hypothesis test example statistics

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Lesson 10 of 24 By Avijeet Biswal

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In today’s data-driven world , decisions are based on data all the time. Hypothesis plays a crucial role in that process, whether it may be making business decisions, in the health sector, academia, or in quality improvement. Without hypothesis & hypothesis tests, you risk drawing the wrong conclusions and making bad decisions. In this tutorial, you will look at Hypothesis Testing in Statistics.

What Is Hypothesis Testing in Statistics?

Hypothesis Testing is a type of statistical analysis in which you put your assumptions about a population parameter to the test. It is used to estimate the relationship between 2 statistical variables.

Let's discuss few examples of statistical hypothesis from real-life - 

• A teacher assumes that 60% of his college's students come from lower-middle-class families.
• A doctor believes that 3D (Diet, Dose, and Discipline) is 90% effective for diabetic patients.

Now that you know about hypothesis testing, look at the two types of hypothesis testing in statistics.

Hypothesis Testing Formula

Z = ( x̅ – μ0 ) / (σ /√n)

• Here, x̅ is the sample mean,
• μ0 is the population mean,
• σ is the standard deviation,
• n is the sample size.

How Hypothesis Testing Works?

An analyst performs hypothesis testing on a statistical sample to present evidence of the plausibility of the null hypothesis. Measurements and analyses are conducted on a random sample of the population to test a theory. Analysts use a random population sample to test two hypotheses: the null and alternative hypotheses.

The null hypothesis is typically an equality hypothesis between population parameters; for example, a null hypothesis may claim that the population means return equals zero. The alternate hypothesis is essentially the inverse of the null hypothesis (e.g., the population means the return is not equal to zero). As a result, they are mutually exclusive, and only one can be correct. One of the two possibilities, however, will always be correct.

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Null Hypothesis and Alternate Hypothesis

The Null Hypothesis is the assumption that the event will not occur. A null hypothesis has no bearing on the study's outcome unless it is rejected.

H0 is the symbol for it, and it is pronounced H-naught.

The Alternate Hypothesis is the logical opposite of the null hypothesis. The acceptance of the alternative hypothesis follows the rejection of the null hypothesis. H1 is the symbol for it.

Let's understand this with an example.

A sanitizer manufacturer claims that its product kills 95 percent of germs on average. 

To put this company's claim to the test, create a null and alternate hypothesis.

H0 (Null Hypothesis): Average = 95%.

Alternative Hypothesis (H1): The average is less than 95%.

Another straightforward example to understand this concept is determining whether or not a coin is fair and balanced. The null hypothesis states that the probability of a show of heads is equal to the likelihood of a show of tails. In contrast, the alternate theory states that the probability of a show of heads and tails would be very different.

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Hypothesis Testing Calculation With Examples

Let's consider a hypothesis test for the average height of women in the United States. Suppose our null hypothesis is that the average height is 5'4". We gather a sample of 100 women and determine that their average height is 5'5". The standard deviation of population is 2.

To calculate the z-score, we would use the following formula:

z = ( x̅ – μ0 ) / (σ /√n)

z = (5'5" - 5'4") / (2" / √100)

z = 0.5 / (0.045)

 We will reject the null hypothesis as the z-score of 11.11 is very large and conclude that there is evidence to suggest that the average height of women in the US is greater than 5'4".

Steps of Hypothesis Testing

Step 1: specify your null and alternate hypotheses.

It is critical to rephrase your original research hypothesis (the prediction that you wish to study) as a null (Ho) and alternative (Ha) hypothesis so that you can test it quantitatively. Your first hypothesis, which predicts a link between variables, is generally your alternate hypothesis. The null hypothesis predicts no link between the variables of interest.

Step 2: Gather Data

For a statistical test to be legitimate, sampling and data collection must be done in a way that is meant to test your hypothesis. You cannot draw statistical conclusions about the population you are interested in if your data is not representative.

Step 3: Conduct a Statistical Test

Other statistical tests are available, but they all compare within-group variance (how to spread out the data inside a category) against between-group variance (how different the categories are from one another). If the between-group variation is big enough that there is little or no overlap between groups, your statistical test will display a low p-value to represent this. This suggests that the disparities between these groups are unlikely to have occurred by accident. Alternatively, if there is a large within-group variance and a low between-group variance, your statistical test will show a high p-value. Any difference you find across groups is most likely attributable to chance. The variety of variables and the level of measurement of your obtained data will influence your statistical test selection.

Step 4: Determine Rejection Of Your Null Hypothesis

Your statistical test results must determine whether your null hypothesis should be rejected or not. In most circumstances, you will base your judgment on the p-value provided by the statistical test. In most circumstances, your preset level of significance for rejecting the null hypothesis will be 0.05 - that is, when there is less than a 5% likelihood that these data would be seen if the null hypothesis were true. In other circumstances, researchers use a lower level of significance, such as 0.01 (1%). This reduces the possibility of wrongly rejecting the null hypothesis.

Step 5: Present Your Results 

The findings of hypothesis testing will be discussed in the results and discussion portions of your research paper, dissertation, or thesis. You should include a concise overview of the data and a summary of the findings of your statistical test in the results section. You can talk about whether your results confirmed your initial hypothesis or not in the conversation. Rejecting or failing to reject the null hypothesis is a formal term used in hypothesis testing. This is likely a must for your statistics assignments.

Types of Hypothesis Testing

To determine whether a discovery or relationship is statistically significant, hypothesis testing uses a z-test. It usually checks to see if two means are the same (the null hypothesis). Only when the population standard deviation is known and the sample size is 30 data points or more, can a z-test be applied.

A statistical test called a t-test is employed to compare the means of two groups. To determine whether two groups differ or if a procedure or treatment affects the population of interest, it is frequently used in hypothesis testing.

Chi-Square 

You utilize a Chi-square test for hypothesis testing concerning whether your data is as predicted. To determine if the expected and observed results are well-fitted, the Chi-square test analyzes the differences between categorical variables from a random sample. The test's fundamental premise is that the observed values in your data should be compared to the predicted values that would be present if the null hypothesis were true.

Hypothesis Testing and Confidence Intervals

Both confidence intervals and hypothesis tests are inferential techniques that depend on approximating the sample distribution. Data from a sample is used to estimate a population parameter using confidence intervals. Data from a sample is used in hypothesis testing to examine a given hypothesis. We must have a postulated parameter to conduct hypothesis testing.

Bootstrap distributions and randomization distributions are created using comparable simulation techniques. The observed sample statistic is the focal point of a bootstrap distribution, whereas the null hypothesis value is the focal point of a randomization distribution.

A variety of feasible population parameter estimates are included in confidence ranges. In this lesson, we created just two-tailed confidence intervals. There is a direct connection between these two-tail confidence intervals and these two-tail hypothesis tests. The results of a two-tailed hypothesis test and two-tailed confidence intervals typically provide the same results. In other words, a hypothesis test at the 0.05 level will virtually always fail to reject the null hypothesis if the 95% confidence interval contains the predicted value. A hypothesis test at the 0.05 level will nearly certainly reject the null hypothesis if the 95% confidence interval does not include the hypothesized parameter.

Simple and Composite Hypothesis Testing

Depending on the population distribution, you can classify the statistical hypothesis into two types.

Simple Hypothesis: A simple hypothesis specifies an exact value for the parameter.

Composite Hypothesis: A composite hypothesis specifies a range of values.

A company is claiming that their average sales for this quarter are 1000 units. This is an example of a simple hypothesis.

Suppose the company claims that the sales are in the range of 900 to 1000 units. Then this is a case of a composite hypothesis.

One-Tailed and Two-Tailed Hypothesis Testing

The One-Tailed test, also called a directional test, considers a critical region of data that would result in the null hypothesis being rejected if the test sample falls into it, inevitably meaning the acceptance of the alternate hypothesis.

In a one-tailed test, the critical distribution area is one-sided, meaning the test sample is either greater or lesser than a specific value.

In two tails, the test sample is checked to be greater or less than a range of values in a Two-Tailed test, implying that the critical distribution area is two-sided.

If the sample falls within this range, the alternate hypothesis will be accepted, and the null hypothesis will be rejected.

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Right Tailed Hypothesis Testing

If the larger than (>) sign appears in your hypothesis statement, you are using a right-tailed test, also known as an upper test. Or, to put it another way, the disparity is to the right. For instance, you can contrast the battery life before and after a change in production. Your hypothesis statements can be the following if you want to know if the battery life is longer than the original (let's say 90 hours):

• The null hypothesis is (H0 <= 90) or less change.
• A possibility is that battery life has risen (H1) > 90.

The crucial point in this situation is that the alternate hypothesis (H1), not the null hypothesis, decides whether you get a right-tailed test.

Left Tailed Hypothesis Testing

Alternative hypotheses that assert the true value of a parameter is lower than the null hypothesis are tested with a left-tailed test; they are indicated by the asterisk "<".

Suppose H0: mean = 50 and H1: mean not equal to 50

According to the H1, the mean can be greater than or less than 50. This is an example of a Two-tailed test.

In a similar manner, if H0: mean >=50, then H1: mean <50

Here the mean is less than 50. It is called a One-tailed test.

Type 1 and Type 2 Error

A hypothesis test can result in two types of errors.

Type 1 Error: A Type-I error occurs when sample results reject the null hypothesis despite being true.

Type 2 Error: A Type-II error occurs when the null hypothesis is not rejected when it is false, unlike a Type-I error.

Suppose a teacher evaluates the examination paper to decide whether a student passes or fails.

H0: Student has passed

H1: Student has failed

Type I error will be the teacher failing the student [rejects H0] although the student scored the passing marks [H0 was true]. 

Type II error will be the case where the teacher passes the student [do not reject H0] although the student did not score the passing marks [H1 is true].

Level of Significance

The alpha value is a criterion for determining whether a test statistic is statistically significant. In a statistical test, Alpha represents an acceptable probability of a Type I error. Because alpha is a probability, it can be anywhere between 0 and 1. In practice, the most commonly used alpha values are 0.01, 0.05, and 0.1, which represent a 1%, 5%, and 10% chance of a Type I error, respectively (i.e. rejecting the null hypothesis when it is in fact correct).

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A p-value is a metric that expresses the likelihood that an observed difference could have occurred by chance. As the p-value decreases the statistical significance of the observed difference increases. If the p-value is too low, you reject the null hypothesis.

Here you have taken an example in which you are trying to test whether the new advertising campaign has increased the product's sales. The p-value is the likelihood that the null hypothesis, which states that there is no change in the sales due to the new advertising campaign, is true. If the p-value is .30, then there is a 30% chance that there is no increase or decrease in the product's sales.  If the p-value is 0.03, then there is a 3% probability that there is no increase or decrease in the sales value due to the new advertising campaign. As you can see, the lower the p-value, the chances of the alternate hypothesis being true increases, which means that the new advertising campaign causes an increase or decrease in sales.

Why is Hypothesis Testing Important in Research Methodology?

