Hypothesis Tests Regarding a Parameter Chapter 10 Hypothesis Tests Regarding a Parameter

The Language of Hypothesis Testing Section 10.1 The Language of Hypothesis Testing

Objectives Determine the null and alternative hypotheses Explain Type I and Type II errors State conclusions to hypothesis tests

Objective 1 Determine the Null and Alternative Hypotheses

A hypothesis is a statement regarding a characteristic of one or more populations. In this chapter, we look at hypotheses regarding a single population parameter.

Examples of Claims Regarding a Characteristic of a Single Population In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today. Source: ReadersDigest.com poll created on 2008/05/02

Examples of Claims Regarding a Characteristic of a Single Population In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today. According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then.

Examples of Claims Regarding a Characteristic of a Single Population In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today. According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then. Using an old manufacturing process, the standard deviation of the amount of wine put in a bottle was 0.23 ounces. With new equipment, the quality control manager believes the standard deviation has decreased.

Hypothesis testing is a procedure, based on sample evidence and probability, used to test statements regarding a characteristic of one or more populations.

Steps in Hypothesis Testing A statement is made regarding the nature of the population. Evidence (sample data) is collected in order to test the statement. The data is analyzed to assess the plausibility of the statement.

The null hypothesis, denoted H0, is a statement to be tested The null hypothesis, denoted H0, is a statement to be tested. The null hypothesis is a statement of no change, no effect or no difference. The null hypothesis is assumed true until evidence indicates otherwise. In this chapter, it will be a statement regarding the value of a population parameter.

The alternative hypothesis, denoted H1, is a statement that we are trying to find evidence to support. In this chapter, it will be a statement regarding the value of a population parameter.

In this chapter, there are three ways to set up the null and alternative hypotheses: Equal versus not equal hypothesis (two-tailed test) H0: parameter = some value H1: parameter ≠ some value

In this chapter, there are three ways to set up the null and alternative hypotheses: Equal versus not equal hypothesis (two-tailed test) H0: parameter = some value H1: parameter ≠ some value Equal versus less than (left-tailed test) H1: parameter < some value

In this chapter, there are three ways to set up the null and alternative hypotheses: Equal versus not equal hypothesis (two-tailed test) H0: parameter = some value H1: parameter ≠ some value Equal versus less than (left-tailed test) H1: parameter < some value Equal versus greater than (right-tailed test) H1: parameter > some value

“In Other Words” The null hypothesis is a statement of “status quo” or “no difference” and always contains a statement of equality. The null hypothesis is assumed to be true until we have evidence to the contrary. The claim that we are trying to gather evidence for determines the alternative hypothesis.

Parallel Example 2: Forming Hypotheses For each of the following claims, determine the null and alternative hypotheses. State whether the test is two-tailed, left-tailed or right-tailed. In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today. According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then. Using an old manufacturing process, the standard deviation of the amount of wine put in a bottle was 0.23 ounces. With new equipment, the quality control manager believes the standard deviation has decreased.

Solution In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today. The hypothesis deals with a population proportion, p. If the percentage participating in charity work is no different than in 2008, it will be 0.62 so the null hypothesis is H0: p=0.62. Since the researcher believes that the percentage is different today, the alternative hypothesis is a two-tailed hypothesis: H1: p≠0.62.

Solution b) According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then. The hypothesis deals with a population mean, . If the mean call length on a cellular phone is no different than in 2006, it will be 3.25 minutes so the null hypothesis is H0: =3.25. Since the researcher believes that the mean call length has increased, the alternative hypothesis is: H1:  > 3.25, a right-tailed test.

Solution c) Using an old manufacturing process, the standard deviation of the amount of wine put in a bottle was 0.23 ounces. With new equipment, the quality control manager believes the standard deviation has decreased. The hypothesis deals with a population standard deviation, . If the standard deviation with the new equipment has not changed, it will be 0.23 ounces so the null hypothesis is H0:  = 0.23. Since the quality control manager believes that the standard deviation has decreased, the alternative hypothesis is: H1:  < 0.23, a left-tailed test.

Practice Get in groups of 2 or 3 and discuss questions 15-22 on page 511 in the book. Only answer part (a) from the directions.

Objective 3 State Conclusions to Hypothesis Tests

CAUTION! We never “accept” the null hypothesis because without having access to the entire population, we don’t know the exact value of the parameter stated in the null. Rather, we say that we do not reject the null hypothesis.

