Question

You measure 32 backpacks' weights, and find they have a mean weight of 53 ounces. Assume the populationstandard deviation is 4.7 ounces. Based on this, what is the maximal margin of error associated with a 99%confidence interval for the true population mean backpack weight.Give your answer as a decimal, to two places+Ounces

Answer 1

Given the mean weight and population standard deviation;

`\begin{gathered} n=32 \\ \sigma=4.7 \\ \bar{x}=53 \end{gathered}`

The margin of error also called the maximum error of the estimate is given as;

`E=Z_{(\alpha)/(2)}*\frac{\sigma}{\sqrt[]{n}}`

The z-value of 99% confidence interval is 2.576.

So,

`\begin{gathered} E=2.576*\frac{4.7}{\sqrt[]{32}} \\ E=2.576*0.8309 \\ E=2.1403 \\ E\cong2.14ounces\ldots\ldots\ldots\ldots..(two\text{ decimal places)} \end{gathered}`