Question

	No Exercise	Exercise	Total
Campus Dormitory	32	58	90
On-Campus Apartment	74	106	180
Off –Campus Apartment	110	40	150
At Home	39	11	50
Total	255	215	470

Chi Square Test

1. State null and alternative hypothesis
2. Determine a= using a=.05
3. Using the table above, compute x2 to determine if there is a difference in living arrangement and exercise status.
4. Find the critical value
5. Reject or fail to reject
6. State conclusion

Answer 1

Chi-Square Independence test - Results

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: H0​: The two variables - Living arrangements and Exercise status are independent Ha​: The two variables - Living arrangements and Exercise status are dependent This corresponds to a Chi-Square test of independence. (2) Degrees of Freedom The number of degrees of freedom is df = (4 - 1) * (2 - 1) = 3 (3) Critical value and Rejection Region Based on the information provided, the significance level is α=0.05, the number of degrees of freedom is df = (4 - 1) * (2 - 1) = 3, so the critical value is 7.8147. Then the rejection region for this test becomes R={χ2:χ2>7.8147}. (4)Test Statistics The Chi-Squared statistic is computed as follows: (5)P-value The corresponding p-value for the test is p=Pr(χ2​>58.5666)=0 (6)The decision about the null hypothesis Since it is observed that χ2=58.5666>χ2_c​rit=7.8147, it is then concluded that the null hypothesis is rejected. (7)Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the two variables - Living arrangements and Exercise status are dependent, at the 0.05 significance level. Conditions: a. The sampling method is simple random sampling. b. The data in the cells should be counts/frequencies c. The levels (or categories) of the variables are mutually exclusive.