Question

In: Statistics and Probability

A nationwide study of American homeowners revealed that 64% have one or more lawn mowers. A...

A nationwide study of American homeowners revealed that 64% have one or more lawn mowers. A lawn equipment manufacturer, located in Omaha, feels the estimate is too low for households in Omaha.

1) Can the value 0.64 be rejected if a survey of 490 homes in Omaha yields 331 with one or more lawn mowers?

2) Find the confidence interval of the proportion of all the one hwo have the lawn mowers.

3) Check the assumption used in the 1)

Expert Solution

Given:

Ho: P = 0.64

Ha: P > 0.64

Sample proportion () = X/n = 331/490 = 0.68

Test statistic:

P-value: 0.0508 ..........From standard Normal Table

P-value > , i.e. 0.0508 > 0.05, That is Reject Ho at 5% level of significance.

There is Not sufficient evidence to warrant rejection of the claim that the proportion with lawn mowers on Omaha is 0.64

b) 95% Confidence Interval:

Where,

Z_/2 = Z _0.05/2 = 1.96 ..........From standard Normal Table

Therefore,

0.67 < P < 0.68

We are 95% Confident that the population proportion (P) lies in that interval.

orchestra answered 2 years ago

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