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8_murik_8 [283]

2 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.

1 answer:

svetoff [14.1K]2 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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