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.