In: Math
The 90 students in a statistics class are categorized by gender and by the year in school. The numbers are listed in the following table:
| Year in School | Freshman | Sophmore | Junior | Senior | 
| Gender | ||||
| Male | 1 | 4 | 8 | 17 | 
| Female | 23 | 17 | 13 | 7 | 
Test the null hypothesis that there is no association between the year in school and the gender using a 1% significance level. Be sure to specify the test statistic with degrees of freedom, the P-value or critical value, and your conclusion. Please no computer software answers! Thank you!
Solution:
We are given that: The 90 students in a statistics class are categorized by gender and by the year in school. The numbers are listed in the following table:
| Year in School | Freshman | Sophmore | Junior | Senior | 
| Gender | ||||
| Male | 1 | 4 | 8 | 17 | 
| Female | 23 | 17 | 13 | 7 | 
We have to test the null hypothesis that there is no association between the year in school and the gender using a 1% significance level.
Step 1) State H0 and H1:
H0: there is no association between the year in school and the gender
Vs
H1: there is association between the year in school and the gender.
Step 2) Test statistic:
Formula:

where Oij = Observed frequencies and Eij = Expected Frequencies

| Year in School | Freshman | Sophmore | Junior | Senior | Row Totals | |
|---|---|---|---|---|---|---|
| Gender | Male | 1 | 4 | 8 | 17 | R1=30 | 
| Female | 23 | 17 | 13 | 7 | R2=60 | |
| Column Totals | C1=24 | C2=21 | C3=21 | C4=24 | N= 90 | |









Thus we need to make following table:
| Oij | Eij | Oij^2 / Eij | 
|---|---|---|
| 1 | 8 | 0.125 | 
| 4 | 7 | 2.285714 | 
| 8 | 7 | 9.142857 | 
| 17 | 8 | 36.125 | 
| 23 | 16 | 33.0625 | 
| 17 | 14 | 20.64286 | 
| 13 | 14 | 12.07143 | 
| 7 | 16 | 3.0625 | 
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Thus Chi-square test statistic is:



Step 3) Chi-square critical value:
df = (R-1)x (C-1) = ( 2-1) x ( 4-1) = 1 x 3 = 3
df = 3
Level of significance = 1% = 0.01

Chi-square critical value = 11.345
Step 4) Decision rule: Reject H0, if Chi-square test statistic value > Chi-square critical value, otherwise we fail to reject H0.
Since Chi-square test statistic value = 
 > Chi-square critical value = 11.345, we reject H0.
Step 5) Conclusion: Since we have rejected H0, we conclude that there is association between the year in school and the gender using a 1% significance level.