Question

a nationwide study of american homeowners revealed that​ 65% own at least one lawn mower. a lawn equipment​ manufacturer, located in​ charlotte, feels that this estimate is too low for households in charlotte. to support his​ claim, he randomly selects 497 homes in charlotte and finds that 340 had one or more lawn mowers. find the​ p-value for testing the claim that the proportion of homeowners owning lawn mowers in charlotte is higher than​ 65%. round your answer to three decimal places. when you calculate the sample proportion keep 4 decimal places.

Answer 1

Answer:

$z=\frac{0.6841 -0.65}{\sqrt{\frac{0.65(1-0.65)}{497}}}=1.5938$  

We have a right tailed test then the p value would be:  

$p_v =P(z>1.5938)=0.0555$  

Step-by-step explanation:

Information provided

n=497 represent the random sample of homes selected

X=340 represent the number of homes with one or more lawn

$\hat p=\frac{340}{497}=0.6841$ estimated proportion of homes with one or more lawn

$p_o=0.65$ is the value that we want to test

z would represent the statistic

$p_v$ represent the p value (variable of interest)  

System of hypothesis

We want to check if proportion of homeowners owning lawn mowers in charlotte is higher than​ 65%m so then the correct hypothesis are .:  

Null hypothesis:$p\leq 0.65$  

Alternative hypothesis:$p > 0.65$  

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

Replacing the info given we got:

$z=\frac{0.6841 -0.65}{\sqrt{\frac{0.65(1-0.65)}{497}}}=1.5938$  

We have a right tailed test then the p value would be:  

$p_v =P(z>1.5938)=0.0555$