Reference no: EM132219564
Question 1
Analysis of variance (ANOVA) is a test for the equality of:
Variances
Means
Proportions
Any two population parameters
None of the above
Question 2
For this and the next 2 questions: A recently hired MBA on the staff of Ace Systems thinks she can improve the training of production workers, who until now have been trained only on the job. She has developed three new alternative programs involving classroom sessions. These are compared, along with the existing procedure, as four methods or treatments:
[1]On-the-job training only.
[2]Classroom training only.
[3]Heavy on-the-job training with light classroom work.
[4]Light on-the-job training with heavy classroom work.
Each method is conducted with a different sample group of 8 randomly chosen trainees, who are given a placement examination upon completion. The investigator uses the scores on that test as the response variable. Statistical analysis provides the following sample means:
X-bar1 = 50.2. X-bar2 = 44.9. X-bar3 = 78.5. X-bar4 = 64.0
The SSE for the study is 8,431.31. Calculate the mean square treatment (MSTR).
8,803.12
8,431.31
301.12
None of the above
Question 3
If the computed F statistic for the study is 6.03, how should the investigator conclude?
At alpha = 0.01, the critical value of F is 4.57. Therefore the null hypothesis that all training methods result in identical outcomes should be rejected
At alpha = 0.01, the critical value of F is 10.85. Therefore the null hypothesis that all training methods result in identical outcomes should be rejected
At alpha = 0.01, the critical value of F is 4.57. Therefore the null hypothesis that all training methods result in identical outcomes should NOT be rejected
At alpha = 0.01, the critical value of F is 10.85. Therefore the null hypothesis that all training methods result in identical outcomes should NOT be rejected
Question 4
At alpha = 0.05, compute the Tukey Criterion appropriate for a pairwise comparison test. Choose the closest answer.
6.1351
23.8044
125.9605
33.6645
Question 5
For this and the next 2 questions. A watch manufacturer markets a particular model with 3 different straps: gold, silver, and leather. The average number of sales in a week of each watch are 12, 15, and 11, respectively. An ANOVA to find whether or not customers prefer any particular strap over the random sample of 38 observations, shows SSE = 1490 and SSTR = 760. How might the null hypothesis be written?
Not all the straps are the same
There is no difference in customer preference for the three different straps
Customer preference is the same for at least two of the three straps
Not all the means are equal
None of the above is correct
Question 6
Compute the F statistic for this hypothesis test.
380
42.571
8.926
None of the above
Question 7
At alpha = 0.05, how do you conclude?
Reject the null hypothesis and conclude that customer preference for the straps is not the same
Do not reject the null hypothesis and conclude that customer preference for the straps is not the same
Reject the null hypothesis and conclude that customers are indifferent across all three straps
Reject Ho. The critical value of F is less than 3.23
None of the above is correct
Question 8
For this and the next 2 questions: A marketing manager wants to determine if the average advertising spending per month of his competitors is equal or not. Data over the last six months reveal the following figures (in thousands of dollars):
A
|
B
|
C
|
$11
|
$9
|
$12
|
17
|
12
|
23
|
27
|
27
|
28
|
35
|
45
|
27
|
43
|
54
|
39
|
38
|
32
|
41
|
Calculate the sum of squares error (SSE). You may use spreadsheet to perform ANOVA.
2,933.67
8.11
2941.78
195.58
None of the above
Question 9
What is the sum of squares treatment (SSTR)?
2,933.67
8.11
2941.78
195.58
None of the above
Question 10
What is the calculated F statistic for the ANOVA?
0.9795
3.682
195.578
4.055
None of the above
Question 11
FOR THIS AND THE NEXT QUESTION. According to The Boston Globe (August 16, 2010), Asian residents in Boston have the highest average life expectancy of any racial or ethnic group. The report shows, in particular, that Asians live a decade longer than black residents. Suppose the sample results below are indicative of the overall results. Calculate the F statistic.
|
Asian
|
Black
|
Latino
|
White
|
Sample mean
|
83.70
|
73.50
|
80.60
|
79.00
|
Sample variance
|
26.30
|
27.50
|
28.20
|
24.80
|
Sample size
|
20
|
20
|
20
|
20
|
79.20
13.67
2.72
26.70
None of the above
Question 12
Suppose the p-value for the above study = 0.00. Which of the following conclusions with respect to the study is correct?
