Anationwide study of american homeowners revealed that 65% have one or more lawn mowers. a lawn equipment manufacturer, located in omaha, feels the estimate is too low for households in omaha. find the p-value for a test of the claim that the proportion with lawn mowers in omaha is higher than 65%. among 497 randomly selected homes in omaha, 340 had one or more lawn mowers.

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 :

A  lawn equipment manufacturer, located in Omaha, feels the estimate is too low for households in Omaha.

Then, his hypothesis will be :-

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=

Test statistic for proportion :-

i.e.

Now, by using the standard normal table for z , we have

P-value (left tailed) =

{3 + i, 3 - i, 3, -1}

step-by-step explanation:

180-142=38 for the acute angle next to the entry

90+38=128

180-128=52

a=52

step-by-step explanation: