Answer: [tex](12.007\ , \ 12.033)[/tex]
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 : [tex]\alpha: 1-0.95=0.05[/tex]
Using Standard normal table , Critical value : [tex]z_{\alpha/2}=1.96[/tex]
Sample mean : [tex]\overline{x}=12.02[/tex]
Standard deviation: [tex]\sigma= 0.04[/tex]
The confidence interval for population means is given by :-
[tex]\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]
i.e.[tex]12.02\pm(1.96)\dfrac{0.04}{\sqrt{36}}[/tex]
[tex]\approx12.02\pm0.013=(12.02-0.013,\ 12.02+0.013)\\\\=(12.007\ , \ 12.033)[/tex]
Hence, the 95 percent confidence interval for the fill rate = [tex](12.007\ , \ 12.033)[/tex]
