Calculate the mean square for blocking

Assignment Help Applied Statistics

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.

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.

132

None of the above

Question 23

Calculate the F statistic for treatment.

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.

Reference no: EM132219564

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