Answer: [tex](3.0,\ 3.2)[/tex]
Step-by-step explanation:
Confidence interval for population mean is given by :-
[tex]\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex] (1)
, where [tex]\overline{x}[/tex] = Sample mean
[tex]z_{\alpha/2}[/tex]= Critical z-value
[tex]\sigma[/tex] = Population standard deviation.
n= Sample size.
As per given , we have
n= 610
[tex]\sigma=1.1[/tex]
[tex]\overline{x}=3.1[/tex]
Significance level for 85% confidence : [tex]\alpha=1-0.85=0.15[/tex]
By z-table critical two tailed z-value : [tex]z_{\alpha/2}=z_{0.075}=1.44[/tex]
Put all values in (1) , we get
[tex]3.1\pm 1.44\dfrac{1.1}{\sqrt{610}}[/tex]
[tex]3.1\pm 1.44\dfrac{1.1}{24.698}[/tex]
[tex]3.1\pm 0.064[/tex]
[tex]=(3.1-0.064,\ 3.1+0.064)=(3.036,\ 3.164)\approx(3.0,\ 3.2)[/tex]
Hence, the 85% confidence interval for the mean consumption of meat among people over age 40. = [tex](3.0,\ 3.2)[/tex]