At the 5% level, do not reject the null hypothesis. Average life expectancy differs across the four racial groups
At the 1% level, reject the null hypothesis. Average life expectancy is the same across the four racial groups
At the 1% level, reject the null hypothesis. Average life expectancy differs across the four racial groups
None of the above
Question 13
FOR THIS AND THE NEXT QUESTION. During a typical PGA tournament, the competing golfers play 4 rounds of golf, where the hole locations are changed for each round. Below are the scores for the top 5 finishers at the 2009 U.S. Open. At the 5% level, can you conclude that average scores produced by the 4 different ROUNDS differ?
|
Round
|
Golfer
|
1
|
2
|
3
|
4
|
Lucas Glover
|
69
|
64
|
70
|
73
|
Phil Mickelson
|
69
|
70
|
69
|
70
|
David Duval
|
67
|
70
|
70
|
71
|
Ricky Barnes
|
67
|
65
|
70
|
76
|
Ross Fisher
|
70
|
78
|
79
|
72
|
There is NO statistical evidence of a difference in the average scores produced by the 4 different rounds. The F critical value of 3.24 exceeds the calculated F value of 1.26.
There is NO statistical evidence of a difference in the average scores produced by the 4 different rounds. The p-value of 0.3215 is less than the significance level for the study.
There is a statistical evidence of a difference in the average scores produced by the 4 different rounds. The p-value is greater than the significance level.
There is NO statistical evidence of a difference in the average scores produced by the 4 different rounds. The F critical value of 3.0556 exceeds the calculated F value of 1.94.
Question 14
For this and the next 3 questions. Standardized stock price indicators in three different countries over a week are listed below. An analyst is interested in knowing if the stock markets of these different countries are dependent on one another. The data set and a partial ANOVA table for this study are provided below.
I
|
II
|
III
|
890
|
900
|
905
|
899
|
900
|
900
|
900
|
887
|
896
|
905
|
906
|
928
|
871
|
893
|
899
|
910
|
900
|
934
|
Source of
variation
|
SS
|
DF
|
MS
|
F
|
Treatment
|
748
|
2
|
374
|
???
|
Error
|
2526
|
???
|
???
|
|
|
|
|
|
|
Total
|
3274
|
???
|
|
|
Compute the MSE and the F statistic
MSE = 374. F = 2.22
MSE = 168.4. F = 2.22
MSE = 2.22. F = 15
MSE = 2,526. F = 2.22
None of the above
Question 15
Suppose the p-value for the test is 0.143. At the 0.05 level of significance, how do you conclude?
Do not reject H0. There is no evidence that the means are significantly different
Do not reject H0. Evidence exists that the means are significantly different
Reject H0. There is no evidence that the means are different
Reject H0. P-value is greater than alpha
None of the above statements is correct
Question 16
Calculate the Tukey Criterion (T) for use in a Tukey pairwise comparisons test. Use alpha of 0.05.
T = 3.67
T = 19.05
T = 19.44
None of the above
Question 17
Which of the pairs of sample means are statistically significant?
Sample means for Groups 1 and 2
Sample means for Groups 1 and 3
Sample means for Groups 2 and 3
None is significant
Question 18
For this and the next 2 questions: A survey is conducted to find whether students like a particular type of music more than another: rock, hip-hop, classical, and jazz. Ratings for each type of music are collected from 50 students (total sample size is 50). The following partial ANOVA results are provided: SSTR = 28,590 and SST = 40,220. Under what circumstance will the null hypothesis for this ANOVA NOT be rejected?
If students have equal preference for all four types of music
If students do not have equal preference for all four types of music
If the p-value associated with the calculated F statistic is less than the level of significance
None of the above statements is correct
Question 19
Calculate the F statistic for this study.