Hypothesis testing is crucial in research methodology for several reasons:

• Provides evidence-based conclusions: It allows researchers to make objective conclusions based on empirical data, providing evidence to support or refute their research hypotheses.
• Supports decision-making: It helps make informed decisions, such as accepting or rejecting a new treatment, implementing policy changes, or adopting new practices.
• Adds rigor and validity: It adds scientific rigor to research using statistical methods to analyze data, ensuring that conclusions are based on sound statistical evidence.
• Contributes to the advancement of knowledge: By testing hypotheses, researchers contribute to the growth of knowledge in their respective fields by confirming existing theories or discovering new patterns and relationships.

Limitations of Hypothesis Testing

Hypothesis testing has some limitations that researchers should be aware of:

• It cannot prove or establish the truth: Hypothesis testing provides evidence to support or reject a hypothesis, but it cannot confirm the absolute truth of the research question.
• Results are sample-specific: Hypothesis testing is based on analyzing a sample from a population, and the conclusions drawn are specific to that particular sample.
• Possible errors: During hypothesis testing, there is a chance of committing type I error (rejecting a true null hypothesis) or type II error (failing to reject a false null hypothesis).
• Assumptions and requirements: Different tests have specific assumptions and requirements that must be met to accurately interpret results.

After reading this tutorial, you would have a much better understanding of hypothesis testing, one of the most important concepts in the field of Data Science . The majority of hypotheses are based on speculation about observed behavior, natural phenomena, or established theories.

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If you have any questions regarding this ‘Hypothesis Testing In Statistics’ tutorial, do share them in the comment section. Our subject matter expert will respond to your queries. Happy learning!

1. What is hypothesis testing in statistics with example?

Hypothesis testing is a statistical method used to determine if there is enough evidence in a sample data to draw conclusions about a population. It involves formulating two competing hypotheses, the null hypothesis (H0) and the alternative hypothesis (Ha), and then collecting data to assess the evidence. An example: testing if a new drug improves patient recovery (Ha) compared to the standard treatment (H0) based on collected patient data.

2. What is hypothesis testing and its types?

Hypothesis testing is a statistical method used to make inferences about a population based on sample data. It involves formulating two hypotheses: the null hypothesis (H0), which represents the default assumption, and the alternative hypothesis (Ha), which contradicts H0. The goal is to assess the evidence and determine whether there is enough statistical significance to reject the null hypothesis in favor of the alternative hypothesis.

Types of hypothesis testing:

• One-sample test: Used to compare a sample to a known value or a hypothesized value.
• Two-sample test: Compares two independent samples to assess if there is a significant difference between their means or distributions.
• Paired-sample test: Compares two related samples, such as pre-test and post-test data, to evaluate changes within the same subjects over time or under different conditions.
• Chi-square test: Used to analyze categorical data and determine if there is a significant association between variables.
• ANOVA (Analysis of Variance): Compares means across multiple groups to check if there is a significant difference between them.

3. What are the steps of hypothesis testing?

The steps of hypothesis testing are as follows:

• Formulate the hypotheses: State the null hypothesis (H0) and the alternative hypothesis (Ha) based on the research question.
• Set the significance level: Determine the acceptable level of error (alpha) for making a decision.
• Collect and analyze data: Gather and process the sample data.
• Compute test statistic: Calculate the appropriate statistical test to assess the evidence.
• Make a decision: Compare the test statistic with critical values or p-values and determine whether to reject H0 in favor of Ha or not.
• Draw conclusions: Interpret the results and communicate the findings in the context of the research question.

4. What are the 2 types of hypothesis testing?

• One-tailed (or one-sided) test: Tests for the significance of an effect in only one direction, either positive or negative.
• Two-tailed (or two-sided) test: Tests for the significance of an effect in both directions, allowing for the possibility of a positive or negative effect.

The choice between one-tailed and two-tailed tests depends on the specific research question and the directionality of the expected effect.

5. What are the 3 major types of hypothesis?

The three major types of hypotheses are:

• Null Hypothesis (H0): Represents the default assumption, stating that there is no significant effect or relationship in the data.
• Alternative Hypothesis (Ha): Contradicts the null hypothesis and proposes a specific effect or relationship that researchers want to investigate.
• Nondirectional Hypothesis: An alternative hypothesis that doesn't specify the direction of the effect, leaving it open for both positive and negative possibilities.

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Avijeet is a Senior Research Analyst at Simplilearn. Passionate about Data Analytics, Machine Learning, and Deep Learning, Avijeet is also interested in politics, cricket, and football.

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Hypothesis Testing

Hypothesis testing is a tool for making statistical inferences about the population data. It is an analysis tool that tests assumptions and determines how likely something is within a given standard of accuracy. Hypothesis testing provides a way to verify whether the results of an experiment are valid.

A null hypothesis and an alternative hypothesis are set up before performing the hypothesis testing. This helps to arrive at a conclusion regarding the sample obtained from the population. In this article, we will learn more about hypothesis testing, its types, steps to perform the testing, and associated examples.

What is Hypothesis Testing in Statistics?

Hypothesis testing uses sample data from the population to draw useful conclusions regarding the population probability distribution . It tests an assumption made about the data using different types of hypothesis testing methodologies. The hypothesis testing results in either rejecting or not rejecting the null hypothesis.

Hypothesis Testing Definition

Hypothesis testing can be defined as a statistical tool that is used to identify if the results of an experiment are meaningful or not. It involves setting up a null hypothesis and an alternative hypothesis. These two hypotheses will always be mutually exclusive. This means that if the null hypothesis is true then the alternative hypothesis is false and vice versa. An example of hypothesis testing is setting up a test to check if a new medicine works on a disease in a more efficient manner.

