Q:

A publisher reports that 65% of their readers own a laptop. A marketing executive wants to test the claim that the percentage is actually different from the reported percentage. A random sample of 340 found that 60% of the readers owned a laptop. Is there sufficient evidence at the 0.01 level to support the executive's claim? Determine the P-value of the test statistic.

Accepted Solution

A:
Answer with explanation:Let p be the population proportion of their readers own a laptop. .By considering the given information , we have [tex]H_0:p=0.65\\\\ H_a:p\neq0.65[/tex]Since [tex]H_a[/tex] is two-tailed , so the hypothesis test is a two-tailed test.Also, we have n= 340[tex]\hat{p}=0.60[/tex]z-statistic : [tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex] [tex]z=\dfrac{0.60-0.65}{\sqrt{\dfrac{0.65(1-0.65)}{340}}}=-1.933[/tex]P-value ( two- tailed test) : 2P(Z>|z|)2P(Z>|-1.933|)=2P(Z>1.933)=2(1-P(z<1.933))         [∵ P(Z>z)=1-P(Z<z)]=2(1-0.9734)= 2(0.0266)=0.0532Decision : Since p-value (0.0532) > significance level (0.01), so we are fail to reject the null hypothesis.[Null hypothesis gets rejected , when p-value < [tex]\alpha[/tex]]Thus , we conclude that we have do not have enough evidence to support the claim at 0.01 significance level to support the executive's claim that the percentage is actually different from the reported percentage.