Question

In a simple random sample of 95 families, 70 had one or more pets at home. At the 1% level of significance, test the claim that the proportion of families with one or more pets differs from 60%.

Answer 1

Answer---->>

given that , In a simple random Sample of 95 families , 70 had one or more pets at home i.e.n = 95 , x = 70

Sample proportion = phat = x/n = 70/95 = 0.7368

Level of significance = $\alpha$ = 0.01

Claim :- The proportion of families with one or more pets differ from 60% i.e. population proportion 'p' is not equal to 0.60

Null and alternative hypothesis is

H0 :- p is equal to 0.60 Vs Ha :- p is not equal to 0.60

Test statistic :- under H0 ,

Z = (phat - p)/√((p*1-p)/n)

= ( 0.7368 - 0.60)/((0.6*0.4)/95)

Z= 2.736

Critical value = Z_$\alpha$/2 = Z_{​​​​​​0.01/2} = 2.58

Decision Rule :- Reject H0 if |Z| > critical value otherwise fail to. Reject H0.

Here , |Z| = 2.736 > critical value = 2.58

Therefore we Reject H0 at 1% level of significance .

Conclusion :-

There is enough evidence to conclude that the proportion of families with one or more pets differ from 60 % .