Null Hypothesis

The null hypothesis is a concise mathematical statement that is used to indicate that there is no difference between two possibilities. In other words, there is no difference between certain characteristics of data. This hypothesis assumes that the outcomes of an experiment are based on chance alone. It is denoted as \(H_{0}\). Hypothesis testing is used to conclude if the null hypothesis can be rejected or not. Suppose an experiment is conducted to check if girls are shorter than boys at the age of 5. The null hypothesis will say that they are the same height.

Alternative Hypothesis

The alternative hypothesis is an alternative to the null hypothesis. It is used to show that the observations of an experiment are due to some real effect. It indicates that there is a statistical significance between two possible outcomes and can be denoted as \(H_{1}\) or \(H_{a}\). For the above-mentioned example, the alternative hypothesis would be that girls are shorter than boys at the age of 5.

Hypothesis Testing P Value

In hypothesis testing, the p value is used to indicate whether the results obtained after conducting a test are statistically significant or not. It also indicates the probability of making an error in rejecting or not rejecting the null hypothesis.This value is always a number between 0 and 1. The p value is compared to an alpha level, \(\alpha\) or significance level. The alpha level can be defined as the acceptable risk of incorrectly rejecting the null hypothesis. The alpha level is usually chosen between 1% to 5%.

Hypothesis Testing Critical region

All sets of values that lead to rejecting the null hypothesis lie in the critical region. Furthermore, the value that separates the critical region from the non-critical region is known as the critical value.

Hypothesis Testing Formula

Depending upon the type of data available and the size, different types of hypothesis testing are used to determine whether the null hypothesis can be rejected or not. The hypothesis testing formula for some important test statistics are given below:

• z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\). \(\overline{x}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation and n is the size of the sample.
• t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\). s is the sample standard deviation.
• \(\chi ^{2} = \sum \frac{(O_{i}-E_{i})^{2}}{E_{i}}\). \(O_{i}\) is the observed value and \(E_{i}\) is the expected value.

We will learn more about these test statistics in the upcoming section.

Types of Hypothesis Testing

Selecting the correct test for performing hypothesis testing can be confusing. These tests are used to determine a test statistic on the basis of which the null hypothesis can either be rejected or not rejected. Some of the important tests used for hypothesis testing are given below.

Hypothesis Testing Z Test

A z test is a way of hypothesis testing that is used for a large sample size (n ≥ 30). It is used to determine whether there is a difference between the population mean and the sample mean when the population standard deviation is known. It can also be used to compare the mean of two samples. It is used to compute the z test statistic. The formulas are given as follows:

• One sample: z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\).
• Two samples: z = \(\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}\).

Hypothesis Testing t Test

The t test is another method of hypothesis testing that is used for a small sample size (n < 30). It is also used to compare the sample mean and population mean. However, the population standard deviation is not known. Instead, the sample standard deviation is known. The mean of two samples can also be compared using the t test.

• One sample: t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\).
• Two samples: t = \(\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}\).

Hypothesis Testing Chi Square

The Chi square test is a hypothesis testing method that is used to check whether the variables in a population are independent or not. It is used when the test statistic is chi-squared distributed.

One Tailed Hypothesis Testing

One tailed hypothesis testing is done when the rejection region is only in one direction. It can also be known as directional hypothesis testing because the effects can be tested in one direction only. This type of testing is further classified into the right tailed test and left tailed test.

Right Tailed Hypothesis Testing

The right tail test is also known as the upper tail test. This test is used to check whether the population parameter is greater than some value. The null and alternative hypotheses for this test are given as follows:

\(H_{0}\): The population parameter is ≤ some value

\(H_{1}\): The population parameter is > some value.

If the test statistic has a greater value than the critical value then the null hypothesis is rejected

Left Tailed Hypothesis Testing

The left tail test is also known as the lower tail test. It is used to check whether the population parameter is less than some value. The hypotheses for this hypothesis testing can be written as follows:

\(H_{0}\): The population parameter is ≥ some value

\(H_{1}\): The population parameter is < some value.

The null hypothesis is rejected if the test statistic has a value lesser than the critical value.

Two Tailed Hypothesis Testing

In this hypothesis testing method, the critical region lies on both sides of the sampling distribution. It is also known as a non - directional hypothesis testing method. The two-tailed test is used when it needs to be determined if the population parameter is assumed to be different than some value. The hypotheses can be set up as follows:

\(H_{0}\): the population parameter = some value

\(H_{1}\): the population parameter ≠ some value

The null hypothesis is rejected if the test statistic has a value that is not equal to the critical value.

Hypothesis Testing Steps

Hypothesis testing can be easily performed in five simple steps. The most important step is to correctly set up the hypotheses and identify the right method for hypothesis testing. The basic steps to perform hypothesis testing are as follows:

• Step 1: Set up the null hypothesis by correctly identifying whether it is the left-tailed, right-tailed, or two-tailed hypothesis testing.
• Step 2: Set up the alternative hypothesis.
• Step 3: Choose the correct significance level, \(\alpha\), and find the critical value.
• Step 4: Calculate the correct test statistic (z, t or \(\chi\)) and p-value.
• Step 5: Compare the test statistic with the critical value or compare the p-value with \(\alpha\) to arrive at a conclusion. In other words, decide if the null hypothesis is to be rejected or not.

Hypothesis Testing Example

The best way to solve a problem on hypothesis testing is by applying the 5 steps mentioned in the previous section. Suppose a researcher claims that the mean average weight of men is greater than 100kgs with a standard deviation of 15kgs. 30 men are chosen with an average weight of 112.5 Kgs. Using hypothesis testing, check if there is enough evidence to support the researcher's claim. The confidence interval is given as 95%.

