SECTION 7.2 Hypothesis Testing for the Mean (s Known) 367 Hypothesis Testing Using a P-Value According to a study of U.S. homes that use heating equipment, the mean indoor temperature at night during winter is 68.3°F. You think this information is incorrect. You randomly select 25 U.S. homes that use heating equipment in the winter and find that the mean indoor temperature at night is 67.2°F. From past studies, the population standard deviation is known to be 3.5°F and the population is normally distributed. Is there enough evidence to support your claim at a = 0.05? Use a P@value. (Adapted from U.S. Energy Information Administration) SOLUTION Because s is known 1s = 3.5°F2, the sample is random, and the population is normally distributed, you can use the z@test. The claim is “the mean is different from 68.3°F.” So, the null and alternative hypotheses are H0: m = 68.3°F and Ha: m ≠ 68.3°F. (Claim) The level of significance is a = 0.05. The standardized test statistic is z = x - m s 1n Because s is known and the population is normally distributed, use the z@test. = 67.2 - 68.3 3.5 225 Assume m = 68.3°F. ≈ -1.57. Round to two decimal places. In Table 4, the area corresponding to z = -1.57 is 0.0582. Because the test is a two-tailed test, the P@value is equal to twice the area to the left of z = -1.57, as shown in the figure. z = −1.57 z 1 2 3 −1 −2 −3 0 The area to the left of z = −1.57 is 0.0582, so P = 2(0.0582) = 0.1164. Two-Tailed Test So, the P@value is P = 2(0.0582) = 0.1164. Because the P@value is greater than a = 0.05, you fail to reject the null hypothesis. Interpretation There is not enough evidence at the 5% level of significance to support the claim that the mean indoor temperature at night during winter is different from 68.3°F for U.S. homes that use heating equipment. TRY IT YOURSELF 5 According to a study of employed U.S. adults ages 18 and over, the mean number of workdays missed due to illness or injury in the past 12 months is 3.6 days. You think this information is incorrect. You randomly select 25 employed U.S. adults ages 18 and over and find that the mean number of workdays missed is 4 days. Assume the population standard deviation is 1.5 days and the population is normally distributed. Is there enough evidence to support your claim at a = 0.01? Use a P@value. (Adapted from U.S. National Center for Health Statistics) Answer: Page A41 See Minitab steps on page 414. EXAMPLE 5
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