Question

Suppose a veterinarian wants to compare the proportion of cat owners who are single to the

proportion of dog owners who are single. Of the pet owners that visit any of the veterinarian

clinics in her city regularly, she randomly surveys 124 cat owners and 151 dog owners and asks

each of them if they are single. Her findings are 80 of the cat owners are not single and 102 of

the dog owners are not single. Is there any evidence to suggest that it is more likely for a cat

owner to be single?

(a) State the null and alternative hypotheses.

(b) Find the test statistic.

(c) Find the P-value.

(d) Use =0.05, state the conclusion (reject or fail to reject the null hypothesis).

(e) Interpret the conclusion in context.

(f) Find and interpret the 95% confidence interval for the difference between proportion of cat

owners who are single and proportion of dog owners who are single.

Answer 1

To Test :-

H0 :- P1 = P2

H1 :- P1 > P2

Test Statistic :-

is the pooled estimate of the proportion P
= ( x1 + x2) / ( n1 + n2)
= ( 80 + 102 ) / ( 124 + 151 )
= 0.6618

Z = -0.53

Test Criteria :-
Reject null hypothesis if

= -0.53 < 1.64, hence we fail to reject the null hypothesis
Conclusion :- We Fail to Reject H0

Decision based on P value
P value = P ( Z > -0.53 )
P value = 0.7019
Reject null hypothesis if P value <
Since P value = 0.7019 > 0.05, hence we fail to reject the null hypothesis
Conclusion :- We Fail to Reject H0

There is insufficient evidence to support the claim that  it is more likely for a cat owner to be single.

n1 = 124
n2 = 151


Lower Limit =
upper Limit =
95% Confidence interval is ( -0.0823 , 0.1429 )

( -0.0823 < ( P2 - P1 ) < 0.1429 )