Step 1: This is an example of a right-tailed test. Set up the null hypothesis as \(H_{0}\): \(\mu\) = 100.

Step 2: The alternative hypothesis is given by \(H_{1}\): \(\mu\) > 100.

Step 3: As this is a one-tailed test, \(\alpha\) = 100% - 95% = 5%. This can be used to determine the critical value.

1 - \(\alpha\) = 1 - 0.05 = 0.95

0.95 gives the required area under the curve. Now using a normal distribution table, the area 0.95 is at z = 1.645. A similar process can be followed for a t-test. The only additional requirement is to calculate the degrees of freedom given by n - 1.

Step 4: Calculate the z test statistic. This is because the sample size is 30. Furthermore, the sample and population means are known along with the standard deviation.

z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\).

\(\mu\) = 100, \(\overline{x}\) = 112.5, n = 30, \(\sigma\) = 15

z = \(\frac{112.5-100}{\frac{15}{\sqrt{30}}}\) = 4.56

Step 5: Conclusion. As 4.56 > 1.645 thus, the null hypothesis can be rejected.

Hypothesis Testing and Confidence Intervals

Confidence intervals form an important part of hypothesis testing. This is because the alpha level can be determined from a given confidence interval. Suppose a confidence interval is given as 95%. Subtract the confidence interval from 100%. This gives 100 - 95 = 5% or 0.05. This is the alpha value of a one-tailed hypothesis testing. To obtain the alpha value for a two-tailed hypothesis testing, divide this value by 2. This gives 0.05 / 2 = 0.025.

Related Articles:

• Probability and Statistics
• Data Handling

Important Notes on Hypothesis Testing

• Hypothesis testing is a technique that is used to verify whether the results of an experiment are statistically significant.
• It involves the setting up of a null hypothesis and an alternate hypothesis.
• There are three types of tests that can be conducted under hypothesis testing - z test, t test, and chi square test.
• Hypothesis testing can be classified as right tail, left tail, and two tail tests.

Examples on Hypothesis Testing

• Example 1: The average weight of a dumbbell in a gym is 90lbs. However, a physical trainer believes that the average weight might be higher. A random sample of 5 dumbbells with an average weight of 110lbs and a standard deviation of 18lbs. Using hypothesis testing check if the physical trainer's claim can be supported for a 95% confidence level. Solution: As the sample size is lesser than 30, the t-test is used. \(H_{0}\): \(\mu\) = 90, \(H_{1}\): \(\mu\) > 90 \(\overline{x}\) = 110, \(\mu\) = 90, n = 5, s = 18. \(\alpha\) = 0.05 Using the t-distribution table, the critical value is 2.132 t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\) t = 2.484 As 2.484 > 2.132, the null hypothesis is rejected. Answer: The average weight of the dumbbells may be greater than 90lbs
• Example 2: The average score on a test is 80 with a standard deviation of 10. With a new teaching curriculum introduced it is believed that this score will change. On random testing, the score of 38 students, the mean was found to be 88. With a 0.05 significance level, is there any evidence to support this claim? Solution: This is an example of two-tail hypothesis testing. The z test will be used. \(H_{0}\): \(\mu\) = 80, \(H_{1}\): \(\mu\) ≠ 80 \(\overline{x}\) = 88, \(\mu\) = 80, n = 36, \(\sigma\) = 10. \(\alpha\) = 0.05 / 2 = 0.025 The critical value using the normal distribution table is 1.96 z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\) z = \(\frac{88-80}{\frac{10}{\sqrt{36}}}\) = 4.8 As 4.8 > 1.96, the null hypothesis is rejected. Answer: There is a difference in the scores after the new curriculum was introduced.
• Example 3: The average score of a class is 90. However, a teacher believes that the average score might be lower. The scores of 6 students were randomly measured. The mean was 82 with a standard deviation of 18. With a 0.05 significance level use hypothesis testing to check if this claim is true. Solution: The t test will be used. \(H_{0}\): \(\mu\) = 90, \(H_{1}\): \(\mu\) < 90 \(\overline{x}\) = 110, \(\mu\) = 90, n = 6, s = 18 The critical value from the t table is -2.015 t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\) t = \(\frac{82-90}{\frac{18}{\sqrt{6}}}\) t = -1.088 As -1.088 > -2.015, we fail to reject the null hypothesis. Answer: There is not enough evidence to support the claim.

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FAQs on Hypothesis Testing

What is hypothesis testing.

Hypothesis testing in statistics is a tool that is used to make inferences about the population data. It is also used to check if the results of an experiment are valid.

What is the z Test in Hypothesis Testing?

The z test in hypothesis testing is used to find the z test statistic for normally distributed data . The z test is used when the standard deviation of the population is known and the sample size is greater than or equal to 30.

What is the t Test in Hypothesis Testing?

The t test in hypothesis testing is used when the data follows a student t distribution . It is used when the sample size is less than 30 and standard deviation of the population is not known.

What is the formula for z test in Hypothesis Testing?

The formula for a one sample z test in hypothesis testing is z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\) and for two samples is z = \(\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}\).

What is the p Value in Hypothesis Testing?

The p value helps to determine if the test results are statistically significant or not. In hypothesis testing, the null hypothesis can either be rejected or not rejected based on the comparison between the p value and the alpha level.

What is One Tail Hypothesis Testing?

When the rejection region is only on one side of the distribution curve then it is known as one tail hypothesis testing. The right tail test and the left tail test are two types of directional hypothesis testing.

What is the Alpha Level in Two Tail Hypothesis Testing?

