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3. Two accounting professors decided to compare the variance of their grading pr

ID: 3309807 • Letter: 3

Question

3. Two accounting professors decided to compare the variance of their grading procedures. To accomplish this, they each graded the same 10 exams, with the following results:

  

Mean Grade

Standard Deviation

Professor 1

79.3

22.4

Professor 2

82.1

12.0

At the 10% level of significance, what is the decision regarding the null hypothesis?

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Reject the null hypothesis and conclude the variances are different.

Fail to reject the null hypothesis and conclude no significant difference in the variances.

Reject the null hypothesis and conclude the variances are the same.

Fail to reject the null hypothesis and conclude the variances are the same.

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5. the following ANOVA table for three treatments each with six observations:

Source

Sum of Squares

df

Mean square

Treatment

1116

Error

1068

Total

2184

What is the computed value of F?

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7.48

7.84

8.84

8.48

16. An ANOVA procedure is applied to data obtained from four distinct populations. The samples, each comprised of 15 observations, were taken from the four populations. The degrees of freedom for the numerator and denominator for the critical value of F are _______.

Multiple Choice

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4 and 60, respectively

4 and 21, respectively

3 and 59, respectively

3 and 56, respectively

19. Two accounting professors decided to compare the variance of their grading procedures. To accomplish this, they each graded the same 10 exams, with the following results:

Mean Grade

Standard Deviation

Professor 1

79.3

22.4

Professor 2

82.1

12.0

What is the critical value of F at the 0.02 level of significance?

Top of Form

5.85

5.35

6.51

4.03

  

Mean Grade

Standard Deviation

Professor 1

79.3

22.4

Professor 2

82.1

12.0

Explanation / Answer

3) for test statistic F =(s1/s2)2 =(22.4/12)2 =3.484

for (n1-1=9) and (n2-1=9) degree of freedom and above test stat ; p value =0.077

as p vlaue is less then 0.1 level ;

Reject the null hypothesis and conclude the variances are different.

5)

from above F value =7.84

16)(

degrees of freedom for the numerator =treatment-1= 4-1 =3

degrees of freedom for the denominator =total values -treatment =60-4 =56

3 and 56, respectively

19)

critical value of F at the 0.02 level of significance and (9,9) degree of freedom =5.35

Source Sum of Squares df Mean square F Treatment 1116 2 558 7.84 Error 1068 15 71.2 Total 2184 17
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