In: Statistics and Probability
on the excel
Annual Income | No H.S Degree | H.S Degree | Some College | Total |
Low | 65 | 45 | 35 | 145 |
Medium | 43 | 66 | 52 | 161 |
High | 34 | 35 | 52 | 121 |
Total | 142 | 146 | 139 | 427 |
1) state the null hypothesis for this study.
2) check whether all chi-square assumptions are met
3) calculate chi-square test statistics for the hypothetical data shown above.
(1) calculate total column percentages.
(2) calculate expected frequencies for all cells.
(3) calculate chi-square for each cell using formula (Oi-Ei)2 Ei
(4) calculate chi-square statistic
4) Identify degree of freedom for the data
5) Identify critical value for the data
6) Calculate the P-value for the data
7) Draw a conclusion referring to test statistic and p-value
Solution:
Here, we have to use chi square test for independence of two categorical variables.
1) State the null hypothesis for this study.
Null hypothesis: H0: The annual income and educational degree are independent from each other.
Alternative hypothesis: Ha: The annual income and educational degree are dependent.
We assume level of significance = α = 0.05
2) Check whether all chi-square assumptions are met
Required assumptions are given as below:
Given data is a categorical data.
Two variables should consist of two or more categorical and independent groups.
3) Calculate chi-square test statistics for the hypothetical data shown above.
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 = 9.487729037
(by using Chi square table or excel)
Calculation tables for test statistic are given as below:
Observed Frequencies |
||||
Column variable |
||||
Row variable |
No Degree |
Degree |
Some college |
Total |
Low |
65 |
45 |
35 |
145 |
Medium |
43 |
66 |
52 |
161 |
High |
34 |
35 |
52 |
121 |
Total |
142 |
146 |
139 |
427 |
Expected Frequencies |
||||
Column variable |
||||
Row variable |
No Degree |
Degree |
Some college |
Total |
Low |
48.22014052 |
49.578454 |
47.20140515 |
145 |
Medium |
53.54098361 |
55.04918 |
52.40983607 |
161 |
High |
40.23887588 |
41.372365 |
39.38875878 |
121 |
Total |
142 |
146 |
139 |
427 |
Calculations |
||
(O - E) |
||
16.77986 |
-4.57845 |
-12.2014 |
-10.541 |
10.95082 |
-0.40984 |
-6.23888 |
-6.37237 |
12.61124 |
(O - E)^2/E |
||
5.83913 |
0.42281 |
3.154022 |
2.075276 |
2.178424 |
0.003205 |
0.967313 |
0.981502 |
4.037787 |
Test statistic = Chi square = ∑[(O – E)^2/E] = 19.65946815
P-value = 0.000582933
(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 annual income and educational degree are dependent.