To get the alpha level in a two tail hypothesis testing divide \(\alpha\) by 2. This is done as there are two rejection regions in the curve.

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Unit 12: Significance tests (hypothesis testing)

About this unit, 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)

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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Example of performing a basic hypothesis test

You can follow six basic steps to correctly set up and perform a hypothesis test. For example, the manager of a pipe manufacturing facility wants to determine whether the average diameter of their pipes is different from 5cm. The manager follows the basic steps for doing a hypothesis test.

You should determine the criteria for the test and the required sample size before you collect the data.

First, the manager formulates the hypotheses. The null hypothesis is: The population mean of all the pipes is equal to 5 cm. Formally, this is written as: H 0 : μ = 5

Because they need to ensure that the pipes are not larger or smaller than 5 cm, the manager chooses the two-sided alternative hypothesis, which states that the population mean of all the pipes is not equal to 5 cm. Formally, this is written as H 1 : μ ≠ 5

• Choose a significance level (also called alpha or α). The manager selects a significance level 0.05, which is the most commonly used significance level.
• Determine the power and sample size for the test. The manager uses a power and sample size calculation to determine how many pipes they need to measure to have a good chance of detecting a difference of 0.1 cm or more from the target diameter.
• Collect the data. They collect a sample of pipes and measure their diameters.
• Compare the p-value from the test to the significance level. After they perform the hypothesis test, the manager obtains a p-value of 0.004. The p-value is less than the significance level of 0.05.
• Decide whether to reject or fail to reject the null hypothesis. The manager rejects the null hypothesis and concludes that the mean pipe diameter of all pipes is not equal to 5cm.
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10.1 - setting the hypotheses: examples.

A significance test examines whether the null hypothesis provides a plausible explanation of the data. The null hypothesis itself does not involve the data. It is a statement about a parameter (a numerical characteristic of the population). These population values might be proportions or means or differences between means or proportions or correlations or odds ratios or any other numerical summary of the population. The alternative hypothesis is typically the research hypothesis of interest. Here are some examples.

Example 10.2: Hypotheses with One Sample of One Categorical Variable Section  

About 10% of the human population is left-handed. Suppose a researcher at Penn State speculates that students in the College of Arts and Architecture are more likely to be left-handed than people found in the general population. We only have one sample since we will be comparing a population proportion based on a sample value to a known population value.

• Research Question : Are artists more likely to be left-handed than people found in the general population?
• Response Variable : Classification of the student as either right-handed or left-handed

State Null and Alternative Hypotheses

• Null Hypothesis : Students in the College of Arts and Architecture are no more likely to be left-handed than people in the general population (population percent of left-handed students in the College of Art and Architecture = 10% or p = .10).
• Alternative Hypothesis : Students in the College of Arts and Architecture are more likely to be left-handed than people in the general population (population percent of left-handed students in the College of Arts and Architecture > 10% or p > .10). This is a one-sided alternative hypothesis.

Example 10.3: Hypotheses with One Sample of One Measurement Variable Section  

A generic brand of the anti-histamine Diphenhydramine markets a capsule with a 50 milligram dose. The manufacturer is worried that the machine that fills the capsules has come out of calibration and is no longer creating capsules with the appropriate dosage.

• Research Question : Does the data suggest that the population mean dosage of this brand is different than 50 mg?
• Response Variable : dosage of the active ingredient found by a chemical assay.
• Null Hypothesis : On the average, the dosage sold under this brand is 50 mg (population mean dosage = 50 mg).
• Alternative Hypothesis : On the average, the dosage sold under this brand is not 50 mg (population mean dosage ≠ 50 mg). This is a two-sided alternative hypothesis.

Example 10.4: Hypotheses with Two Samples of One Categorical Variable Section  

Many people are starting to prefer vegetarian meals on a regular basis. Specifically, a researcher believes that females are more likely than males to eat vegetarian meals on a regular basis.

• Research Question : Does the data suggest that females are more likely than males to eat vegetarian meals on a regular basis?
• Response Variable : Classification of whether or not a person eats vegetarian meals on a regular basis
• Explanatory (Grouping) Variable: Sex
• Null Hypothesis : There is no sex effect regarding those who eat vegetarian meals on a regular basis (population percent of females who eat vegetarian meals on a regular basis = population percent of males who eat vegetarian meals on a regular basis or p females = p males ).
• Alternative Hypothesis : Females are more likely than males to eat vegetarian meals on a regular basis (population percent of females who eat vegetarian meals on a regular basis > population percent of males who eat vegetarian meals on a regular basis or p females > p males ). This is a one-sided alternative hypothesis.

Example 10.5: Hypotheses with Two Samples of One Measurement Variable Section  

Obesity is a major health problem today. Research is starting to show that people may be able to lose more weight on a low carbohydrate diet than on a low fat diet.

• Research Question : Does the data suggest that, on the average, people are able to lose more weight on a low carbohydrate diet than on a low fat diet?
• Response Variable : Weight loss (pounds)
• Explanatory (Grouping) Variable : Type of diet
• Null Hypothesis : There is no difference in the mean amount of weight loss when comparing a low carbohydrate diet with a low fat diet (population mean weight loss on a low carbohydrate diet = population mean weight loss on a low fat diet).
• Alternative Hypothesis : The mean weight loss should be greater for those on a low carbohydrate diet when compared with those on a low fat diet (population mean weight loss on a low carbohydrate diet > population mean weight loss on a low fat diet). This is a one-sided alternative hypothesis.