49
37.69
2.46
Incomplete information
Question 20
If the p-value for this study is less than alpha, then one may conclude that:
At least one of the "treatment means" is different
Evidence exists that students do not like all the types of music equally
There is the need for further analysis using, for example, the Tukey Criterion for one factor ANOVA
All of the above is correct
Only statements (a) and (c) are correct
Question 21
You are given the following ANOVA table, with n = 55. Compute the F statistic for the interaction term.
Source of Variation
|
Sum of Squares
|
Degrees of Freedom
|
Mean Square
|
F
|
Factor A
|
127
|
c-1 = 3
|
|
|
Factor B
|
784
|
r-1 = 7
|
|
|
Interaction
|
253
|
|
|
|
Error
|
5761
|
|
|
|
Total
|
|
|
|
|
12.0476
3.608
1.611
0.048
none of the above
Question 22
For this and next 2 questions: A randomized block design is used to test the null hypothesis that mean responses are identical under five treatments. Using four levels for the blocking factor, the following data are obtained: SSTR = 84; SSBL = 132; TSS = 288. Calculate the mean square for blocking (MSBL). Hint: Set up the ANOVA table first.
84
132
21
44
None of the above
Question 23
Calculate the F statistic for treatment.
21
6
3.5
7.33
None of the above
Question 24
Should the null hypothesis of identical population means for the treatment factor be accepted or rejected at alpha of 0.01?
At alpha of 0.01, F(4, 12) critical value = 5.41. Do NOT reject H0. There is no evidence that treatment means are different.
At alpha of 0.01, F(3, 12) critical value = 5.95. Do NOT reject H0. There is evidence of interaction.
At alpha of 0.01, F(3, 12) critical value = 5.95. Reject H0. There is evidence that treatment means are different.
None of the above statements is correct
Question 25
Excel Spreadsheet Exercise: A financial analyst for Investments Concepts is evaluating three rules for triggering changes in variable rate mortgages. Using randomly generated market interest rates scenarios as the blocking variable, the analyst obtained the following sample data from computer simulations.
|
Loan Portfolio Yield
|
Block
|
A
|
B
|
C
|
1
|
11.3
|
12.4
|
10.9
|
2
|
14.2
|
14.3
|
13.7
|
3
|
14.9
|
15.2
|
14.7
|
4
|
12.2
|
11.3
|
12.5
|
At alpha of 0.05, how should the analyst conclude regarding the mean portfolio yield?
The null hypothesis should NOT be rejected. There is NO evidence that mean portfolio yields differ
The null hypothesis should be REJECTED. There is evidence that mean portfolio yields differ. P-value is less than 0.05
The null hypothesis should NOT be rejected. P-value for blocking is 0.0011
None of the above statements is correct
Question 26
ONE WAY ANOVA. During a typical PGA tournament, the competing golfers play 4 rounds of golf, where the hole locations are changed for each round. Below are the scores for the top 5 finishers at the 2009 U.S. Open. At the 5% level, can you conclude that average scores produced by the 5 different PLAYERS differ?
|
Round
|
Golfer
|
1
|
2
|
3
|
4
|
Lucas Glover
|
69
|
64
|
70
|
73
|
Phil Mickelson
|
69
|
70
|
69
|
70
|
David Duval
|
67
|
70
|
70
|
71
|
Ricky Barnes
|
67
|
65
|
70
|
76
|
Ross Fisher
|
70
|
78
|
79
|
72
|
There is NO statistical evidence of a difference in the average scores produced by the 5 different players. The F critical value of 3.24 exceeds the calculated F value of 1.26.
There is strong evidence of a difference in the average scores produced by the 5 different players. The F critical value of 3.0556 exceeds the calculated F value of 1.94.
There is NO statistical evidence of a difference in the average scores produced by the 5 different players. The p-value of 0.1552 exceeds the significance level for the study.
There is a statistical evidence of a difference in the average scores produced by the 5 different players. The p-value is greater than the significance level for the study.