The following data are the semester tuition charges ($000) for a
sample of private colleges in various regions of the United States.
At the .05 significance level, can we conclude there is a difference
in the mean tuition rates for the various regions?
Northeast Southeast West
($000) ($000) ($000)
10 8 7
11 9 8
12 10 6
10 8 7
12 6
(a) State the null and the alternate hypotheses.
(b) What is the decision rule?
(c) Develop an ANOVA table.
(d) What is the value of the test statistic?
(e) What is your decision regarding the null hypothesis?
(a) Null hypothesis (H0): There is no difference in the mean tuition rates for the various regions.
Alternate hypothesis (Ha): There is a difference in the mean tuition rates for the various regions.
(b) The decision rule is to reject the null hypothesis if the calculated F statistic is greater than the critical F value.
(c) To develop an ANOVA table:
- Calculate the mean of each region (Northeast, Southeast, West)
- Calculate the overall mean
- Calculate the sum of squares within groups (SSW)
- Calculate the sum of squares between groups (SSB)
- Calculate the degrees of freedom for between groups and within groups
- Calculate the mean square for between groups (MSB) and within groups (MSW)
- Calculate the F statistic
(d) Calculating the F statistic, we will use the formula F= MSB/MSW and compare it to the critical F value from the F distribution table.
(e) Based on the calculated F statistic and comparing it to the critical F value, we would then make a decision regarding the null hypothesis. If the calculated F statistic is greater than the critical F value, we would reject the null hypothesis and conclude that there is a difference in the mean tuition rates for the various regions.
The following data are the semester tuition charges ($000) for a
sample of private colleges in various regions of the United States.
At the .05 significance level, can we conclude there is a difference
in the mean tuition rates for the various regions?
Northeast Southeast West
($000) ($000) ($000)
10 8 7
11 9 8
12 10 6
10 8 7
12 6
(a) State the null and the alternate hypotheses.
(b) What is the decision rule?
(c) Develop an ANOVA table.
(d) What is the value of the test statistic?
(e) What is your decision regarding the null hypothesis?
(a) Null hypothesis (H0): There is no difference in the mean tuition rates for the various regions.
Alternate hypothesis (Ha): There is a difference in the mean tuition rates for the various regions.
(b) The decision rule is to use the F-test in ANOVA. We will reject the null hypothesis if the calculated F-value is greater than the critical F-value at a significance level of 0.05.
(c) ANOVA Table:
| Source | SS | df | MS | F |
|-------------|-------|----|------|------|
| Between | 12.167| 2 | 6.084| 5.526|
| Within | 3.000 | 12 | 0.250| |
Total SS = 15.167
Total df = 14
(d) The value of the F-test statistic is 5.526.
(e) Since the calculated F-value of 5.526 is greater than the critical F-value at a significance level of 0.05, we reject the null hypothesis. Therefore, we can conclude that there is a significant difference in the mean tuition rates for the various regions.