1. Two accounting professors decided to compare the variation of their grading procedures. To accomplish this they each graded the same 10 exams with the following results: Answer these questions....

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1. Two accounting professors decided to compare the variation of their grading procedures. To accomplish this they each graded the same 10 exams with the following results:
Answer these questions.
What is H0?
A)

B)

C)

D)

What is H1?
What are the degrees of freedom for the numerator of the F ratio?
What are the degrees of freedom for the denominator of the F ratio?
What is the critical value of F at the 0.01 level of significance?
The calculated F ratio is?
At the 1% level of significance, what is the decision?
A) Reject the null hypothesis and conclude the variance is different.
B) Fail to reject the null hypothesis and conclude the variance is different.
C) Reject the null hypothesis and conclude the variance is the same.
D) Fail to reject the null hypothesis and conclude the variance is the same.
2. Given the following Analysis of Variance table for three treatments each with six observations.
Answer these questions.
What are the degrees of freedom for the numerator and denominator?
What is the critical value of F at the 5% level of significance?
What is the mean square for treatments?
What is the computed value of F?
A) 7.48
B) 7.84
C) 8.84
D) 8.48
What is the decision?
A) Reject H0 -- there is a difference in treatment means
B) Fail to reject H0 -- there is a difference in treatment means
C) Reject H0 -- there is a difference in errors
D) Fail to reject H0 -- there is a difference in errors
3. What is the null hypothesis for an ANOVA?
4. In ANOVA, when we do not reject the null hypothesis, what inference do we make about the population means?

5. ANOVA requires that the populations should be ______________________________, __________________________________, and __________________________________.
6. What is the shape of the F distribution?
7. A large department store examined a sample of the 18 credit card sales and recorded the amounts charged for each of three types of credit cards: MasterCard, Visa and Discover. Six MasterCard sales, seven Visa and five Discover sales were recorded. The store used ANOVA to test if the mean sales for each credit card were equal. What are the degrees of freedom for the F statistic?
8. A bottle cap manufacture with four machines and three operators wants to see if variation in hourly production is due to the machines and/or the operators or an interaction effect of machine and operator. Each operator is assigned to each machine and the production of caps from 3 randomly selected hours is recorded. The analysis shows the following Analysis of Variance table.
Answer these questions.
What are the degrees of freedom for the machines, operators, interaction and error?
What is the critical value of F for the machine effect at the 1% level of significance?
What is the mean square for machines, operators, interaction and error?
What is the computed value of F for the machines?
What is the computed value of F for the operators?
Using a 1% significance level, what is the decision for the machines?
9. A preliminary study of hourly wages paid to unskilled employees in three metropolitan areas was conducted. Seven employees were included from Area A, 9 from Area B and 12 from Area C. The test statistic was computed to be 4.91. What can we conclude at the 0.05 level?
A) Mean hourly wages of unskilled employees all areas are equal
B) Mean hourly wages in at least 2 metropolitan areas are different
C) More degrees of freedom are needed
D) None of these is correct
10. An electronics company wants to compare the quality of their cell phones to the cell phones from three competitors. They sample 10 phones from each company and count the number of defects for each phone. If ANOVA were used to compare the average number of defects, the treatments would be defined as:
A) the number of cell phones sampled.
B) the average number of defects.
C) The total number of phones
D) The four companies.
11. Recently, students in a marketing research class were interested in the driving behavior of students. Specifically, the marketing students were interested if exceeding the speed limit was related to social activity. They collected the following responses from 100 randomly selected students:
The null hypothesis for the analysis is:
A) There is no relationship between gender and driving behavior.
B) The correlation between driving behavior and gender is zero.
C) As driving behavior increases, gender increases.
D) The mean of driving behavior equals the mean of gender.
The degrees of freedom for the analysis is:
A) 1
B) 2
C) 3
D) 4
Using 0.05 as the significance level, what is the critical value for the test statistic?
A) 9.488
B) 5.991
C) 7.815
D) 3.841
What is the value of the test statistic?
A) 83.67
B) 9.89
C) 50
D) 4.94
Based on the analysis, what can be concluded?
A) driving behavior and gender are correlated.
B) driving behavior and gender are not related.
C) driving behavior and gender are related.
D) No conclusion is possible.
12. A survey of the opinions of property owners about a street widening project was taken to determine whether the resulting opinion was related to the distance of front footage. A randomly selected sample of 100 property owners was contacted and the results are shown below.
Answer these questions.
What is the critical value at the 5% level of significance?
What is the expected frequency for people who are undecided about the project and have property front-footage between 45 and 120 feet?
What is the null hypothesis?
What is the computed value of chi-square?
If the computed chi-square is 8.5, what is your decision at the 5% level of significance?
13. If the assumptions for the paired t test cannot be met, what is the nonparametric alternative test?
A) Rank correlation
B) Kruskal-Wallis
C) Wilcoxon signed rank
D) Median
14. The chi-square distribution is
A) positively skewed.
B) negatively skewed.
C) normally distributed.
D) negatively or positively skewed.
15. Which of the following assumptions is necessary to apply a nonparametric test of hypothesis using the chi-square distribution?
A) Normal population is required
B) Interval scale of measurement is required
C) Population variance must be known
D) Both "a" and "c"
E) None of the above
16. The chi-square has
A) one distribution.
B) two distributions.
C) a family of distribution.
D) a uniform distribution.
17. The computed value of chi-square statistic is always positive because the difference between the observed frequencies and the expected frequencies is _______________.
Answered Same DayDec 22, 2021

