CHAPTER

9

Hypothesis Tests about the Mean and Proportion

CHAPTER OUTLINE 9.1 Hypothesis Tests about a Population Mean for a Large Sample: p-value Approach 9.2 Hypothesis Tests about a Population Mean for a Large Sample: Critical Value Approach

9.3 Hypothesis Tests about a Population Mean: Small Samples 9.4 Hypothesis Tests about a Population Proportion: Large Samples

9.1 HYPOTHESIS TESTS ABOUT A POPULATION MEAN FOR A LARGE SAMPLE: P-VALUE APPROACH By definition, the p-value is the smallest significance level at which the null hypothesis is rejected. In this section, we will use Excel to set up a hypothesis test and calculate this value. Example 9-1 The management of Priority Health Club claims that its members lose an average of 10 pounds or more within the first month after joining the club. A consumer agency that wanted to check this claim took a random sample of 36 members of this health club and found that they lost an average of 9.2 pounds within the first month of membership with a standard deviation of 2.4 pounds. Find the p-value for this test. Solution: The claim to be tested is that µ ≥ 10. Also, the sample statistics are: n = 36, x = 9.2, and s = 2.4.

119

120 Chapter 9 Hypothesis Tests about the Mean and Proportion Step 1. Enter the null and alternative hypotheses into an Excel spreadsheet. Note that the claim is the null hypothesis, since it includes equality. (To get the Greek letter, µ, first just type it in as an “m.” Then double-click on the cell to edit it, highlight just the “m,” go to Format>Cells, and select Symbol for the Font.) Step 2. Note that this is a left-tailed test, since the alternate hypothesis contains the “ 50352. Step 1. Enter the null and alternate hypotheses and note that the claim is the alternate hypothesis since it does not include equality. Step 2. Note that this is a right-tailed test since the alternate hypothesis has the symbol “>” in it. Step 3. Determine the critical value corresponding to the significance level of α = .01. Since this is a right-tailed test, the area of the right tail is .01. To find the associated z-score, use NORMSINV(.01) to get the negative z-score that has cumulative area of .01. In this case of a right-tailed test, we want the positive value: 2.326342. Step 4. Calculate the value of the test statistic. We want the z-score associated with the sample statistics. In other words, we want to standardize the value of our statistic. Insert the STANDARDIZE function into a blank cell, with inputs 51750 for X, 50352 for Mean, and 5240/SQRT(200) for Standard_dev. (Remember that you can click on the cells containing these values instead of typing them in if you wish.) You should obtain a value of 3.7730354. Step 5. Since the test statistic of approximately 3.77 is well beyond the critical value of approximately 2.33, we reject the null hypothesis and conclude that evidence supports the claim (the alternate hypothesis) at the 1% significance level.

124 Chapter 9 Hypothesis Tests about the Mean and Proportion Again, this test statistic is rather extreme. It is more than 3 standard deviations above the mean. The associated p-value can be found by typing in: =1-NORMSDIST(3.77) which results in 8.066×10-5, or 0.00008066. This area to the right of z = 3.77 is definitely less than α = 0.01!

Figure 9.4 Setting up and calculating the critical value (=–NORMSINV(.01)), the test statistic (=STANDARDIZE(51750, 50352, 5240/SQRT(200)), and the p-value (see formula bar above) for this right-tailed hypothesis test. Reject Ho and accept the claim.

Example 9-5 The mayor of a large city claims that the average net worth of families living in this city is at least $300,000. A random sample of 100 families selected from this city produced a mean net worth of $288,000 with a standard deviation of $80,000. Using the 2.5% significance level, test the mayor’s claim. Solution: In order to set up the test, identify and enter the sample statistics into a blank Excel worksheet: x = 288000, s = 80000, and n = 100. Note that the claim to be tested is that µ > 300000. Step 1. Enter the null and alternate hypotheses and note that the claim is the null hypothesis since it does include equality. Step 2. Note that this is a left-tailed test since the alternate hypothesis has the symbol “ 65. Step 1. Enter the null and alternate hypotheses and note that the claim is the null hypothesis since it includes equality. Step 2. Note that this is a left-tailed test, since the alternate hypothesis has the symbol “