Data for gas mileage (in mpg) for different vehicles was entered into a software package and part of the ANOVA table is shown below: Source DF SS MS Vehicle 6 412 206.00 Error 7 312 44.57 Total 13 724 Determine the p-value for the data. a) 0.0324 b) 0.4721 c) 0.0462 d) 0.0162 e) 0.0108 Question 13 Your answer is CORRECT. If the P -value is smaller than the level of significance α , then the researcher should __________ at level α . a) Fail to reject H 0 b) Reject H 0
7/27/16, 12 : 37 PM Print Test Page 7 of 9 c) Accept H 0 Question 14 Your answer is CORRECT. The one-sample t statistic for a test of H 0 : μ = 14 vs. H a : μ < 14 based on n = 16 observations has the test statistic value of t = − 1.68. What is the p -value for this test? a) 0.057 b) 0.114 c) 0.000 d) 0.943 e) 0.357 Question 15 Your answer is CORRECT. This is a written question, worth 9 points. DO NOT place the problem code on the answer sheet. A proctor will fill this out after exam submission. Show all steps (work) on your answer sheet for full credit. Problem Code: 1561 The amount of lateral expansion (in mils) was determined for a sample of n = 9 pulsed-power gas metal arc welds used in LNG ship containment tanks. The resulting sample standard deviation was s = 2.83 mils. Assuming normality, derive a 95% confidence interval for σ , the population standard deviation. (Hint: You will first need to come up with a 95% confidence interval for the variance) a) I have placed my work and my answer on my answer sheet. b) I want to have points deducted from my test for not working this problem. Question 16 Your answer is CORRECT. This is a written question, worth 9 points. DO NOT place the problem code on the answer sheet. A proctor will fill this out after exam submission. Show all steps (work) on your answer sheet for full credit. Problem Code: 1662
7/27/16, 12 : 37 PM Print Test Page 8 of 9 Consumer Reports rated 77 cereals on a scale of 0 to 100. The number of grams of sugar contained in each serving of the corresponding cereals was also recorded. Using sugar as the explanatory variable and the Consumer Reports rating as the dependent variable, computer output of the data is as follows (the p- values are intentionally left blank): Predictor Coef StDev T P Constant 58.93 1.847 30.58 --- Sugars − 2.56 0.28 − 9.98 --- S = 9.204 R-Sq = 61.2% R-Sq(adj) = 59.9% Part a: What is the regression equation?
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