Question

A food snack manufacturer samples 15 bags of pretzels off the assembly line and weighed their contents. If the sample mean is 9.9 and the sample standard deviation is 0.30, find the 95% confidence interval for the true mean.

Answer 1

Answer:

$9.9-2.14\frac{0.30}{\sqrt{15}}=9.734$    

$9.9+2.14\frac{0.30}{\sqrt{15}}=10.066$    

Step-by-step explanation:

Information given

$\bar X= 9.9$ represent the sample mean

$\mu$ population mean (variable of interest)

s=0.3 represent the sample standard deviation

n=15 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:

$df=n-1=15-1=14$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and the critical value wuld be $t_{\alpha/2}=2.14$

Now we have everything in order to replace into formula (1):

$9.9-2.14\frac{0.30}{\sqrt{15}}=9.734$    

$9.9+2.14\frac{0.30}{\sqrt{15}}=10.066$