Question

You are working at a Silver Mountain Tea bottling plant! The bottles of iced tea are supposed to be filled with 12 ounces of liquid. A recent sample of 36 bottles returned an average fill level of 12.02 ounces. Historically, the process has a standard deviation of .04 ounces. Create a 95 percent confidence interval for the fill rate!

Answer 1

Answer: $(12.007\ , \ 12.033)$

Step-by-step explanation:

Given : Sample size : n= 36 , which is a large sample (n>30) so the test applied here is z-test.

Significance level : $\alpha: 1-0.95=0.05$

Using Standard normal table , Critical value : $z_{\alpha/2}=1.96$

Sample mean : $\overline{x}=12.02$

Standard deviation: $\sigma= 0.04$

The confidence interval for population means is given by :-

$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

i.e.$12.02\pm(1.96)\dfrac{0.04}{\sqrt{36}}$

$\approx12.02\pm0.013=(12.02-0.013,\ 12.02+0.013)\\\\=(12.007\ , \ 12.033)$

Hence, the 95 percent confidence interval for the fill rate = $(12.007\ , \ 12.033)$