effect), you can simply summarize the data you have collected. For example, if Sarah and Mike wanted to know which teaching method was the best, they could simply compare the performance achieved by the two groups of students – the group of students that took lectures and seminar classes, and the group of students that took lectures by themselves – and conclude that the best method was the teaching method which resulted in the highest performance. However, this is generally of only limited appeal because the conclusions could only apply to students in this study. However, if those students were representative of all statistics students on a graduate management degree, the study would have wider appeal.

In statistics terminology, the students in the study are the sample and the larger group they represent (i.e., all statistics students on a graduate management degree) is called the population . Given that the sample of statistics students in the study are representative of a larger population of statistics students, you can use hypothesis testing to understand whether any differences or effects discovered in the study exist in the population. In layman's terms, hypothesis testing is used to establish whether a research hypothesis extends beyond those individuals examined in a single study.

Another example could be taking a sample of 200 breast cancer sufferers in order to test a new drug that is designed to eradicate this type of cancer. As much as you are interested in helping these specific 200 cancer sufferers, your real goal is to establish that the drug works in the population (i.e., all breast cancer sufferers).

As such, by taking a hypothesis testing approach, Sarah and Mike want to generalize their results to a population rather than just the students in their sample. However, in order to use hypothesis testing, you need to re-state your research hypothesis as a null and alternative hypothesis. Before you can do this, it is best to consider the process/structure involved in hypothesis testing and what you are measuring. This structure is presented on the next page .

A Level Maths

Maths A-Level Resources for AQA, OCR and Edexcel

Hypothesis Testing

• Hypothesis testing is just a method for testing a claim or hypothesis about a parameter in a population, using data measured in a sample.

• Four steps of hypothesis testing are: (i) We state the Hypothesis (ii) Set the criteria for a decision (iii) Compute the test statistic (iv) Make a decision

What is Hypothesis Testing?

We use samples because it allows us to measure behaviors and to learn more about the behavior in populations that are often too large or inaccessible.

Hypothesis testing is just a method for testing a claim or hypothesis about a parameter in a population, using data measured in a sample. In this method, we test some hypothesis by determining the likelihood that a sample statistic could have been selected, if the hypothesis regarding the population parameter were true.

A few types of Hypothesis:

Some of the examples of null hypotheses that are generally accepted as being true are:

(i) DNA is shaped like a double helix. (ii) There are 8 planets in the solar system (excluding Pluto).

The four steps of hypothesis testing:

Step 1: We state the Hypothesis. We start by assuming that the hypothesis or claim we are testing is true. This is stated in the null hypothesis.

Step 3: Compute the test statistic. The test statistic is a mathematical formula that allows researchers to determine the likelihood of obtaining sample outcomes if the null hypothesis were true. The value of the test statistic is used to make a decision regarding the null hypothesis.

Step 4: Make a decision. We use the value of the test statistic to make a decision about the null hypothesis. The decision is based on the probability of obtaining a sample mean, given that the value stated in the null hypothesis is true. There are two decisions a researcher can make; either reject the null hypothesis or retain the null hypothesis.

When testing a hypothesis of a proportion, we use the z-test and the formula for this is:

Q. The average score of all sixth graders in school District A on a math aptitude exam is 75 with a standard deviation of 8.1. A random sample of 100 students in one school was taken. The mean score of these 100 students was 71. Does this indicate that the students of this school are significantly less skilled in their mathematical abilities than the average student in the district? (Use a 5% level of significance.)

Firstly we write down the data provided to us in the question:

We know the mean for the population µ = 75 and standard deviation for the population σ = 8.1

Thus, we are testing the sample mean against the population mean with a population standard deviation which is known to us.

We will use the z-test here.

Now we carry out the above steps in order to come to a conclusion.

Step 1: We state the null hypothesis and the alternate hypothesis:

Step 2: We select the level of significance which is stated in the problem as 5% or α = 0.05

Step 3: Compute the test statistics. We first identify the test to be used. In this case we are using the z-test because is known and the sample is n=100 is a large sample.

Step 4: Make a decision. Since the alternate hypothesis states µ < 75, this is a one-tailed test to the left.

For α = 0.05, z in the normal curve table that gives a probability of 0.05 to the left of z.

Hence, the critical value after looking at the table gives a value of 0.5 – 0.05 = 0.45 or z = -1.645. That is P(z < -1.645) = 0.05.

Because 0.4500 is exactly halfway between 0.4495 and 0.4505, we get half way between 1.640 and 1.650 to get z = 1.645. Since 71 is to the left of 75, we have z = -1.645.

That is P(z < -1.645) = 0.05.

Since the computed z = -4.938 < -1.645 (critical z value), we reject the null hypothesis that the students in the school are not less skilled in mathematical ability.

• http://cfcc.edu/faculty/cmoore/0801-HypothesisTests.pdf
• https://www.sagepub.com/sites/default/files/upm-binaries/40007_Chapter8.pdf
• https://www.a-levelmaths.com/Summary%20Handouts/Hypothesis%20Testing%20Summary.pdf