Question

Solve the problem. During its manufacture, a product is subjected to four different tests in sequential order. An efficiency expert claims that the fourth (and last) test is unnecessary since its results can be predicted based on the first three tests. To test this claim, multiple regression will be used to model Test4 score , as a function of Test1 score , Test 2 score , and Test3 score [Note: All test scores range from 200 to 800, with higher scores indicative of a higher quality product.] Consider the model: E(y) = β1 + β1x1 + β2x2 + β3x3 The first-order model was fit to the data for each of 12 units sampled from the production line. The results are summarized in the printout. _____________________________________________________________________ ROOT MSE 52.72 R-SQUARE 0.872 DEP MEAN 645.8 ADJ R-SQ 0.824 Compute a 95% confidence interval for β 3. See Image

Solve the problem. During its manufacture, a product is subjected to four different tests in sequential order. An efficiency expert claims that the fourth (and last) test is unnecessary since its results can be predicted based on the first three tests. To test this claim, multiple regression will be used to model Test4 scoret), as a function of Test1 score (x1), Test 2 score (x2), and Test score (23). [Note: All test scores range from 200 to 800, with higher scores indicative of a higher quality product.] Consider the model: ELV) = B1 + B1X1 +B2X2 + B3X3 The first-order model was fit to the data for each of 12 units sampled from the production line. The results are summarized in the printout. SOURCE DF SS MS F VALUE PROB>F 18.16 .0075 MODEL ERROR TOTAL 3 S 12 151417 50472 22231 2779 173648 ROOT MSE 52.72 DEP MEAN 645.8 0.872 R-SQUARE ADJ R-SQ 0.824 PARAMETER STANDARD T FORO VARIABLE ESTIMATE ERROR PARAMETER 1 PROB> ITI INTERCEPT 11.98 50.50 0.15 0.585 X1(TESTI) 0.2745 0.1111 247 0.039 X2(TEST) 0.3762 0.0986 3.82 0.005 X3 TEST3) 0.3265 Q.OSOS 4.04 0.004 Compute a 95% confidence interval for B 3.

Answer 1

confidence interval for slope                      
                      
n =   12   
alpha,α =    0.05                  
estimated slope=   0.3265                  
std error =    0.0808                  
                      
Df = 8                  
t critical value =    2.3060   [excel function: =t.inv.2t(α,df) ]              
                      
margin of error ,E = t*std error =    2.3060   *   0.0808   =   0.18633  
                      
95%   confidence interval is ß3 ± E                   
lower bound = estimated slope - margin of error =    0.3265   -   0.1863   =   0.1402  
upper bound = estimated slope + margin of error =    0.3265   +   0.1863   =   0.5128  

CI (   0.1402 , 0.5128)

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