In: Statistics and Probability
A researcher is interested to learn if there is a relationship between the level of interaction a women in her 20s has with her mother and her life satisfaction ranking. Below is a list of women who fit into each of four level of interaction. Conduct a One-Way ANOVA on the data to determine if a relationship exists. State whether or not a relationship exists and why or why not. Explain in as much detail as possible.
|
No Interaction |
Low Interaction |
Moderate Interaction |
High Interaction |
|
2 |
3 |
3 |
9 |
|
4 |
3 |
10 |
10 |
|
4 |
5 |
2 |
8 |
|
4 |
1 |
1 |
5 |
|
7 |
2 |
2 |
8 |
|
8 |
2 |
3 |
4 |
|
1 |
7 |
10 |
9 |
|
1 |
8 |
8 |
4 |
|
8 |
6 |
4 |
1 |
|
4 |
5 |
3 |
8 |
answer:
Hypotheses are:
H0: Mean interaction of the four groups are same.
Ha: Mean interaction of the four groups are not same.
Follwoing table shows the mean, sd, and total of all the four grades:
| G1 | G2 | G3 | G4 | |
| 2 | 3 | 3 | 9 | |
| 4 | 3 | 10 | 10 | |
| 4 | 5 | 2 | 8 | |
| 4 | 1 | 1 | 5 | |
| 7 | 2 | 2 | 8 | |
| 8 | 2 | 3 | 4 | |
| 1 | 7 | 10 | 9 | |
| 1 | 8 | 8 | 4 | |
| 8 | 6 | 4 | 1 | |
| 4 | 5 | 3 | 8 | |
| Total | 43 | 42 | 46 | 66 |
| Mean | 4.3 | 4.2 | 4.6 | 6.6 |
| SD | 2.626785 | 2.347576 | 3.405877 | 2.91357 |
Let G shows the grade. Follwoing table shows the calculation for ANOVA:
| G | G^2 | |
| 2 | 4 | |
| 4 | 16 | |
| 4 | 16 | |
| 4 | 16 | |
| 7 | 49 | |
| 8 | 64 | |
| 1 | 1 | |
| 1 | 1 | |
| 8 | 64 | |
| 4 | 16 | |
| 3 | 9 | |
| 3 | 9 | |
| 5 | 25 | |
| 1 | 1 | |
| 2 | 4 | |
| 2 | 4 | |
| 7 | 49 | |
| 8 | 64 | |
| 6 | 36 | |
| 5 | 25 | |
| 3 | 9 | |
| 10 | 100 | |
| 2 | 4 | |
| 1 | 1 | |
| 2 | 4 | |
| 3 | 9 | |
| 10 | 100 | |
| 8 | 64 | |
| 4 | 16 | |
| 3 | 9 | |
| 9 | 81 | |
| 10 | 100 | |
| 8 | 64 | |
| 5 | 25 | |
| 8 | 64 | |
| 4 | 16 | |
| 9 | 81 | |
| 4 | 16 | |
| 1 | 1 | |
| 8 | 64 | |
| Total | 197 | 1301 |
Here we have 
Now

Now

Now

Degree of freedom:
Since there are four groups so we have k = 4 therefore

And since total number of observations are 40 so N = 40 therefore




For F = 1.5703 and degree of freedom 3, 36 P-value of the test is 0.2134. Since P-value is greater than 0.05 so we fail to reject the null hypothesis. That is on the basis of sample evidence we cannot conclude that there is a relationship between the level of interaction a women in her 20s has with her mother and her life satisfaction ranking.
Follwoing is the completed ANOVA table:
| ANOVA | ||||||
| Source of Variation | SS | df | MS | F | P-value | F crit |
| Between Groups | 38.275 | 3 | 12.7583333 | 1.57025641 | 0.21343404 | 2.86626556 |
| Within Groups | 292.5 | 36 | 8.125 | |||
| Total | 330.775 | 39 |
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