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.