You have an experiment with 4 groups. The problem with comparing each mean with each other mean using a student's t test is:

false
true
If you make several comparisons, you have an increased chance of a Type I error.
false
The t test assumes normality, which does not occur with more than two groups.
false
The assumption of independence is violated.
With four groups, there are 6 comparisons among means. Even if the population means are the same,
the probability is well over 0.05 that at least one comparison will be significant at the 0.05 level.
A pairwise comparison is

false
false
comparison of two pieces of fruit.
false
a comparison of two levels of intelligence.
true
a comparison of two means.
false
a comparison of two variances.
A pariwse comparison is a comparison between a pair of means.
Assume that you do an experiment with 8 groups and the population means for all 8 are equal. If you make all pairwise comparisons
among the means using the 0.05 level, the chance that 1 or more comparisons will be significant is about:

false
false
0.05
false
0.10
false
0.20
true
0.50
As you can see in Figure 2, it is about 0.50.
Assume you did an experiment with 3 groups and 16 subjects per group. The sample
variances in the three groups were 14, 16, and 18. The value of MSE would be

16
0.0
The MSE is the mean of the sample variances.
Assume you did an experiment with 3 groups and 16 subjects per group. The sample
variances in the three groups were 14, 16, and 18. Using Tukey's test to compare the means,
what would be the value of Q for a comparison of the first mean (14) with the last mean (18)?

4
0.0001
The MSE is 16. The denominator of the formula is the square root of MSE/n = square root of (16/16) = 1. The difference between means is 4.
Q = the difference between means (4) divided by the denominator (1) = 4.
Assume you did an experiment with 3 groups and 16 subjects per group. The sample
variances in the three groups were 14, 16, and 18. Using Tukey's test to compare the means,
what would be the value of Q for a comparison of the first mean (14) with the last mean (18)? What would the df
for the test be?

45
0.0
The degrees of freedom is equal to the total number of subjects (48) minus the number of groups (3) which = 45.
Assume you did an experiment with 3 groups and 16 subjects per group. The sample
variances in the three groups were 14, 16, and 18. Using Tukey's test to compare the means,
what would be the value of Q for a comparison of the first mean (14) with the last mean (18)? What would the two-tailed
probability value be?

.0187
0.005
Use the studentized range calculator. Q = 4, df = 45, number of means = 3. The number of means is the total number of means. p = 0.0187.