Explanation Given info: The data represents the values of the variables abrasive wear loss ( y ) and retained austenite content ( x ) for a sample of 17 observations using the first abrasive method. Furthermore SAS output of regression analysis is also provided for the first abrasive method. The summary statistic using the second abrasive method for a sample of 15 observations is S S x 2 = 7 , 152.5578 , S S E 2 = 0.51350 and γ ^ 1 = 0.006845 . The level of significance is 0.05. Calculation: Linear regression model: A linear regression model is given as y ^ = b 0 + b 1 x where y ^ be the predicted values of response variable and x be the predictor variable. b 1 be the slope and b 0 be the intercept of the line. A linear regression model is given as y ^ = β ^ 0 + β ^ 1 x where y ^ be the predicted values of response variable and x be the predictor variable. The β ^ 1 be the estimate of slope and β ^ 0 be the estimate of intercept of the line. Here, y ^ = β 0 + β 1 x + ε is the regression model for the variables abrasive wear loss ( y ) and retained austenite content ( x ) suing the first abrasive method. Here, y = γ 0 + γ 1 x + ε is the regression model for the variables abrasive wear loss ( y ) and retained austenite content ( x ) suing the second abrasive method. From the accompanying SAS-output, the slope coefficient of the regression equation is β ^ 1 = − 0.074676 . Here, β ^ 1 be the slope of the sample regression line of first abrasive method and β 1 is the slope of the population regression line of the first abrasive method. The slope parameter of the second abrasive method is γ ^ 1 = 0.006845 Here, γ ^ 1 be the slope of the sample regression line of second abrasive method and γ 1 is the slope of the population regression line of the second abrasive method. The test hypotheses are given below: Null hypothesis: H 0 : β 1 = γ 1 That is, the expected change in wear loss associated with 1% increase in austenite content is identical for the two types of abrasive. Alternative hypothesis: H 1 : β 1 ≠ γ 1 That is, the expected change in wear loss associated with 1% increase in austenite content significantly differs for the two types of abrasive. Test statistic: The test statistic is, T = β ^ 1 − γ ^ 1 σ ^ 1 S S x 1 + 1 S S x 2 ∼ t ( n 1 + n 1 − 4 ) Critical value: For 0.05 level of significance, α = 0.05 α 2 = 0.025 Degrees of freedom: The degrees of freedom is, d . f = n 1 + n 1 − 4 = 17 + 15 − 4 = 28 Thus, the degree of freedom is 28. From Table A.5 of the t- distribution in Appendix A, the critical value corresponding to the right tail area 0.025 and 28 degrees of freedom is 2.048. Thus, the critical value is ( t α ) = 2.048 . The value of S S x for the first abrasive method: The value of S S x 1 is obtained as follows: i x 1 i x 1 i 2 1 4.6 21.16 2 17.0 289 3 17.4 302.76 4 18.0 324 5 18.5 342.25 6 22.4 501.76 7 26.5 702.25 8 30.0 900 9 34.0 1156 10 38.8 1505.44 11 48.2 2323.24 12 63.5 4032.25 13 65.8 4329.64 14 73.9 5461.21 15 77.2 5959

Probability and Statistics for Engineering and the Sciences STAT 400 - University Of Maryland

9th Edition

ISBN: 9781305764477

Author: Jay L. Devore

Publisher: Cengage Learning

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Chapter 12, Problem 87SE

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Test whether there is enough evidence to conclude that the slope coefficient of the variable austenite content is identical for the two types of abrasive.

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Probability and Statistics for Engineering and the Sciences STAT 400 - University Of Maryland

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