The hypotheses tested are,
H0: Gender of a professor is independent of the department.
Against
H1: gender of a professor is not independent of the department.
We first determine the expected count for each cell using the formula below,
Eij=(ri∗cj)/n, i=1,2 & j=1,2,3,4,5, where ri is the corresponding row total for each cell and cj is the corresponding column total for each cell. n=258 is the sample size.
The expected counts are as follows,
E11=(r1∗c1)/n=(63∗205)/258=50.06
E12=(r1∗c2)/n=(81∗205)/258=64.36
E13=(r1∗c3)/n=(46∗205)/258=36.55
E14=(r1∗c4)/n=(23∗205)/258=18.28
E15=(r1∗c5)/n=(45∗205)/258=35.76
E21=(r2∗c1)/n=(63∗53)/258=12.94
E22=(r2∗c2)/n=(81∗53)/258=16.64
E23=(r2∗c3)/n=(46∗53)/258=9.45
E24=(r2∗c4)/n=(23∗53)/258=4.72
E25=(r2∗c5)/n=(45∗53)/258=9.24
Next is to determine the test statistic given as,
χ∗2=i=1∑2j=1∑5(Oij−Eij)2/Eij
Now,
χ∗2=(58−50.06)2/50.05+(74−64.36)2/64.36+(36−36.55)2/36.55+(12−18.28)2/18.28+(25−35.76)2/35.76+(5−12.94)2/12.94+(7−16.64)2/16.64+(10−9.45)2/9.45+(11−4.72)2/4.72+(20−9.24)2/9.24=39.4412(4dp)
χ∗2 is compared with the table value at α level of significance with (r−1)∗(c−1)=(2−1)∗(5−1)=1∗4=4 degrees of freedom.
The table value is χα=0.01,42=13.2767 and the null hypothesis is rejected if, χ∗2>χ0.01,42.
Since χ∗2=39.4412>χ0.01,42=13.2767, we reject the null hypothesis and conclude that there is no sufficient evidence to show that the gender of a professor is independent of the department at 1% significance level.
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