Spring 2000 Exam 2 Solutions MSCI3710 No. Solution Version 1 ( 7 ) According to the ANOVA table, The between treatment variation is the value of mean square term, which is 18.75 However because of the terminology of the question both the variance (mean square) and the sum of squares are correctly measures of variation. D Or A 2 ( 8 ) SSE= 134.64(190.92-56.25), MSE=16.83(134.64/8), F value= 1.11(18.75/16.83) B 3 ( 9 ) The calculated value =1.11, F critical value= 4.07, therefore we Fail to Reject the null. C 4 ( 1 ) We want to test P1 is greater than P2, therefore, Ho : P1-P2 ≤ 0, Ha: P1-P2 0 C 5 ( 2 ) According to the t table, when df is greater than 120 and one tail test, we can use the Z test, and the critical value is 1.645 ( α =0.05) A 6 ( 3 ) The expected value for the Male/Local is 45.83(110*100/240) A 7 ( 4 ) According to the Chi-square table, when α =0.01, df=1 The critical value =6.635 A 8 ( 5 ) If the calculated value is 7.17, this value is greater than the critical
This is the end of the preview. Sign up
access the rest of the document.