Respuesta :
Answer:
95% confidence interval for the mean consumption of meat among people over age 30 is [2.9 pounds , 3.1 pounds].
Step-by-step explanation:
We are given that a sample of 2092 people over age 30 was drawn and the mean meat consumption was 3 pounds. Assume that the population standard deviation is known to be 1.4 pounds.
Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;
P.Q. = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\bar X[/tex] = sample mean meat consumption = 3 pounds
s = population standard deviation = 1.4 pounds
n = sample of people = 2092
[tex]\mu[/tex] = population mean consumption of meat
Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.
So, 95% confidence interval for the population mean, [tex]\mu[/tex] is ;
P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level
of significance are -1.96 & 1.96}
P(-1.96 < [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < 1.96) = 0.95
P( [tex]-1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]-1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ) = 0.95
P( [tex]\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] , [tex]\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ]
= [ [tex]3-1.96 \times {\frac{1.4}{\sqrt{2092} } }[/tex] , [tex]3+1.96 \times {\frac{1.4}{\sqrt{2092} } }[/tex] ]
= [2.9 pounds , 3.1 pounds]
Therefore, 95% confidence interval for the mean consumption of meat among people over age 30 is [2.9 pounds , 3.1 pounds].
