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

This preview shows pages 1–3. Sign up to view the full content.

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:

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 6

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online