Parallel Example 4: Stating the Conclusion According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then. Suppose the sample evidence indicates that the null hypothesis should be rejected. State the wording of the conclusion. Suppose the sample evidence indicates that the null hypothesis should not be rejected. State the wording of the conclusion.

Solution Suppose the sample evidence indicates that the null hypothesis should be rejected. State the wording of the conclusion. The statement in the alternative hypothesis is that the mean call length is greater than 3.25 minutes. Since the null hypothesis (=3.25) is rejected, we conclude that there is sufficient evidence to conclude that the mean length of a phone call on a cell phone is greater than 3.25 minutes.

Solution b) Suppose the sample evidence indicates that the null hypothesis should not be rejected. State the wording of the conclusion. Since the null hypothesis (=3.25) is not rejected, we conclude that there is insufficient evidence to conclude that the mean length of a phone call on a cell phone is greater than 3.25 minutes. In other words, the sample evidence is consistent with the mean call length equaling 3.25 minutes.

Practice Get in groups of 2 or 3 and discuss questions 23-34 on page 512 in the book.

Objective 2 Explain Type I and Type II Errors

Four Outcomes from Hypothesis Testing We reject the null hypothesis when the alternative hypothesis is true. This decision would be correct.

Four Outcomes from Hypothesis Testing We reject the null hypothesis when the alternative hypothesis is true. This decision would be correct. We do not reject the null hypothesis when the null hypothesis is true. This decision would be correct.

Four Outcomes from Hypothesis Testing We reject the null hypothesis when the alternative hypothesis is true. This decision would be correct. We do not reject the null hypothesis when the null hypothesis is true. This decision would be correct. We reject the null hypothesis when the null hypothesis is true. This decision would be incorrect. This type of error is called a Type I error.

Four Outcomes from Hypothesis Testing We reject the null hypothesis when the alternative hypothesis is true. This decision would be correct. We do not reject the null hypothesis when the null hypothesis is true. This decision would be correct. We reject the null hypothesis when the null hypothesis is true. This decision would be incorrect. This type of error is called a Type I error. We do not reject the null hypothesis when the alternative hypothesis is true. This decision would be incorrect. This type of error is called a Type II error.

Parallel Example 3: Type I and Type II Errors For each of the following claims, explain what it would mean to make a Type I error. What would it mean to make a Type II error? In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today. According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then.

Solution In 2008, 62% of American adults regularly volunteered their time for charity work. A researcher believes that this percentage is different today. A Type I error is made if the researcher concludes that p≠0.62 when the true proportion of Americans 18 years or older who participated in some form of charity work is currently 62%. A Type II error is made if the sample evidence leads the researcher to believe that the current percentage of Americans 18 years or older who participated in some form of charity work is still 62% when, in fact, this percentage differs from 62%.

Solution b) According to a study published in March, 2006 the mean length of a phone call on a cellular telephone was 3.25 minutes. A researcher believes that the mean length of a call has increased since then. A Type I error occurs if the sample evidence leads the researcher to conclude that >3.25 when, in fact, the actual mean call length on a cellular phone is still 3.25 minutes. A Type II error occurs if the researcher fails to reject the hypothesis that the mean length of a phone call on a cellular phone is 3.25 minutes when, in fact, it is longer than 3.25 minutes.

 = P(Type I Error) = P(rejecting H0 when H0 is true)  = P(Type II Error) = P(not rejecting H0 when H1 is true)

The probability of making a Type I error, , is chosen by the researcher before the sample data is collected. The level of significance, , is the probability of making a Type I error.

“In Other Words” As the probability of a Type I error increases, the probability of a Type II error decreases, and vice-versa.

Practice Get in groups of 2 or 3 and discuss questions 15-22 on page 511 in the book. Now answer part (b) and (c) from the directions.

Section 10.2 Hypothesis Tests for a Population Mean-Population Standard Deviation Known

Objectives Explain the logic of hypothesis testing Test the hypotheses about a population mean with  known using the classical approach Test hypotheses about a population mean with  known using P-values Test hypotheses about a population mean with  known using confidence intervals Distinguish between statistical significance and practical significance.

Objective 1 Explain the Logic of Hypothesis Testing

To test hypotheses regarding the population mean assuming the population standard deviation is known, two requirements must be satisfied: A simple random sample is obtained. The population from which the sample is drawn is normally distributed or the sample size is large (n≥30). If these requirements are met, the distribution of is normal with mean  and standard deviation .

