In: Math
Television advertisers base their investment decisions regarding the promotion of their products and services on demographic information about television viewers. The age of the viewers is a key factor in their process. The following table shows the number of hours that a random sample of individuals watched television during the week. The individuals are grouped according to their ages. Minitab is required. You will need to enter the data into Minitab.
Age Group |
|||
18-24 |
25-34 |
35-49 |
50-64 |
49 |
41 |
44 |
39 |
33 |
40 |
19 |
14 |
33 |
33 |
27 |
15 |
39 |
35 |
36 |
17 |
71 |
21 |
49 |
20 |
a. At the 0.05 level of significance, determine if there is a difference in the mean number of hours of television watched by age group. State your hypotheses and show all 7 steps clearly. (14 points)
b. Give and interpret the p-value. (3 points)
c. Should Tukey pairwise comparisons be conducted? Why or why not? (3 points)
d. If appropriate, use Minitab to produce Tukey pairwise comparison. Write a few sentences with your conclusions from those comparisons. (4 points)
e. Use Levene’s test to determine if the assumption of homogeneity of variances is valid. Give the hypotheses, test statistic, p-value, decision, and conclusion. Use the 0.05 level of significance. (8 points)
f. Verify with Minitab by attaching or including relevant output. (6 points)
a. At the 0.05 level of significance, determine if there is a difference in the mean number of hours of television watched by age group. State your hypotheses and show all 7 steps clearly. (14 points)
One-way ANOVA: 18-24, 25-34, 35-49, 50-64
Method
Null hypothesis |
All means are equal |
Alternative hypothesis |
Not all means are equal |
Significance level |
α = 0.05 |
Equal variances were assumed for the analysis.
Factor Information
Factor |
Levels |
Values |
Factor |
4 |
18-24, 25-34, 35-49, 50-64 |
Analysis of Variance
Source |
DF |
Adj SS |
Adj MS |
F-Value |
P-Value |
Factor |
3 |
1454 |
484.6 |
3.38 |
0.044 |
Error |
16 |
2296 |
143.5 |
||
Total |
19 |
3750 |
Model Summary
S |
R-sq |
R-sq(adj) |
R-sq(pred) |
11.9791 |
38.77% |
27.29% |
4.33% |
Means
Factor |
N |
Mean |
StDev |
95% CI |
18-24 |
5 |
45.00 |
15.94 |
(33.64, 56.36) |
25-34 |
5 |
34.00 |
8.00 |
(22.64, 45.36) |
35-49 |
5 |
35.00 |
12.23 |
(23.64, 46.36) |
50-64 |
5 |
21.00 |
10.32 |
(9.64, 32.36) |
Pooled StDev = 11.9791
Tukey Pairwise Comparisons
Grouping Information Using the Tukey Method and 95% Confidence
Factor |
N |
Mean |
Grouping |
|
18-24 |
5 |
45.00 |
A |
|
35-49 |
5 |
35.00 |
A |
B |
25-34 |
5 |
34.00 |
A |
B |
50-64 |
5 |
21.00 |
B |
Means that do not share a letter are significantly different.
Tukey Simultaneous Tests for Differences of Means
Difference of |
Difference |
SE of |
95% CI |
T-Value |
Adjusted |
25-34 - 18-24 |
-11.00 |
7.58 |
(-32.70, 10.70) |
-1.45 |
0.487 |
35-49 - 18-24 |
-10.00 |
7.58 |
(-31.70, 11.70) |
-1.32 |
0.564 |
50-64 - 18-24 |
-24.00 |
7.58 |
(-45.70, -2.30) |
-3.17 |
0.027 |
35-49 - 25-34 |
1.00 |
7.58 |
(-20.70, 22.70) |
0.13 |
0.999 |
50-64 - 25-34 |
-13.00 |
7.58 |
(-34.70, 8.70) |
-1.72 |
0.348 |
50-64 - 35-49 |
-14.00 |
7.58 |
(-35.70, 7.70) |
-1.85 |
0.288 |
Individual confidence level = 98.87%
Critical value of F(3,16) at 0.05 level of significance = 3.239
Calculated F=3.38 > critical F value 3.239. Ho is rejected.
Calculated F=3.38, P=0.044 which is < 0.05 level of significance. Ho is rejected. We conclude that there is a difference in the mean number of hours of television watched by age group.
b. Give and interpret the p-value. (3 points)
P=0.044.
P value is observing this much difference between groups or more when the null hypothesis is true.
c. Should Tukey pairwise comparisons be conducted? Why or why not? (3 points)
since ANOVA is significant pairwise comparisons to be conducted to see which pairs are significant.
d. If appropriate, use Minitab to produce Tukey pairwise comparison. Write a few sentences with your conclusions from those comparisons. (4 points)
Tukey Simultaneous Tests for Differences of Means
Difference of |
Difference |
SE of |
95% CI |
T-Value |
Adjusted |
25-34 - 18-24 |
-11.00 |
7.58 |
(-32.70, 10.70) |
-1.45 |
0.487 |
35-49 - 18-24 |
-10.00 |
7.58 |
(-31.70, 11.70) |
-1.32 |
0.564 |
50-64 - 18-24 |
-24.00 |
7.58 |
(-45.70, -2.30) |
-3.17 |
0.027 |
35-49 - 25-34 |
1.00 |
7.58 |
(-20.70, 22.70) |
0.13 |
0.999 |
50-64 - 25-34 |
-13.00 |
7.58 |
(-34.70, 8.70) |
-1.72 |
0.348 |
50-64 - 35-49 |
-14.00 |
7.58 |
(-35.70, 7.70) |
-1.85 |
0.288 |
Individual confidence level = 98.87%
The age groups 50-64 and 18-24 are significant. All other pairs are not significant.
e. Use Levene’s test to determine if the assumption of homogeneity of variances is valid. Give the hypotheses, test statistic, p-value, decision, and conclusion. Use the 0.05 level of significance. (8 points)
Null hypothesis |
All variances are equal |
Alternative hypothesis |
At least one variance is different |
Significance level |
α = 0.05 |
Tests
Method |
Test |
P-Value |
Multiple comparisons |
— |
0.650 |
Levene |
0.45 |
0.721 |
Calculated test value 0.45, P=0.721 which is > 0.05 level. Ho is not rejected.
We conclude that assumption of homogeneity of variances is valid.