Question

The ANOVA summary table for an experiment with six groups, with five values in each group, is shown to the right. Complete parts (a) through (d) below. Source | Degrees of Freedom | Sum of Squares | Mean Square (Variance) | F Among groups | c - 1 = 5 | SSA = 300 | MSA = 60 | FSTAT = 3.00 Within groups | n - c = 24 | SSW = 480 | MSW = 20 Total | n - 1 = 29 | SST = 780 a. At the 0.01 level of significance, state the decision rule for testing the null hypothesis that all six groups have equal population means. Determine the hypotheses. Choose the correct answer below. A. H0: ?1 = ?2 = ... = ?6 H1: ?1 ? ?2 ? ... ? ?6 B. H0: ?1 = ?2 = ... = ?5 H1: Not all the means are equal. C. H0: ?1 = ?2 = ... = ?6 H1: Not all the means are equal. D. H0: ?1 = ?2 = ... = ?5 H1: ?1 ? ?2 ? ... ? ?5 Find the test statistic. FSTAT = (Round to two decimal places as needed.) Now find the critical value. b. What is your statistical decision? For the given level of significance, ?, reject the null hypothesis if the FSTAT test statistic is greater than the upper-tail critical value, F?. Otherwise, do not reject the null hypothesis. Use this information to make a statistical decision about the equality of the group population means. c. At the 0.01 level of significance, what is the upper-tail critical value from the Studentized range distribution? Find the upper-tail critical value from the Studentized range distribution having c degrees of freedom in the numerator and n - c degrees of freedom in the denominator. d. To perform the Tukey-Kramer procedure, what is the critical range? The critical range for the Tukey-Kramer procedure is given by the formula below, where Q? is the upper-tail critical value from a Studentized range distribution having c degrees of freedom in the numerator and n - c degrees of freedom in the denominator, MSW is the mean square within groups, and nj is the number of values in group j.

          The ANOVA summary table for an experiment with six groups, with five values in each group, is shown to the right. Complete parts (a) through (d) below.

Source | Degrees of Freedom | Sum of Squares | Mean Square (Variance) | F
Among groups | c - 1 = 5 | SSA = 300 | MSA = 60 | FSTAT = 3.00
Within groups | n - c = 24 | SSW = 480 | MSW = 20
Total | n - 1 = 29 | SST = 780

a. At the 0.01 level of significance, state the decision rule for testing the null hypothesis that all six groups have equal population means.
Determine the hypotheses. Choose the correct answer below.

A. H0: ?1 = ?2 = ... = ?6
   H1: ?1 ? ?2 ? ... ? ?6
B. H0: ?1 = ?2 = ... = ?5
   H1: Not all the means are equal.
C. H0: ?1 = ?2 = ... = ?6
   H1: Not all the means are equal.
D. H0: ?1 = ?2 = ... = ?5
   H1: ?1 ? ?2 ? ... ? ?5

Find the test statistic.
FSTAT = (Round to two decimal places as needed.)

Now find the critical value.

b. What is your statistical decision?
For the given level of significance, ?, reject the null hypothesis if the FSTAT test statistic is greater than the upper-tail critical value, F?. Otherwise, do not reject the null hypothesis.

Use this information to make a statistical decision about the equality of the group population means.

c. At the 0.01 level of significance, what is the upper-tail critical value from the Studentized range distribution?

Find the upper-tail critical value from the Studentized range distribution having c degrees of freedom in the numerator and n - c degrees of freedom in the denominator.

d. To perform the Tukey-Kramer procedure, what is the critical range?
The critical range for the Tukey-Kramer procedure is given by the formula below, where Q? is the upper-tail critical value from a Studentized range distribution having c degrees of freedom in the numerator and n - c degrees of freedom in the denominator, MSW is the mean square within groups, and nj is the number of values in group j.

The ANOVA summary table for an experiment with six groups, with five values in each group, is shown to the right. Complete parts (a) through (d) below.

Source | Degrees of Freedom | Sum of Squares | Mean Square (Variance) | F
Among groups | c - 1 = 5 | SSA = 300 | MSA = 60 | FSTAT = 3.00
Within groups | n - c = 24 | SSW = 480 | MSW = 20
Total | n - 1 = 29 | SST = 780

a. At the 0.01 level of significance, state the decision rule for testing the null hypothesis that all six groups have equal population means.
Determine the hypotheses. Choose the correct answer below.

A. H0: ?1 = ?2 = ... = ?6
H1: ?1 ? ?2 ? ... ? ?6
B. H0: ?1 = ?2 = ... = ?5
H1: Not all the means are equal.
C. H0: ?1 = ?2 = ... = ?6
H1: Not all the means are equal.
D. H0: ?1 = ?2 = ... = ?5
H1: ?1 ? ?2 ? ... ? ?5

Find the test statistic.
FSTAT = (Round to two decimal places as needed.)

Now find the critical value.

b. What is your statistical decision?
For the given level of significance, ?, reject the null hypothesis if the FSTAT test statistic is greater than the upper-tail critical value, F?. Otherwise, do not reject the null hypothesis.

Use this information to make a statistical decision about the equality of the group population means.

c. At the 0.01 level of significance, what is the upper-tail critical value from the Studentized range distribution?

Find the upper-tail critical value from the Studentized range distribution having c degrees of freedom in the numerator and n - c degrees of freedom in the denominator.

d. To perform the Tukey-Kramer procedure, what is the critical range?
The critical range for the Tukey-Kramer procedure is given by the formula below, where Q? is the upper-tail critical value from a Studentized range distribution having c degrees of freedom in the numerator and n - c degrees of freedom in the denominator, MSW is the mean square within groups, and nj is the number of values in group j.

Added by Josep R.

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