420 Chapter 9 Hypothesis Testing Step 2: Choose a level of significance. The level of significance is �� = 0.05. Step 3: Compute the test statistic. Because the sample size is large (n = 100), the sample mean ̄x is approximately normally distributed. The test statistic is the z-score for ̄x. To find the z-score, we first need to find the mean and standard deviation of ̄x. Because we are assuming that H0 is true, we assume that the mean of ̄x is �� = 69.7. We know that the population standard deviation is �� = 3. The standard deviation of ̄x is therefore �� √ n = 3 √ 100 = 0.3 It follows that ̄x is normally distributed with mean 69.7 and standard deviation 0.3. We observed a value of ̄x = 69.9. The z-score is z = ̄x − ��0 ��∕ √ n = 69.9 − 69.7 0.3 = 0.67 Explain It Again Using technology: In Example 9.13, the P-value is the area to the right of z = 0.67. This area can be found with technology. The following display illustrates the normalcdf command on the TI-84 Plus calculator. We enter the √ values x̄ =69.9, ��0 =69.7, and ��∕ n=0.3. The result given by the calculator differs slightly from the result found by using Table A.2. Step 4: Compute the P-value. Since the alternate hypothesis is �� > 69.7, this is a right-tailed test. The P-value is the area under the curve to the right of z = 0.67. Using Table A.2, we see that the area to the left of z = 0.67 is 0.7486. Therefore, the area to the right of z = 0.67 is 1 − 0.7486 = 0.2514. The P-value is 0.2514. See Figure 9.8. z = 0.67 Area = 0.2514 Figure 9.8 Step 5: Interpret the P-value. The P-value of 0.2514 says that ifH0 is true, the probability of observing a test statistic as large or larger than 0.67 is 0.2514. This is not unusual; it will happen for about one out of every four samples. Therefore, this sample does not strongly disagree with H0. In particular, P > 0.05, so we do not reject the null hypothesis at the �� = 0.05 level. Step 6: State a conclusion. There is not enough evidence to conclude that male executives have a greater mean height than adult males in general. The mean height of male executives may be the same as the mean height of adult males in general. EXAMPLE 9.14 Perform a two-tailed hypothesis test At a large company, the attitudes of workers are regularly measured with a standardized test. The scores on the test range from 0 to 100, with higher scores indicating greater satisfaction with their jobs. The mean score over all of the company’s employees was 74, with a standard deviation of �� = 8. Some time ago, the company adopted a policy of telecommuting. Under this policy, workers could spend one day per week working from home. After the policy had been in place for some time, a random sample of 80 workers was given the test to determine whether their mean level of satisfaction had changed since the policy was put into effect. The sample mean was 76. Assume the standard deviation is still �� = 8. Can we conclude that the mean level of satisfaction is different since the policy change at the �� = 0.05 level? Solution We first check the assumptions. We have a simple random sample, the sample size is large (n > 30), and the population standard deviation is known. The assumptions are satisfied.
navidi_monk_elementary_statistics_2e_ch7-9
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