Question

A study reveals that 65% of homeowners in Omaha own a lawn mower. To test this claim, you form a sample of 500 homeowners in Omaha, of which, 68% owned a lawnmower. Is there enough evidence to reject the claim using a level of significance of 5%?

Answer 1

Answer:

There is no enough evidence to reject the claim using a level of significance of 5%

Step-by-step explanation:

Let p be the true proportion of homeowners in Omaha who own a lawnmower. We want to test  

$H_{0}: p = p_{0} = 0.65$ vs $H_{1}: p\neq 0.65$ (two-tailed alternative).

We have a large sample of size n = 500 and a point estimate for p is $\hat{p}=0.68$. The test statistic is given by

$Z=\frac{\hat{p}-p_{0}}{\sqrt{p_{0}(1-p_{0})/n}}=\frac{0.68-0.65}{\sqrt{0.65(1-0.65)/500}}$ where $\sqrt{0.65(1-0.65)/500}$ is the standard error of $\hat{p}$ when $H_{0}$ is true. The observed value is $z_{0}$ = 1.4064 which comes from a standard normal distribution. Because we want to use a level of significance of 5%, our rejection region is RR={z | z < -1.96 or z > 1.96} where -1.96 and 1.96 are the 2.5th and 97.5th quantiles respectively. Because 1.4064 does not belong to RR, we fail to reject the null hypothesis.