416 Chapter 9 Hypothesis Testing SUMMARY ∙ The P-value is the probability, assuming that H0 is true, of observing a value for the test statistic that is as extreme as or more extreme than the value actually observed. ∙ The smaller the P-value, the stronger the evidence against H0. EXAMPLE 9.9 Find and interpret a P-value A test is made of H0 : �� = 10 versus H1: �� > 10. The value of the test statistic is z = 2.25. Find the P-value and interpret it. Solution The alternate hypothesis is H1: �� > 10, so this is a right-tailed test. Therefore, values of z greater than our observed value of 2.25 are more extreme than our value is. The P-value is the area under the normal curve to the right of the test statistic z = 2.25. Using Table A.2, we see that the area to the left of z = 2.25 is 0.9878. Therefore, the area to the right of z = 2.25 is 1 − 0.9878 = 0.0122. See Figure 9.5. Area = 0.0122 z = 2.25 Figure 9.5 The P-value of 0.0122 tells us that if H0 is true, then the probability of observing a test statistic of 2.25 or more is only 0.0122. This result is fairly unusual if we assume H0 to be true. Therefore, this is fairly strong evidence against H0. EXAMPLE 9.10 Find and interpret a P-value A test is made of H0 : �� = 5 versus H1: �� < 5. The value of the test statistic is z = −0.63. Find the P-value and interpret it. Area = 0.2643 z = −0.63 Figure 9.6 Solution The alternate hypothesis is H1: �� < 5, so this is a left-tailed test. Therefore, values of z less than our value of −0.63 are more extreme than our value is. The P-value is the area under the normal curve to the left of the test statistic z = −0.63. Using Table A.2, we see that the area to the left of z = −0.63 is 0.2643. See Figure 9.6. The P-value of 0.2643 tells us that if H0 is true, then the probability of observing a test statistic of −0.63 or less is 0.2643. This is not particularly unusual, so this is not strong evidence against H0. EXAMPLE 9.11 Find and interpret a P-value A test is made of H0 : �� = 20 versus H1: �� ≠ 20. The value of the test statistic is z = −2.70. Find the P-value and interpret it. Solution The alternate hypothesis is H1: �� ≠ 20, so this is a two-tailed test. Therefore, values of z less than our value of −2.70 and values greater than 2.70 are both more extreme than our value is. The P-value is the sum of the areas under the normal curve to the right of z = 2.70 and to the left of z = −2.70. Using Table A.2, we see that the area to the left of z = −2.70 is 0.0035. The area to the right of z = 2.70 is also 0.0035. The sum of the areas is therefore 0.0035 + 0.0035 = 0.0070. See Figure 9.7. Area = 0.0035 Area = 0.0035 z = −2.70 z = 2.70 Figure 9.7 The P-value of 0.0070 tells us that if H0 is true, then the probability of observing a test statistic greater than 2.70 or less than −2.70 is only 0.0070. This result is quite unusual if we assume H0 to be true. Therefore, this is very strong evidence against H0.
navidi_monk_elementary_statistics_2e_ch7-9
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