Standardized stock price indicators in three different countries over a week are listed below. An analyst...

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Standardized stock price indicators in three different countriesover a week are listed below. An analyst is interested in knowingif the stock markets of these different countries are dependent onone another. The data set and a partial ANOVA table for this studyare provided below. I II III 890 900 905 899 900 900 900 887 896905 906 928 871 893 899 910 900 934 Source of variation SS DF MS FTreatment 748 2 374 ??? Error 2526 ??? ??? Total 3274 ???

Compute the MSE andthe 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

Suppose the p-valuefor the test is 0.143. At the 0.05 level of significance, how doyou 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 noevidence that the means are different

Reject H0. P-value isgreater than alpha

None

Calculate the TukeyCriterion (T) for use in a Tukey pairwise comparisons test. Usealpha of 0.05.

T = 3.67

T = 19.05

T = 19.44

None of the above

of the abovestatements is correct

Which of the pairs ofsample means are statistically significant?

Sample means forGroups 1 and 2

Sample means forGroups 1 and 3

Sample means forGroups 2 and 3

None issignificant

Answer & Explanation Solved by verified expert
3.9 Ratings (547 Votes)

Source SS df MS F
treatment 748 2 374 2.22
error 2526 15 168.4
total 3274 17

1)

MSE = 168.4. F = 2.22

2)

Do not reject H0. There is no evidence that the means are significantly different

3)

pooled standard deviation                                                                             =Sp =?MSE      = 12.976
critical value of q with 0.05 level and at k=3 and N-k=15 degree of freedom= 3.673
Tukey's (HSD) for group i and j =                  (q/?2)*(sp*?(1/ni+1/nj)         = 19.44

4)

None is significant


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