Answer To: 1. Two accounting professors decided to compare the variation of their grading procedures. To...

Robert answered on Dec 22 2021
135 Votes
HW 6 Week 7 (due week 8) 6 points
1. Two accounting professors decided to compare the variation 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
Answer these questions.
What is H0?
A)
2
2
1
2
σ
σ
=
B)
2
2
1
2
σ
σ
¹
C)
2
1
μ
μ
=
D)
2
1
μ
μ
¹

What is H1?
Alternative Hypothesis (H1): σ1² ≠ σ2²
What are the degrees of freedom for the numerator of the F ratio?
Degrees of freedom (numerator) = 10 – 1 = 9
What are the degrees of freedom for the denominator of the F ratio?
Degrees of freedom (denominator) = 10 – 1 = 9
What is the critical value of F at the 0.01 level of significance?
F (0.01, 9, 9) = 5.351
The calculated F ratio is?
F = s12/s22
= (22.4)2/ (12)2
= 3.484
At the 1% level of significance, what is the decision?
A) Reject the null hypothesis and conclude the variance is different.
B) Fail to reject the null hypothesis and conclude the variance is different.
C) Reject the null hypothesis and conclude the variance is the same.
D) Fail to reject the null hypothesis and conclude the variance is the same.
Since calculated F is less than critical value, we fail to reject the null hypothesis.
2. Given the following Analysis of Variance table for three treatments each with six observations.
Source

Sum of Squares

df

Mean Square

Treatments

1116
Error

1068
Total

2184
Answer these questions.
What are the degrees of freedom for the numerator and denominator?
Degrees of freedom (numerator) = k – 1
= 3 – 1
= 2
Degrees of freedom (denominator) = n – k
= 6*3 – 3
= 15
What is the critical value of F at the 5% level of significance?
F (0.05, 2, 15) = 3.682
What is the mean square for treatments?
MS (treatment) = SS/ df
= 1116/2
= 558
What is the computed value of F?
A) 7.48
B) 7.84
C) 8.84
D) 8.48
F = MS (treatment)/ MS (error)
= 558/71.2
= 7.84
What is the decision?
A) Reject H0 -- there is a difference in treatment means
B) Fail to reject H0 -- there is a difference in treatment means
C) Reject H0 -- there is a difference in errors
D) Fail to reject H0 -- there is a difference in errors
Since computed F is greater than the critical value, we reject Ho.
3. What is the null hypothesis for an ANOVA?
 The null hypothesis in ANOVA is that the means of the groups are equal.
4. In ANOVA, when we do not reject the null hypothesis, what inference do we make about the population means?
The population means are equal.
5. ANOVA requires that the populations should be ______________________________, __________________________________, and __________________________________.
ANOVA requires that the populations should be _normally distributed_____________________________, ____have same variance______________________________, and _____independent samples_____________________________.
6. What is the shape of the F distribution?
The shape of the F distribution is rightly skewed.
7. A large department store examined a sample of the 18 credit card sales and recorded the amounts charged for each of three types of credit cards: MasterCard, Visa and Discover. Six MasterCard sales, seven Visa and five...
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