Recall the researcher who believes that the mean length of a cell phone call has increased from its March, 2006 mean of 3.25 minutes. Suppose we take a simple random sample of 36 cell phone calls. Assume the standard deviation of the phone call lengths is known to be 0.78 minutes. What is the sampling distribution of the sample mean? Answer: is normally distributed with mean 3.25 and standard deviation .

Suppose the sample of 36 calls resulted in a sample mean of 3 Suppose the sample of 36 calls resulted in a sample mean of 3.56 minutes. Do the results of this sample suggest that the researcher is correct? In other words, would it be unusual to obtain a sample mean of 3.56 minutes from a population whose mean is 3.25 minutes? What is convincing or statistically significant evidence?

When observed results are unlikely under the assumption that the null hypothesis is true, we say the result is statistically significant. When results are found to be statistically significant, we reject the null hypothesis.

The Logic of the Classical Approach One criterion we may use for sufficient evidence for rejecting the null hypothesis is if the sample mean is too many standard deviations from the assumed (or status quo) population mean. For example, we may choose to reject the null hypothesis if our sample mean is more than 2 standard deviations above the population mean of 3.25 minutes.

Recall that our simple random sample of 36 calls resulted in a sample mean of 3.56 minutes with standard deviation of 0.13. Thus, the sample mean is standard deviations above the hypothesized mean of 3.25 minutes. Therefore, using our criterion, we would reject the null hypothesis and conclude that the mean cellular call length is greater than 3.25 minutes.

Why does it make sense to reject the null hypothesis if the sample mean is more than 2 standard deviations above the hypothesized mean?

If the null hypothesis were true, then 1-0. 0228=0. 9772=97 If the null hypothesis were true, then 1-0.0228=0.9772=97.72% of all sample means will be less than 3.25+2(0.13)=3.51.

Because sample means greater than 3 Because sample means greater than 3.51 are unusual if the population mean is 3.25, we are inclined to believe the population mean is greater than 3.25.

The Logic of the P-Value Approach A second criterion we may use for sufficient evidence to support the alternative hypothesis is to compute how likely it is to obtain a sample mean at least as extreme as that observed from a population whose mean is equal to the value assumed by the null hypothesis.

We can compute the probability of obtaining a sample mean of 3 We can compute the probability of obtaining a sample mean of 3.56 or more using the normal model.

Recall So, we compute The probability of obtaining a sample mean of 3.56 minutes or more from a population whose mean is 3.25 minutes is 0.0087. This means that fewer than 1 sample in 100 will give us a mean as high or higher than 3.56 if the population mean really is 3.25 minutes. Since this outcome is so unusual, we take this as evidence against the null hypothesis.

Premise of Testing a Hypothesis Using the P-value Approach Assuming that H0 is true, if the probability of getting a sample mean as extreme or more extreme than the one obtained is small, we reject the null hypothesis.

Objective 2 Test Hypotheses about a Population Mean with  Known Using the Classical Approach

Testing Hypotheses Regarding the Population Mean with σ Known Using the Classical Approach To test hypotheses regarding the population mean with  known, we can use the steps that follow, provided that two requirements are satisfied: The sample is obtained using simple random sampling. The sample has no outliers, and the population from which the sample is drawn is normally distributed or the sample size is large (n ≥ 30).

Step 1: Determine the null and alternative. hypotheses Step 1: Determine the null and alternative hypotheses. Again, the hypotheses can be structured in one of three ways:

Step 2: Select a level of significance, , based on the seriousness of making a Type I error.

Step 3: Provided that the population from which the Step 3: Provided that the population from which the sample is drawn is normal or the sample size is large, and the population standard deviation, , is known, the distribution of the sample mean, , is normal with mean and standard deviation . Therefore, represents the number of standard deviations that the sample mean is from the assumed mean. This value is called the test statistic.

Step 4: The level of significance is used to determine the critical value. The critical region represents the maximum number of standard deviations that the sample mean can be from 0 before the null hypothesis is rejected. The critical region or rejection region is the set of all values such that the null hypothesis is rejected.

(critical value) Two-Tailed

(critical value) Left-Tailed

Right-Tailed (critical value)

Step 5: Compare the critical value with the test statistic:

Step 6: State the conclusion.

