19. Two accounting profes sors decided to compare the variance of their grading

Question

19. Two accounting profes sors decided to compare the variance of their grading procedure Professor 1 has a larger variance than the exams with the following results: grading by To accomplish this they each graded the same 10 s to find out if the Professor 1 Professor 2 Mean Grade 79.3 82.1 Sample Standard Deviation 22.4 12.0 At the 5% level of significance, what is the decision regarding the null hypothesis? Show your solution Test statistic value value Critical Decision A) 3.48 Reject the null hypothesis and conclude the variance is different. B) 1.86 C) 3.48 2.98 Reject the null hypothesis and conclude the variance is different. E) 1.86 3.18Fail to reject the null hypothesis and conclude no significant difference in the variance 2.98 Reject the null hypothesis and conclude the variance is the same 2.98 Fail to reject the null hypothesis and conclude no significant difference in the D) 1.86 variance. 20. Given the following ANOVA table for three treatments each with six observations: Source Treatment Sum of Squares df Mean Square Error Total 1116 1068 2184 What is the "critical value" ofF at the 5% level ofsignificance? Show your solution A) 3.29 B) 3.68 C) 3.59 D) 3.20 E) 6.35

Explanation / Answer

Question 19

Critical test statistic

F = s12 /s22 = (22.4/12.0)2 = 3.4844

Here critical F = F0.05, 9,9 = 3.18

so we shall reject the null hypothesis and concllude that variance is different.

Option A is correct.

Q.30

Here In ANOVA

MSW (Treatments) = SST/df = 1116/(3-1)= 1116/2 = 558

Here,

MSE = SSE/dF(error) = 1068/ (3 * 5) = 71.2

so F = MSW/MSW = 558/71.2 = 7.837

Here critical value of F is = F0.05, 2,15 = 3.68

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