Download Regression Analysis: Weight, Fuel Consumption, Oxygen Content, and Weld Strength - Prof. J and more Assignments Statistics in PDF only on Docsity! Practice Problems # 11 - Solutions 1. This is a continuation of exercise # 4 in Practice Problem set # 9. Use the same data. a. What proportion of observed variation in mileage can be attributed to the simple linear regression relationship between weight and mileage? 2 8.859128 0.8923 9.928171 R = = b. Does the fitted model provide enough evidence for us to say that there is a significant relationship between weight and fuel consumption? That is, test the hypothesis that the slope is zero using the F test. Sum of Mean Prob Power Source DF Squares Square F-Ratio Level (5%) Slope 1 8.859128 8.859128 41.434844 0.001345 0.998745 Error 5 1.069043 0.2138087 Adj. Total 6 9.928171 1.654695 To test the hypothesis that β1 = 0 we use the F test from the above ANOVA table. H0: β1 = 0 Ha: β1 ≠ 0 fobs = 41.434844 with an associated p-value of 0.001345. Since this is a very small p-value, we would reject the null hypothesis. c. One way to monitor fuel consumption and environmental emissions for an entire group of vehicle types is to pay attention to the slope of the fitted regression line relating weight and fuel consumption (the less steep the slope of the line, the less differential there is between weight increase and fuel consumption). Does the data provide evidence that the slope is greater than -0.16? Test a hypothesis about β1. To test this hypothesis we must use the t test. H0: β1 = -0.16 Ha: β1 > -0.16 ( ) ( )1 10 1 ˆ 0.155131 0.16 0.20203 ˆ 0.024100 t s β β β − − −− = = = p-value = ( ) ( )( )52 [ 0.20203] 2 0.4239 0.8478P T⋅ > = = . If our significance level α is smaller than this value then we would fail to reject the null hypothesis that β1 = -0.16. If we set α = 0.05, then since this is a two tailed test, the rejection regions and critical values are given by: Reject H0: β1 = -0.16 if and only if 2, 2obs nt tα −> . This critical value is given by 2.01505 . Since tobs = 0.20203 < 2.01505, we would fail to reject the null hypothesis. d. Estimate the true average change in mileage with a change of 1 ton in the weight of the vehicle, by constructing a 99% confidence interval on the parameter. What is needed is a confidence interval on the slope of the regression function. ( ) ( )1 2, 2 0.2138087ˆ 0.155131 4.032143 0.155131 0.097175 0.25231, 0.05796 368.125n xx MSEt sα β −± = − ± = − ± = − − e. Explain how your answer to part d conveys information about precision and reliability. It is a 99% confidence interval; hence the reliability of our inference is high. Since the confidence interval is reasonably narrow, our answer seems to be quite precise. 2. In a study of the relationship between oxygen content (x) and the ultimate testing strength (y) of welds, the data presented in the following table were obtained for n = 29 welds. Here, oxygen content is measured in parts per thousand, and strength is measured in ksi. The following three exercises are to be completed in the context of fitting a simple linear regression to this set of data. Oxygen Content Strength Oxygen Content Strength Oxygen Content Strength 1.08 63.00 1.16 68.00 1.17 73.00 1.19 76.00 1.32 79.67 1.40 81.00 1.57 66.33 1.61 71.00 1.69 75.00 1.72 79.67 1.70 81.00 1.71 75.33 1.80 72.5 1.69 68.65 1.63 73.70 1.65 78.4 1.78 84.40 1.70 91.20 1.50 72.00 1.50 75.05 1.60 79.55 1.60 83.20 1.70 84.45 1.60 73.95 1.20 71.85 1.30 70.25 1.30 66.05 1.80 87.15 1.40 68.05 a. Calculate a 95% confidence interval for the true mean strength of all welds whose oxygen content is 1.7 parts per thousand.
