Question

A researcher is interested in investigating whether living situation and pet ownership are dependent. The table below shows the results of a survey. Frequencies of Living Situation and Pet Single Family Couple Dog 75 83 Cat 96 87 Various 38 54 33 None What can be concluded at the a-0.05 significance level? a. What is the correct statistical test to use? Independence Homogeneity Goodness-of-Fit Paired t-test b. What are the null and alternative hypotheses? HO: Pet ownership and living situation are independent. The distribution of pets is not the same for each living situation The distribution of pets is the same for each living situation Pet ownership and living situation are dependent

Answer 1

1)

SINGLE FAMILY COUPLE Total DOG 75 83 79 237 251 CAT 96 87 68 VARIOUS 38 53 48 139 173 NONE 54 49 70 Total 263 272 265 800

FAMILY Expected Values SINGLE COUPLE Total 263x237 800 272x237 800 265x237 800 DOG 77.914 - 80.58 78.506 237 263x251 800 265x251 800 272x251 800 = 83.144 CAT = 82.516 = 85.34 251 263x139 800 272x139 800 265x139 800 VARIOUS = 45.696 - 47.26 46.044 139 272x173 58.82 263x173 800 265x173 800 56.874 173 NONE 57.306 800 Total 272 265 800 263

COUPLE Squared Distances SINGLE FAMILY (83-80.58) 80.58 (75-77.914) (79-78.506)2 78.506 DOG - 0.109 = 0.073 0.003 77.914 (96-82.516) (68-83.144) 83.144 (87-85.34)2 CAT =2.203 0.032 2.758 82.516 85.34 (38-45.696)2 45.696 (53-47.26)2 47.26 (48-46.044) 46.044 VARIOUS 1.296 = 0.697 0.083 (54-56.874) 56.874 (70-57.306) 57.306 (49-58.82) 58.82 NONE 1.639 - 0.145 - 2.812

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

H0: Pet ownership and living situation are independent

Ha: Pet ownership and living situation are dependent

This corresponds to a Chi-Square test of independence.

Rejection Region

Based on the information provided, the significance level is α=0.05 , the number of degrees of freedom is df=(4−1)×(3−1)=6, so then the rejection region for this test is R={χ2:χ2 >12.592}.

Test Statistics

The Chi-Squared statistic is computed as follows:

п Е;)? (Оj Σ 0.1092.203 1.2960.145 +0.073 0.032 0.6971.639 0.003 2.758 0.083 2.812 11.852 Е, i=1

Decision about the null hypothesis

Since it is observed that χ2=11.852≤χc2=12.592, it is then concluded that the null hypothesis is not rejected.

Conclusion:It is concluded that the null hypothesis Ho is not rejected. Therefore, there is NOT enough evidence to claim that the two variables are dependent, at the 0.05 significance level.

The corresponding p-value for the test is p=Pr(χ2 >11.852)=0.0654.

P value > 0.05

Fail to reject H0.

There is insufficient evidence to conclude that  Pet ownership and living situation are dependent.

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