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8_murik_8 [283]
3 years ago
10

a nationwide study of american homeowners revealed that​ 65% own at least one lawn mower. a lawn equipment​ manufacturer, locate

d 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.
Mathematics
1 answer:
svetoff [14.1K]3 years ago
4 0

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  

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