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)`

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)(\sigma)/(√(n))`

i.e.
`12.02\pm(1.96)(0.04)/(√(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)`