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

Answer:

$75-2.58\frac{13.3}{\sqrt{33}}=69.027$    

$75+2.58\frac{13.3}{\sqrt{33}}=80.793$    

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

Step-by-step 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:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (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:

$75-2.58\frac{13.3}{\sqrt{33}}=69.027$    

$75+2.58\frac{13.3}{\sqrt{33}}=80.793$    

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