466 Chapter 9 Hypothesis Testing Step 4: This is a left-tailed test, so the power is the area to the left of ̄x∗ = 2.475. To find this area, we find the z-score for 2.475, using the value ��1 = 2. z = 2.475 − 2 √ 75 2.66∕ = 1.55 The power is the area under the normal curve to the left of z = 1.55. This area is 0.9394, as the following figure shows. 2 2.475 z = 1.55 Area = 0.9394 Figure 9.18 presents two distributions for ̄x. The curve on the right is the distribution under the assumption that H0 is true. This distribution has mean ��0 = 2.98. The curve on the left is the distribution under the assumption that the mean is equal to the alternate value ��1 = 2. The critical region is the region to the left of ̄x∗ = 2.475. The area of the critical region under the null hypothesis distribution is the significance level �� = 0.05. The area under the alternate distribution is the power, 0.9394. Distribution of x ¯ when μ = 2 2 2.475 2.98 Distribution of x when H0 is true The area of the critical region when μ = 2 is the power. The power is 0.9394. The area of the critical region when H0 is true is α = 0.05. ¯ Figure 9.18 The critical region is shaded. The area of the critical region is �� = 0.05 when the curve for H0 is used, and is equal to the power, 0.9394, when the curve with the alternate mean of 2 is used. Check Your Understanding 1. For testing the hypotheses H0 : �� = 10 versus H1: �� < 10 at the �� = 0.05 level, assume the population standard deviation is �� = 3 and the sample size is n = 50. Find the power against the alternative ��1 = 9. 0.7611 Tech: 0.7618 2. For testing the hypotheses H0 : �� = 35 versus H1: �� > 35 at the �� = 0.01 level, assume the population standard deviation is �� = 10 and the sample size is n = 100. Find the power against the alternative ��1 = 40. 0.9962 Answers are on page 468.
navidi_monk_elementary_statistics_2e_ch7-9
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