Correlation
Solution-10:
Step-1: Select the appropriate test
We are testing 3 means so we shall use ANOVA.
Step-2: State your research hypothesis and your null hypothesis
H0: There are no differences in mean happiness between the three types of residences.
H1: There are differences in the mean happiness between the three types of residences.
Step-3: Describe the NULL distribution
The variance between the means expected by chance is 0.75.
Step-4: Compute your test statistic
The actual variance between the three means is 34.75.
The AVOVA source table is given as follows:
|
Source |
SS |
df |
MS |
F |
|
Between |
69.5 |
2 |
34.75 |
46.33 |
|
Within |
6.75 |
9 |
0.75 |
|
|
Total |
76.25 |
11 |
|
|
Step-5: Determine your critical value
Step-6: Compare the test stat to the critical value and make your decision
As F > critical value, we reject null H0
From our study, we conclude that there are differences in mean happiness between the three types of residences, at significance level α = 0.05.
Solution-12:
Step-1: Select the appropriate test
We are testing 3 means so we shall use ANOVA.
Step-2: State your research hypothesis and your null hypothesis
H0: There are no differences in mean frequency of visits between the three types of medical insurance coverage.
H1: There are differences in the mean frequency of visits between the three types of medical insurance coverage.
Step-3: Describe the NULL distribution
The variance between the means expected by chance is 4.53.
Step-4: Compute your test statistic
The actual variance between the three means is 31.67.
The AVOVA source table is given as follows:
|
Source |
SS |
df |
MS |
F |
|
Between |
63.33 |
2 |
31.67 |
6.99 |
|
Within |
54.4 |
12 |
4.53 |
|
|
Total |
117.73 |
14 |
|
|
Step-5: Determine your critical value
Step-6: Compare the test stat to the critical value and make your decision
As F > critical value, we reject null H0
From our study, we conclude that there are differences in mean frequency of visits between the three types of medical insurance coverage, at significance level α = 0.05.
Solution-19:
Step-1: Select the appropriate test
We are testing 4 means so we shall use ANOVA.
Step-2: State your research hypothesis and your null hypothesis
H0: There are no differences in mean self-reported depression between the four groups.
H1: There are differences in the mean self-reported depression between the four groups.
Step-3: Describe the NULL distribution
The variance between the means expected by chance is 4.90.
Step-4: Compute your test statistic
The actual variance between the three means is 7.78.
The AVOVA source table is given as follows:
|
Source |
SS |
df |
MS |
F |
|
Between |
23.35 |
3 |
7.78 |
1.59 |
|
Within |
78.4 |
16 |
4.9 |
|
|
Total |
101.75 |
19 |
|
|
Step-5: Determine your critical value
Step-6: Compare the test stat to the critical value and make your decision
As F < critical value, we retain null H0
From our study, we conclude that there are no differences in mean self-reported depression between the four groups, at significance level α = 0.05.
Solution-23:
Step-1: Select the appropriate test
We are testing 3 means so we shall use ANOVA.
Step-2: State your research hypothesis and your null hypothesis
H0: There are no differences in mean driving test score between the three groups.
H1: There are differences in the mean driving test score between the three groups.
Step-3: Describe the NULL distribution
The variance between the means expected by chance is 2.27.
Step-4: Compute your test statistic
The actual variance between the three means is 25.87.
The AVOVA source table is given as follows:
|
Source |
SS |
df |
MS |
F |
|
Between |
51.73 |
2 |
25.87 |
11.41 |
|
Within |
27.2 |
12 |
2.27 |
|
|
Total |
78.93 |
14 |
|
|
Step-5: Determine your critical value
Step-6: Compare the test stat to the critical value and make your decision
As F > critical value, we reject null H0
From our study, we conclude that there are differences in mean driving test score between the three groups, at significance level α = 0.05.