Example 10.6: Hypotheses about the relationship between Two Categorical Variables Section  

• Research Question : Do the odds of having a stroke increase if you inhale second hand smoke ? A case-control study of non-smoking stroke patients and controls of the same age and occupation are asked if someone in their household smokes.
• Variables : There are two different categorical variables (Stroke patient vs control and whether the subject lives in the same household as a smoker). Living with a smoker (or not) is the natural explanatory variable and having a stroke (or not) is the natural response variable in this situation.
• Null Hypothesis : There is no relationship between whether or not a person has a stroke and whether or not a person lives with a smoker (odds ratio between stroke and second-hand smoke situation is = 1).
• Alternative Hypothesis : There is a relationship between whether or not a person has a stroke and whether or not a person lives with a smoker (odds ratio between stroke and second-hand smoke situation is > 1). This is a one-tailed alternative.

This research question might also be addressed like example 11.4 by making the hypotheses about comparing the proportion of stroke patients that live with smokers to the proportion of controls that live with smokers.

Example 10.7: Hypotheses about the relationship between Two Measurement Variables Section  

• Research Question : A financial analyst believes there might be a positive association between the change in a stock's price and the amount of the stock purchased by non-management employees the previous day (stock trading by management being under "insider-trading" regulatory restrictions).
• Variables : Daily price change information (the response variable) and previous day stock purchases by non-management employees (explanatory variable). These are two different measurement variables.
• Null Hypothesis : The correlation between the daily stock price change (\$) and the daily stock purchases by non-management employees (\$) = 0.
• Alternative Hypothesis : The correlation between the daily stock price change (\$) and the daily stock purchases by non-management employees (\$) > 0. This is a one-sided alternative hypothesis.

Example 10.8: Hypotheses about comparing the relationship between Two Measurement Variables in Two Samples Section  

• Research Question : Is there a linear relationship between the amount of the bill (\$) at a restaurant and the tip (\$) that was left. Is the strength of this association different for family restaurants than for fine dining restaurants?
• Variables : There are two different measurement variables. The size of the tip would depend on the size of the bill so the amount of the bill would be the explanatory variable and the size of the tip would be the response variable.
• Null Hypothesis : The correlation between the amount of the bill (\$) at a restaurant and the tip (\$) that was left is the same at family restaurants as it is at fine dining restaurants.
• Alternative Hypothesis : The correlation between the amount of the bill (\$) at a restaurant and the tip (\$) that was left is the difference at family restaurants then it is at fine dining restaurants. This is a two-sided alternative hypothesis.

Statistics Made Easy

4 Examples of Hypothesis Testing in Real Life

In statistics, hypothesis tests are used to test whether or not some hypothesis about a population parameter is true.

To perform a hypothesis test in the real world, researchers will obtain a random sample from the population and perform a hypothesis test on the sample data, using a null and alternative hypothesis:

• Null Hypothesis (H 0 ): The sample data occurs purely from chance.
• Alternative Hypothesis (H A ): The sample data is influenced by some non-random cause.

If the p-value of the hypothesis test is less than some significance level (e.g. α = .05), then we can reject the null hypothesis and conclude that we have sufficient evidence to say that the alternative hypothesis is true.

The following examples provide several situations where hypothesis tests are used in the real world.

Example 1: Biology

Hypothesis tests are often used in biology to determine whether some new treatment, fertilizer, pesticide, chemical, etc. causes increased growth, stamina, immunity, etc. in plants or animals.

For example, suppose a biologist believes that a certain fertilizer will cause plants to grow more during a one-month period than they normally do, which is currently 20 inches. To test this, she applies the fertilizer to each of the plants in her laboratory for one month.

She then performs a hypothesis test using the following hypotheses:

• H 0 : μ = 20 inches (the fertilizer will have no effect on the mean plant growth)
• H A : μ > 20 inches (the fertilizer will cause mean plant growth to increase)

If the p-value of the test is less than some significance level (e.g. α = .05), then she can reject the null hypothesis and conclude that the fertilizer leads to increased plant growth.

Example 2: Clinical Trials

Hypothesis tests are often used in clinical trials to determine whether some new treatment, drug, procedure, etc. causes improved outcomes in patients.

For example, suppose a doctor believes that a new drug is able to reduce blood pressure in obese patients. To test this, he may measure the blood pressure of 40 patients before and after using the new drug for one month.

He then performs a hypothesis test using the following hypotheses:

• H 0 : μ after = μ before (the mean blood pressure is the same before and after using the drug)
• H A : μ after < μ before (the mean blood pressure is less after using the drug)

If the p-value of the test is less than some significance level (e.g. α = .05), then he can reject the null hypothesis and conclude that the new drug leads to reduced blood pressure.

Example 3: Advertising Spend

Hypothesis tests are often used in business to determine whether or not some new advertising campaign, marketing technique, etc. causes increased sales.

For example, suppose a company believes that spending more money on digital advertising leads to increased sales. To test this, the company may increase money spent on digital advertising during a two-month period and collect data to see if overall sales have increased.

They may perform a hypothesis test using the following hypotheses:

• H 0 : μ after = μ before (the mean sales is the same before and after spending more on advertising)
• H A : μ after > μ before (the mean sales increased after spending more on advertising)

If the p-value of the test is less than some significance level (e.g. α = .05), then the company can reject the null hypothesis and conclude that increased digital advertising leads to increased sales.

Example 4: Manufacturing

Hypothesis tests are also used often in manufacturing plants to determine if some new process, technique, method, etc. causes a change in the number of defective products produced.

For example, suppose a certain manufacturing plant wants to test whether or not some new method changes the number of defective widgets produced per month, which is currently 250. To test this, they may measure the mean number of defective widgets produced before and after using the new method for one month.

