Question

According to a study by the American Pet Food Dealers association, 63 percent of U.S. households own pets. A report is being prepared for an editorial in the San Francisco Chronicle. As a part of the editorial, a random sample of 300 households showed 202 own pets. Can we conclude that more than 63 percent of the population in San Francisco own pets? Use a .05 level of significance. Please show all work.

Answer 1

According to a study by the American Pet Food Dealers Association,63 percent of the US households own pets.

A random sample of 300 households showed that 202 of them own pets.

So,here we have to perform a one-sample test for population proportion.

Null hypothesis :- p=0.63

Alternate hypothesis :- p>0.63

p_s=202/300

_=0.6733

p_exp=0.63

standard error of p_exp

=sqrt(63*37/300)

=2.7

So,our test statistic

z

=(p_s-p_exp)/0.027

=4.33/2.7

=1.553

Now,this is a right tailed test.So,our p-value is

p

=P(Z>1.553)

which when calculated from the normal table becomes

p=0.1215

Now,as p>0.05

So,we can safely accept the null hypothesis.

So,at 0.05 level of significance,we cannot conclude that more than 63% of the households own pets.