Question

You measure 46 textbooks' weights, and find they have a mean weight of 77 ounces. Assume the population is normally distributed with a standard deviation of 12.2 ounces. Based on this, construct a 99% confidence interval for the true population mean textbook weight. Give your answers correct to at least two decimal places. Lower Limit = Upper Limit =

Answer 1

Using the z-distribution, the 99% confidence interval for the true population mean textbook weight is (72.37, 81.63).

What is a z-distribution confidence interval?

The confidence interval is:

$\overline{x} \pm z\frac{\sigma}{\sqrt{n}}$

In which:

• $\overline{x}$ is the sample mean.

• z is the critical value.

• n is the sample size.

• $\sigma$ is the standard deviation for the population.

In this problem, we have a 99% confidence level, hence$\alpha = 0.99$, z is the value of Z that has a p-value of $\frac{1+0.99}{2} = 0.995$, so the critical value is z = 2.575.

The other parameters are given as follows:

$\overline{x} = 77, \sigma = 12.2, n = 46$

Hence the lower and upper bound of the interval are given, respectively, by:

• $\overline{x} - z\frac{\sigma}{\sqrt{n}} = 77 - 2.575\frac{12.2}{\sqrt{46}} = 72.37$

• $\overline{x} + z\frac{\sigma}{\sqrt{n}} = 77 + 2.575\frac{12.2}{\sqrt{46}} = 81.63$

More can be learned about the z-distribution at brainly.com/question/25890103

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