In: Statistics and Probability
Becoming more interested in what students do with their free time, a researcher collects several different samples of students and also notes the gender of the student. The results are below. Test to see whether or not activity is independent of gender at an alpha level of .01.
Movie | Party | Dancing | Dinner | Tv | Reading | Other | |
Males | 65 | 82 | 81 | 24 | 30 | 41 | 27 |
Females | 12 | 55 | 30 | 12 | 30 | 48 | 33 |
State the hypothesis
H0: activity is independent of gender
H1: activity is not independent of gender
Observed Frequencies
B1 | B2 | B3 | B4 | B5 | B6 | B7 | Total | |
A1 | 65 | 82 | 81 | 24 | 30 | 41 | 27 | 350 |
A2 | 12 | 55 | 30 | 12 | 30 | 48 | 33 | 220 |
Total | 77 | 137 | 111 | 36 | 60 | 89 | 60 | 570 |
Expected Frequencies
B1 | B2 | B3 | B4 | B5 | B6 | B7 | Total | |
A1 | 47.2807 | 84.1228 | 68.1579 | 22.1053 | 36.8421 | 54.6491 | 36.8421 | 350 |
A2 | 29.7193 | 52.8772 | 42.8421 | 13.8947 | 23.1579 | 34.3509 | 23.1579 | 220 |
Total | 77 | 137 | 111 | 36 | 60 | 89 | 60 | 570 |
Compute Chi-square
Chi square = 42.9707
Compute the degrees of freedom (df).
df=(2-1)⋅(7-1)=6
P value
for 6 df, p(χ2≥42.9707)=0
Decision :
Since the p-value(0) < α(0.01)
we reject the null hypothesis H0.
Conclusion :
alpha = 0.01 L.O.S there is no enough evidennce to claim that activity is independent of gender .