They can then perform a hypothesis test using the following hypotheses:

• H 0 : μ after = μ before (the mean number of defective widgets is the same before and after using the new method)
• H A : μ after ≠ μ before (the mean number of defective widgets produced is different before and after using the new method)

If the p-value of the test is less than some significance level (e.g. α = .05), then the plant can reject the null hypothesis and conclude that the new method leads to a change in the number of defective widgets produced per month.

Additional Resources

Introduction to Hypothesis Testing Introduction to the One Sample t-test Introduction to the Two Sample t-test Introduction to the Paired Samples t-test

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Test statistics | Definition, Interpretation, and Examples

Published on July 17, 2020 by Rebecca Bevans . Revised on June 22, 2023.

The test statistic is a number calculated from a statistical test of a hypothesis. It shows how closely your observed data match the distribution expected under the null hypothesis of that statistical test.

The test statistic is used to calculate the p value of your results, helping to decide whether to reject your null hypothesis.

Table of contents

What exactly is a test statistic, types of test statistics, interpreting test statistics, reporting test statistics, other interesting articles, frequently asked questions about test statistics.

A test statistic describes how closely the distribution of your data matches the distribution predicted under the null hypothesis of the statistical test you are using.

The distribution of data is how often each observation occurs, and can be described by its central tendency and variation around that central tendency. Different statistical tests predict different types of distributions, so it’s important to choose the right statistical test for your hypothesis.

The test statistic summarizes your observed data into a single number using the central tendency, variation, sample size, and number of predictor variables in your statistical model.

Generally, the test statistic is calculated as the pattern in your data (i.e., the correlation between variables or difference between groups) divided by the variance in the data (i.e., the standard deviation ).

• Null hypothesis ( H 0 ): There is no correlation between temperature and flowering date.
• Alternate hypothesis ( H A or H 1 ): There is a correlation between temperature and flowering date.

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Below is a summary of the most common test statistics, their hypotheses, and the types of statistical tests that use them.

Different statistical tests will have slightly different ways of calculating these test statistics, but the underlying hypotheses and interpretations of the test statistic stay the same.

In practice, you will almost always calculate your test statistic using a statistical program (R, SPSS, Excel, etc.), which will also calculate the p value of the test statistic. However, formulas to calculate these statistics by hand can be found online.

• a regression coefficient of 0.36
• a t value comparing that coefficient to the predicted range of regression coefficients under the null hypothesis of no relationship

The t value of the regression test is 2.36 – this is your test statistic.

For any combination of sample sizes and number of predictor variables, a statistical test will produce a predicted distribution for the test statistic. This shows the most likely range of values that will occur if your data follows the null hypothesis of the statistical test.

The more extreme your test statistic – the further to the edge of the range of predicted test values it is – the less likely it is that your data could have been generated under the null hypothesis of that statistical test.

The agreement between your calculated test statistic and the predicted values is described by the p value . The smaller the p value, the less likely your test statistic is to have occurred under the null hypothesis of the statistical test.

Because the test statistic is generated from your observed data, this ultimately means that the smaller the p value, the less likely it is that your data could have occurred if the null hypothesis was true.

Test statistics can be reported in the results section of your research paper along with the sample size, p value of the test, and any characteristics of your data that will help to put these results into context.

Whether or not you need to report the test statistic depends on the type of test you are reporting.

By surveying a random subset of 100 trees over 25 years we found a statistically significant ( p < 0.01) positive correlation between temperature and flowering dates ( R 2 = 0.36, SD = 0.057).

In our comparison of mouse diet A and mouse diet B, we found that the lifespan on diet A  ( M = 2.1 years; SD = 0.12) was significantly shorter than the lifespan on diet B ( M = 2.6 years; SD = 0.1), with an average difference of 6 months ( t (80) = -12.75; p < 0.01).

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See an example

If you want to know more about statistics , methodology , or research bias , make sure to check out some of our other articles with explanations and examples.

• Confidence interval
• Descriptive statistics
• Measures of central tendency
• Correlation coefficient

Methodology

• Cluster sampling
• Stratified sampling
• Types of interviews
• Cohort study
• Thematic analysis

Research bias

• Implicit bias
• Cognitive bias
• Survivorship bias
• Availability heuristic
• Nonresponse bias
• Regression to the mean

A test statistic is a number calculated by a  statistical test . It describes how far your observed data is from the  null hypothesis  of no relationship between  variables or no difference among sample groups.

The test statistic tells you how different two or more groups are from the overall population mean , or how different a linear slope is from the slope predicted by a null hypothesis . Different test statistics are used in different statistical tests.

The formula for the test statistic depends on the statistical test being used.

Generally, the test statistic is calculated as the pattern in your data (i.e. the correlation between variables or difference between groups) divided by the variance in the data (i.e. the standard deviation ).

The test statistic you use will be determined by the statistical test.

You can choose the right statistical test by looking at what type of data you have collected and what type of relationship you want to test.

The test statistic will change based on the number of observations in your data, how variable your observations are, and how strong the underlying patterns in the data are.

For example, if one data set has higher variability while another has lower variability, the first data set will produce a test statistic closer to the null hypothesis , even if the true correlation between two variables is the same in either data set.

Statistical significance is a term used by researchers to state that it is unlikely their observations could have occurred under the null hypothesis of a statistical test . Significance is usually denoted by a p -value , or probability value.

Statistical significance is arbitrary – it depends on the threshold, or alpha value, chosen by the researcher. The most common threshold is p < 0.05, which means that the data is likely to occur less than 5% of the time under the null hypothesis .

When the p -value falls below the chosen alpha value, then we say the result of the test is statistically significant.

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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

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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.

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