Answer to Question #148343 in Statistics and Probability for John

Question

A telephone company wants to advertise that more than 30% of all its customers have more than two telephones. to support this ad, the company selects a sample of 200 customers and finds that 72 have more than two telephones. does the evidence support the ad? Use a = 0.05

Answer 1

$H_0:P=0.3$

$H_1:P>0.3$

Test statistic$=\frac{\hat{P}-P}{\sqrt{\frac{P(1-P)}{n}}}$

$=\frac{0.36-0.3}{\sqrt{\frac{0.3×0.7}{200}}}=1.852$

Critical value=$Z_{0.05}=1.645$

Since the test statistic 1.852 is greater than the critical value 1.645, we reject the null hypothesis and conclude that there is enough evidence to support the claim that the more than 30% of its customers customers have more than two telephones.