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
[tex]H_{0}: p = p_{0} = 0.65[/tex] vs [tex]H_{1}: p\neq 0.65[/tex] (two-tailed alternative).
We have a large sample of size n = 500 and a point estimate for p is [tex]\hat{p}=0.68[/tex]. The test statistic is given by
[tex]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}}[/tex] where [tex]\sqrt{0.65(1-0.65)/500}[/tex] is the standard error of [tex]\hat{p}[/tex] when [tex]H_{0}[/tex] is true. The observed value is [tex]z_{0}[/tex] = 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.
