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A survey of 76 randomly selected homeowners find that they spend amount of $68 per month on home maintenance. Construct a 99% confidence interval for the mean amount of money spent per month at home maintenance by all homeowners. Assume that the population standard deviation is $15 per month. Round to the nearest cent

A survey of 76 randomly selected homeowners find that they spend amount of $68 per-example-1

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Step-by-step explanation:

We will use the following formula to construct the confidence interval:


\begin{gathered} CI\text{ = }\bar{x}\text{ }\pm\text{ MSE} \\ We\text{ are told that }\bar{x}\text{ = \$68 per month} \end{gathered}
\begin{gathered} MSE\text{ = z *}(\sigma)/(√(n)) \\ At\text{ a 99\% level of confidence, z = 2.576} \\ We\text{ are given that }\sigma\text{ = \$15 and n = 76} \\ MSE\text{ = 2.576 *}(15)/(√(76)) \\ MSE\text{ = 4.4323....} \end{gathered}

Answer:

CI = ($63.57 ; $72.43)

Lower endpoint: $63.57

Upper endpoint: $72.43

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