Use an app called “PhotoMath” just take a picture of it and it’ll show you the answer. :)
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 : ![p=0.65](https://tex.z-dn.net/?f=p%3D0.65)
A lawn equipment manufacturer, located in Omaha, feels the estimate is too low for households in Omaha.
Then, his hypothesis will be :-
![H_0:p=0.65\\\\ H_1: p](https://tex.z-dn.net/?f=H_0%3Ap%3D0.65%5C%5C%5C%5C%20H_1%3A%20p%3C0.65)
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=![\hat{p}=\dfrac{340}{497}\approx0.68](https://tex.z-dn.net/?f=%5Chat%7Bp%7D%3D%5Cdfrac%7B340%7D%7B497%7D%5Capprox0.68)
Test statistic for proportion :-
![z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7B%5Chat%7Bp%7D-p%7D%7B%5Csqrt%7B%5Cdfrac%7Bp%281-p%29%7D%7Bn%7D%7D%7D)
i.e. ![z=\dfrac{0.68-0.65}{\sqrt{\dfrac{0.65(1-0.65)}{497}}}\approx1.40](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7B0.68-0.65%7D%7B%5Csqrt%7B%5Cdfrac%7B0.65%281-0.65%29%7D%7B497%7D%7D%7D%5Capprox1.40)
Now, by using the standard normal table for z , we have
P-value (left tailed) = ![P(z](https://tex.z-dn.net/?f=P%28z%3C1.40%29%3D0.9192433)
Answer: these are cordnets
Step-by-step explanation: