Question

Suppose that textbook weights are normally distributed. You measure 33 textbooks' weights, and find they have a mean weight of 75 ounces. Assume the population standard deviation is 13.3 ounces. Based on this, construct a 99% confidence interval for the true population mean textbook weight. Round answers to 2 decimal places.

Answer 1

`75-2.58(13.3)/(√(33))=69.027`

`75+2.58(13.3)/(√(33))=80.793`

And the 95% confidence interval would be between (69.027;80.793)

Explanation:

Information given

`\bar X=75` represent the sample mean

`\mu` population mean (variable of interest)

`\sigma=13.3` represent the population standard deviation

n=33 represent the sample size

Confidence interval

The confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

The degrees of freedom are given by:

`df=n-1=33-1=32`

The Confidence level is 0.99 or 99%, the significance would be
`\alpha=0.01` and
`\alpha/2 =0.005`, the critical value for this case would be
`z_(\alpha/2)=2.58`

And replacing we got:

`75-2.58(13.3)/(√(33))=69.027`

`75+2.58(13.3)/(√(33))=80.793`

And the 95% confidence interval would be between (69.027;80.793)