The procedure is robust, which means that minor departures from normality will not adversely affect the results of the test. However, for small samples, if the data have outliers, the procedure should not be used.

Parallel Example 2: The Classical Approach to Hypothesis Testing A can of 7-Up states that the contents of the can are 355 ml. A quality control engineer is worried that the filling machine is miscalibrated. In other words, she wants to make sure the machine is not under- or over-filling the cans. She randomly selects 9 cans of 7-Up and measures the contents. She obtains the following data. 351 360 358 356 359 358 355 361 352 Is there evidence at the =0.05 level of significance to support the quality control engineer’s claim? Prior experience indicates that =3.2ml. Source: Michael McCraith, Joliet Junior College

Solution The quality control engineer wants to know if the mean content is different from 355 ml. Since the sample size is small, we must verify that the data come from a population that is approximately normal with no outliers.

5 Number Summary and Outlier Test 7-Up (mll) 351 5 Number Summary and Outlier Test 352 Minimum 355 Q1 353.5 356 Median 358 Q3 359.5 Max 361 359 IQR 6 360 Q1-1.5IQR 344.5 Q3+1.5IQR 368.5 Assumption of normality appears reasonable.

No outliers.

Solution Step 1: H0: =355 versus H1: ≠355 Step 2: The level of significance is =0.05. Step 3: The sample mean is calculated to be 356.667. The test statistic is then The sample mean of 356.667 is 1.56 standard deviations above the assumed mean of 355 ml.

Solution Step 4: Since this is a two-tailed test, we determine the critical values at the =0.05 level of significance to be -z0.025= -1.96 and z0.025=1.96 Step 5: Since the test statistic, z0=1.56, is less than the critical value 1.96, we fail to reject the null hypothesis. Step 6: There is insufficient evidence at the =0.05 level of significance to conclude that the mean content differs from 355 ml.

Practice Get in groups of 2 or 3 and work on questions 11-13 on page 527 in the book.

Objective 3 Test Hypotheses about a Population Mean with  Known Using P-values.

A P-value is the probability of observing a sample statistic as extreme or more extreme than the one observed under the assumption that the null hypothesis is true.

Testing Hypotheses Regarding the Population Mean with σ Known Using P-values To test hypotheses regarding the population mean with  known, we can use the steps that follow to compute the P-value, provided that two requirements are satisfied: The sample is obtained using simple random sampling. The sample has no outliers, and the population from which the sample is drawn is normally distributed or the sample size is large (n ≥ 30).

Step 1: A claim is made regarding the population. mean Step 1: A claim is made regarding the population mean. The claim is used to determine the null and alternative hypotheses. Again, the hypothesis can be structured in one of three ways:

Step 2: Select a level of significance, , based on the seriousness of making a Type I error.

Step 3: Compute the test statistic,

Step 4: Determine the P-value

Step 5: Reject the null hypothesis if the. P- Step 5: Reject the null hypothesis if the P- value is less than the level of significance, . The comparison of the P-value and the level of significance is called the decision rule. Step 6: State the conclusion.

Parallel Example 3: The P-Value Approach to Hypothesis Parallel Example 3: The P-Value Approach to Hypothesis Testing: Left-Tailed, Large Sample The volume of a stock is the number of shares traded in the stock in a day. The mean volume of Apple stock in 2007 was 35.14 million shares with a standard deviation of 15.07 million shares. A stock analyst believes that the volume of Apple stock has increased since then. He randomly selects 40 trading days in 2008 and determines the sample mean volume to be 41.06 million shares. Test the analyst’s claim at the =0.10 level of significance using P-values.

Solution Step 1: The analyst wants to know if the stock volume has increased. This is a right-tailed test with H0: =35.14 versus H1: >35.14. We want to know the probability of obtaining a sample mean of 41.06 or more from a population where the mean is assumed to be 35.14.

Solution Step 2: The level of significance is =0.10. Step 3: The test statistic is

Solution Step 4: P(Z > z0)=P(Z > 2.48)=0.0066. The probability of obtaining a sample mean of 41.06 or more from a population whose mean is 35.14 is 0.0066.

Solution Step 5: Since the P-value= 0.0066 is less than the level of significance, 0.10, we reject the null hypothesis. Step 6: There is sufficient evidence to reject the null hypothesis and to conclude that the mean volume of Apple stock is greater than 35.14 million shares.

