Answer: 0.9192433
Step-by-step explanation:
Given : A nationwide study of American homeowners revealed that 65% have one or more lawn mowers.
i.e. Population proportion : [tex]p=0.65[/tex]
A lawn equipment manufacturer, located in Omaha, feels the estimate is too low for households in Omaha.
Then, his hypothesis will be :-
[tex]H_0:p=0.65\\\\ H_1: p<0.65[/tex]
Since, the alternative hypothesis is left tailed , so the test is a left-tailed test.
Sample size : n= 497
The sample proportion of people ad one or more lawn mowers=[tex]\hat{p}=\dfrac{340}{497}\approx0.68[/tex]
Test statistic for proportion :-
[tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]
i.e. [tex]z=\dfrac{0.68-0.65}{\sqrt{\dfrac{0.65(1-0.65)}{497}}}\approx1.40[/tex]
Now, by using the standard normal table for z , we have
P-value (left tailed) = [tex]P(z<1.40)=0.9192433[/tex]
