Question

SHOWICE 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 83 98 92 Cat 112 82 54 Various 46 52 58 None 59 55 41 What can be concluded at the a = 0.01 significance level? a. What is the correct statistical test to use? Homogeneity Paired t-test Goodness-of-Fit Independence b. What are the null and alternative hypotheses? HO: The distribution of pets is not the same for each living situation. Pet ownership and living situation are independent Pet ownership and living situation are dependent, The distribution of pets is the same for each living situation. H: Pet ownership and living situation are dependent. Pet ownership and living situation are independent. $ ) 4 5 6 7 8 + 11

Answer 1

a)

Independence

b)

Ho:

Pet ownership and living situation are independent.

Ha:

Pet ownership and living situation are dependent.

c)

Applying chi square test of independence:

Expected	Ei=row total*column total/grand total	Single	Family	Couple	Total
	Dog	98.4375	94.1719	80.3906	273
	Cat	89.4231	85.5481	73.0288	248
	Various	56.2500	53.8125	45.9375	156
	None	55.8894	53.4675	45.6430	155
	total	300	287	245	832
chi square    χ2	=(Oi-Ei)2/Ei	Single	Family	Couple	Total
	Dog	2.421	0.156	1.677	4.2531
	Cat	5.700	0.147	4.958	10.8055
	Various	1.868	0.061	3.167	5.0963
	None	0.173	0.044	0.472	0.6894
	total	10.1620	0.4077	10.2746	20.8442
test statistic X2=		20.844

d)

p value =

0.0020

from excel: chidist(20.8442,6)

e_)

p value is less than alpha

f)

reject the null

g)

There is sufficient evidence to conclude that pet ownership and living situation are dependent.