Question

In: Statistics and Probability

No Exercise Exercise Total Campus Dormitory 32 58 90 On-Campus Apartment 74 106 180 Off –Campus...

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

You have now decided to collapse your data into on-campus and off-campus living (use data from #7) 2x2 table. Determine if there is a difference between living on-campus or off-campus and exercise status (yes exercise, no exercise).


1. state the null and alternative hypothesis

2. determine a

3. select chi square tool

4. Fine the critical value

5. reject or fail to reject

6. State the conclusion

Solutions

Expert Solution

Assume the level of significance to be 5%.

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 on-campus or off-campus and exercise status are independent
Ha​: The two variables - living on-campus or off-campus 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 on-campus or off-campus 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.

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