Students in grade school were asked whether they believed Good Grades, Popularity, or Athletic Ability was the most important to them. The students responses are divided into three school locations: Rural, Suburban, or Urban. The resulting contingency table was produced from the frequencies of observations. Run a hypothesis test to determine if the goals of student are independent of the school location using α = 0.05.
Describe the dependencies if they exist.
School Area |
|||
Goals |
Rural |
Suburban |
Urban |
Grades |
57 |
87 |
24 |
Popular |
50 |
42 |
6 |
Sports |
42 |
22 |
5 |
Chi-square test of independence
Solution:
Here, we have to use chi square test for independence of two categorical variables.
Null hypothesis: H0: The goals of student are independent of the school location.
Alternative hypothesis: Ha: The goals of student are not independent of the school location.
We are given level of significance = α = 0.05
Test statistic formula is given as below:
Chi square = ∑[(O – E)^2/E]
Where, O is observed frequencies and E is expected frequencies.
E = row total * column total / Grand total
We are given
Number of rows = r = 3
Number of columns = c = 3
Degrees of freedom = df = (r – 1)*(c – 1) = 2*2 = 4
α = 0.05
Critical value = 9.487729
(by using Chi square table or excel)
Calculation tables for test statistic are given as below:
Observed Frequencies |
||||
Column variable |
||||
Row variable |
Rural |
Suburban |
Urban |
Total |
Grades |
57 |
87 |
24 |
168 |
Popular |
50 |
42 |
6 |
98 |
Sports |
42 |
22 |
5 |
69 |
Total |
149 |
151 |
35 |
335 |
Expected Frequencies |
||||
Column variable |
||||
Row variable |
Rural |
Suburban |
Urban |
Total |
Grades |
74.72239 |
75.72537313 |
17.55224 |
168 |
Popular |
43.58806 |
44.17313433 |
10.23881 |
98 |
Sports |
30.68955 |
31.10149254 |
7.208955 |
69 |
Total |
149 |
151 |
35 |
335 |
Calculations |
||
(O - E) |
||
-17.7224 |
11.27463 |
6.447761 |
6.41194 |
-2.17313 |
-4.23881 |
11.31045 |
-9.10149 |
-2.20896 |
(O - E)^2/E |
||
4.203332 |
1.678661 |
2.368565 |
0.943217 |
0.106909 |
1.754841 |
4.168397 |
2.663447 |
0.676864 |
Chi square = ∑[(O – E)^2/E] = 18.56423
P-value = 0.000957
(By using Chi square table or excel)
P-value < α = 0.05
So, we reject the null hypothesis
There is sufficient evidence to conclude that the goals of student are not independent of the school location.
There is insufficient evidence to conclude that the goals of student are independent of the school location.
Students in grade school were asked whether they believed Good Grades, Popularity, or Athletic Ability was...