In: Statistics and Probability
Chi-Square -> C2 = S[(fo – fe)2/fe]
| Observed frequency table | ||||
| Low | Average | High | Total | |
| Male | 20 | 40 | 20 | 80 |
| Female | 20 | 60 | 40 | 120 |
| Total | 40 | 100 | 60 | 200 |
| E1= RT*CT/GT= 80*44/200 =16 | ||||
| Expected frequency table | ||||
| Low | Average | High | Total | |
| Male | 16 | 40 | 24 | 80 |
| Female | 24 | 60 | 36 | 120 |
| Total | 40 | 100 | 60 | 200 |
| H0: Attitudes toward the police differ insignificantly by gender. | ||||
| H1: Attitudes toward the police differ significantly by gender. |
| CALCULATION TABLE | ||||
| fo | fe | fo-fe | (fo-fe)2/fe | |
| Male L | 20 | 16 | 4 | 1 |
| Female L | 20 | 24 | -4 | 0.66666667 |
| Male A | 40 | 40 | 0 | 0 |
| Female A | 60 | 60 | 0 | 0 |
| Male H | 20 | 24 | -4 | 0.66666667 |
| Female H | 40 | 36 | 4 | 0.44444444 |
| SUM = | 200 | 200 | 0 | 2.77777778 |
| df=(3-1)(2-1)=2 | fo=OBSERVED | |||
| level=0.05OR 5% | fe= EXPECTED |
Test statiatics Chi square
=2.778
Crtical value of
at 5% for 2df = 5.99
and The P-Value = 0.24935
Statistical conclusion : Test statiatics Chi square
=2.778 <
5.99
or he P-Value ( 0.24935) > 0.05
Fail to reject Ho
So we conclude that Attitudes toward the police differ insignificantly by gender.