In: Statistics and Probability
Religion A |
Religion B |
Religion C |
Religion D |
3 |
2 |
6 |
3 |
3 |
3 |
1 |
3 |
1 |
3 |
2 |
2 |
2 |
5 |
4 |
1 |
4 |
1 |
3 |
2 |
2 |
1 |
1 |
2 |
3 |
2 |
2 |
4 |
Solution:
We are given the following data set:
Religion A |
Religion B |
Religion C |
Religion D |
3 |
2 |
6 |
3 |
3 |
3 |
1 |
3 |
1 |
3 |
2 |
2 |
2 |
5 |
4 |
1 |
4 |
1 |
3 |
2 |
2 |
1 |
1 |
2 |
3 |
2 |
2 |
4 |
Part(a)
We have to test whether the mean number of children differs by religion.
Our Null hypothesis is, H0: The mean number of children does not differ by religion.
The Alternative hypothesis is, H1: The mean number of children differs by religion.
So, we will test the mentioned hypothesis using One-Way Single-factor ANOVA.
So, we perform One-Way Single-factor ANOVA on the given data set and obtain the following output:
ANOVA: Single Factor |
||||
SUMMARY |
||||
Groups |
Count |
Sum |
Average |
Variance |
Religion A |
7 |
18 |
2.571429 |
0.952381 |
Religion B |
7 |
17 |
2.428571 |
1.952381 |
Religion C |
7 |
19 |
2.714286 |
3.238095 |
Religion D |
7 |
17 |
2.428571 |
0.952381 |
ANOVA |
||||||
Source of Variation |
SS |
df |
MS |
F |
P-value |
F crit |
Between Groups |
0.392857 |
3 |
0.130952 |
0.073826 |
0.973461 |
3.008787 |
Within Groups |
42.57143 |
24 |
1.77381 |
|||
Total |
42.96429 |
27 |
We see that the value of the F statistics is, F = 0.073826.
The critical F value at 0.05 level of significance is, F crit = 3.008787.
So, F = 0.073826 < 3.008787 = F crit.
Hence, we fail to reject the null hypothesis, " H0: The mean number of children does not differ by religion " at 0.05 level of significance.
Part(b)
The Between Groups sum of squares is = 0.392857.
The Within Groups sum of squares is = 42.57143.
The total sum of squares is = 42.96429.
The coefficient of determination is =
So, it tells that only 0.9144% of the variability is explained by the model.