Question

2. In the study reported in question 2, there were 25 students in each group.

1. Calculate and report the standard error of the mean.

SE_x-bar = sqrt(MSe/n_g)

n_g is the number per group

2. Conduct the three post-hoc Tukey HSD tests.

(mean₁ – mean₂) / SE_x-bar

3. Interpret the results of these tests.

(THIS WAS THE PREVIOS QUESTION)

2. Suppose I want to know whether students from Iowa, California, and Arizona differ in their average GPA in high school. I collect my data and find the following statistics from three groups of 25 students each (total N=75):

Average GPA: Iowa = 2.10, California = 2.80, Arizona = 3.35

SS_b = 4.20                       SS_e= 49.25                                                                            

Set your alpha level at .05. Then…

1. Report your degrees of freedom for between groups and for within groups;

df between = K – 1

df within = N – K

2. Calculate and report your MS_b and your MS_e

MS_B= SSb/K-1

MS_E = SSe/N-K

3. Calculate and report your F-ratio:

F_{(2, 72)} = MSb/MSe

4. Decide whether you have a statistically significant result. Give your reasoning.

5. Interpret your results. What do you now know?

Answer 1

Given N = 75, k = 3 and alpha = 0.05

a) dfbetween = K-1 = 3-1 = 2
   dfwithin = N - k = 75-3 = 72

b) MSB = SSb/k-1 = 4.20 / 2 = 2.1
   MSE = SSe/N-k = 49.25 / 72 = 0.684

c) F-Ratio = MSB / MSE = 2.1 / 0.684 = 3.0702

d) F critical value = 3.124 at 5% df and (2,72) df

since F-Ratio is less than F-Critical value so we accept H0
i.e. there is no statistically significant

e) Thus we conclude that there is no significant difference among the means of three groups

ANOVA Table: