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**Unformatted text preview: **9/23/11 Chapter 14 11 O bj1 0 Studying Job Satisfaction Case Study II Suppose job satisfaction scores follow a Normal distribution with standard deviation σ = 60. Data from 18 workers gave a sample mean score of 17. If the null hypothesis of no average difference in job satisfaction is true, the test statistic would be: 1.20 18 60 17 ≈- =- = n σ μ x z 9/23/11 Chapter 14 22 O bj1 0 2 Studying Job Satisfaction Case Study II For test statistic z = 1.20 and alternative hypothesis Ha: μ ≠ 0, the P-value would be: P-value = P ( Z < -1.20 or Z > 1.20) = 2 P ( Z < -1.20) = 2 P ( Z > 1.20) = (2)(0.1151) = 0.2302 If H0 is true, there is a 0.2302 (23.02%) chance that we would see results at least as extreme as those in the sample; thus, since we saw results that are likely if H0 is true, 9/23/11 Chapter 14 33 O bj1 0 3 Studying Job Satisfaction Case Study II 9/23/11 Test Statistics Large test statistics (positive or negative, depending on the side of the test) are evidence that Ho is not true. Also, small P-values are evidence that Ho is not true. (Because it’s the probability that we get what we got, if Ho is true!) Large P-values fail to give such evidence. 44 O bj1 0 4 Chapter 14 9/23/11 P-values Remember that a small P-value is strong evidence that our null hypothesis if false. We want our P-value to be as small as possible, because if we reject Ho, our P-value is the probability that we did the wrong thing. Think of it as the probability that we threw an innocent man into prison....

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