Parallel Example 4: The P-Value Approach of Hypothesis Parallel Example 4: The P-Value Approach of Hypothesis Testing: Two-Tailed, Small Sample A can of 7-Up states that the contents of the can are 355 ml. A quality control engineer is worried that the filling machine is miscalibrated. In other words, she wants to make sure the machine is not under- or over-filling the cans. She randomly selects 9 cans of 7-Up and measures the contents. She obtains the following data. 351 360 358 356 359 358 355 361 352 Use the P-value approach to determine if there is evidence at the =0.05 level of significance to support the quality control engineer’s claim. Prior experience indicates that =3.2ml. Source: Michael McCraith, Joliet Junior College

Solution The quality control engineer wants to know if the mean content is different from 355 ml. Since we have already verified that the data come from a population that is approximately normal with no outliers, we will continue with step 1.

Solution Step 1: The quality control engineer wants to know if the content has changed. This is a two-tailed test with H0: =355 versus H1: ≠355.

Solution Step 2: The level of significance is =0.05. Step 3: Recall that the sample mean is 356.667. The test statistic is then

Solution Step 4: Since this is a two-tailed test, P-value = P(Z < -1.56 or Z > 1.56) = 2*(0.0594)=0.1188. The probability of obtaining a sample mean that is more than 1.56 standard deviations away from the assumed mean of 355 ml is 0.1188.

Solution Step 5: Since the P-value= 0.1188 is greater than the level of significance, 0.05, we fail to reject the null hypothesis. Step 6: There is insufficient evidence to conclude that the mean content of 7-Up cans differs from 355 ml.

One advantage of using P-values over the classical approach in hypothesis testing is that P-values provide information regarding the strength of the evidence. Another is that P-values are interpreted the same way regardless of the type of hypothesis test being performed. the lower the P-value, the stronger the evidence against the statement in the null hypothesis.

Practice Get in groups of 2 or 3 and work on questions 15-17 on page 527 in the book.

Objective 4 Test Hypotheses about a Population Mean with  Known Using Confidence Intervals

When testing H0:  = 0 versus H1:  ≠ 0, if a (1-)·100% confidence interval contains 0, we do not reject the null hypothesis. However, if the confidence interval does not contain 0, we have sufficient evidence that supports the statement in the alternative hypothesis and conclude that  ≠ 0 at the level of significance, .

Parallel Example 6: Testing Hypotheses about a Population Parallel Example 6: Testing Hypotheses about a Population Mean Using a Confidence Interval Test the hypotheses presented in Parallel Examples 2 and 4 at the =0.05 level of significance by constructing a 95% confidence interval about , the population mean can content.

Solution Lower bound: Upper bound: We are 95% confident that the mean can content is between 354.6 ml and 358.8 ml. Since the mean stated in the null hypothesis is in this interval, there is insufficient evidence to reject the hypothesis that the mean can content is 355 ml.

Objective 5 Distinguish between Statistical Significance and Practical Significance

When a large sample size is used in a hypothesis test, the results could be statistically significant even though the difference between the sample statistic and mean stated in the null hypothesis may have no practical significance.

Practical significance refers to the idea that, while small differences between the statistic and parameter stated in the null hypothesis are statistically significant, the difference may not be large enough to cause concern or be considered important.

Parallel Example 7: Statistical versus Practical Significance In 2003, the average age of a mother at the time of her first childbirth was 25.2. To determine if the average age has increased, a random sample of 1200 mothers is taken and is found to have a sample mean age of 25.5. Assuming a standard deviation of 4.8, determine whether the mean age has increased using a significance level of =0.05.

Solution Step 1: To determine whether the mean age has increased, this is a right-tailed test with H0: =25.2 versus H1: >25.2. Step 2: The level of significance is =0.05. Step 3: Recall that the sample mean is 25.5. The test statistic is then

Solution Step 4: Since this is a right-tailed test, P-value = P(Z > 2.17) = 0.015. The probability of obtaining a sample mean that is more than 2.17 standard deviations above the assumed mean of 25.2 is 0.015. Step 5: Because the P-value is less than the level of significance, 0.05, we reject the null hypothesis.

Solution Step 6: There is sufficient evidence at the 0.05 significance level to conclude that the mean age of a mother at the time of her first childbirth is greater than 25.2. Although we found the difference in age to be significant, there is really no practical significance in the age difference (25.2 versus 25.5). Large sample sizes can lead to statistically significant results while the difference between the statistic and parameter is not enough to be considered practically significant.