Question

Part A

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

1. MSE = 374. F = 2.22
2. MSE = 168.4. F = 2.22
3. MSE = 2.22. F = 15
4. MSE = 2,526. F = 2.22
5. None of the above

Part B

Suppose the p-value for the test is 0.143. At the 0.05 level of significance, how do you conclude?

1. Do not reject H0. There is no evidence that the means are significantly different
2. Do not reject H0. Evidence exists that the means are significantly different
3. Reject H0. There is no evidence that the means are different
4. Reject H0. P-value is greater than alpha
5. None of the above statements is correct

Part C

Calculate the Tukey Criterion (T) for use in a Tukey pairwise comparisons test. Use alpha of 0.05.

1. T = 3.67
2. T = 19.05
3. T = 19.44
4. None of the above

Part D

Which of the pairs of sample means are statistically significant?

1. Sample means for Groups 1 and 2
2. Sample means for Groups 1 and 3
3. Sample means for Groups 2 and 3
4. None is significant

Answer 1

The number of columns, k = 3

The Total observations, N = 18

DF Treatment = n - 1 = 2

DF Error = N - k = 18 - 3 = 15

DF Total = DF treatment + DF error = 2 + 15 = 17

MSTR = SS treatment/DF treatment = 748/2 = 374

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PART A: Option b

MSE = SS Error/DF error = 2526/15 = 168.4

F statistic = MSTR/MSE = 374/168.4 = 2.22

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PART B: Option a

If p value is < $\alpha$, then Reject H0

Since p value (0.143) is > $\alpha$ (0.05), Do not Reject H0. There is no evidence that the means are significantly different.

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PART C: Option a

Tukeys Crirerior for pairwise comparison at $\alpha$ = 0.05

For k = 3 and DF error = 15, the Tukey Value T (from tables) = 3.67

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Part D: Option d, none is significant.

It is given by

where Mi and Mj are the 2 means being compared and their positive difference is taken. n = number of replicates in each sample. Here n = 6, MSerror 168.4

$\sqrt{\frac{MS_{error}}{n}} = \sqrt{\frac{168.4}{6}} = 5.28$

The Rule is that if Tukeys Observed is > Tukeys Critical, then there is a significant difference between the groups.

M1 = 895.83, M2 = 897.67 and M3 = 910.33

Group 1 and Group 2 : M1 - M2 = Absolute (895.83 - 897.67) = 1.84

Tukeys Observed = 1.84/5.28 = 0.348.

Since Tukeys observed (0.348) is < Tukeys Critical (3.67), there isn't a significant difference between the 2 groups.

Group 1 and Group 3 : M1 - M3 = Absolute (895.83 - 910.33) = 14.5

Tukeys Observed = 14.5/5.28 = 2.746.

Since Tukeys observed (2.746) is < Tukeys Critical (3.67), there isn't a significant difference between the 2 groups.

Group 2 and Group 3 : M2 - M3 = Absolute (897.67 - 910.33) = 12.66

Tukeys Observed = 12.66/5.28 = 2.398.

Since Tukeys observed (2.398) is < Tukeys Critical (3.67), there isn't a significant difference between the 2 groups.

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