HW10 - STAT 500 HW 10 1. Problem 6.8 The samples are...

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STAT 500 HW 10 1. Problem 6.8 The samples are independent and since they are small when need to check if they follow a normal distribution. Data Percent 1.5 1.4 1.3 1.2 1.1 1.0 0.9 99 95 90 80 70 60 50 40 30 20 10 5 1 Mean 0.549 0.997 0.03057 10 0.222 0.765 StDev N AD P 1.236 0.06363 10 0.285 Variable type I type II Probability Plot of type I , type I I Normal - 95% CI The samples follow a normal distribution. Two-sample T for type I vs type II N Mean StDev SE Mean type I 10 1.2360 0.0636 0.020 type II 10 0.9970 0.0306 0.0097 Difference = mu (type I) - mu (type II) Estimate for difference: 0.239000 99% lower bound for difference: 0.179151 T-Test of difference = 0 (vs >): T-Value = 10.71 P-Value = 0.000 DF = 12 Since 0.0636/0.0306 = 2.08, we use separate variance t-test. Hypothesis:
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Ho: μ1 – μ2 = 0 Ha: μ1 – μ2 > 0 right tailed test α= 0.01 Since P-value = 0, we reject the null hypothesis. We conclude that the mean level of emission for type I devices is greater than the mean emission for type II devices at α= 0.01 2. Problem 6.16 a. t= - 4.04 b. t’= - 3.90 c. Ho: μ1 = μ2 Ha: μ1 ≠ μ2 Both statistics to test the research hypothesis lead us to reject the null hypothesis (p=0.0002 and 0.0001) at α= 0.05. and α= 0.01. Therefore, there is no difference in bonus percentages for gender, and the conclusion to reject the null hypothesis is the same at 5% and 1%, regardless of which is used to test the research hypothesis. 3
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HW10 - STAT 500 HW 10 1. Problem 6.8 The samples are...

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