Question

A marketing research company desires to know the mean consumption of meat per week among people over age 30. 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. Construct the 95% confidence interval for the mean consumption of meat among people over age 30. Round your answers to one decimal place.

Answer 1

95% confidence interval for the mean consumption of meat among people over age 30 is [2.9 pounds , 3.1 pounds].

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. =
`(\bar X-\mu)/((\sigma)/(√(n) ) )` ~ N(0,1)

where,
`\bar X` = sample mean meat consumption = 3 pounds

s = population standard deviation = 1.4 pounds

n = sample of people = 2092

`\mu` = 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,
`\mu` 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 <
`(\bar X-\mu)/((\sigma)/(√(n) ) )` < 1.96) = 0.95

P(
`-1.96 * {(\sigma)/(√(n) ) }` <
`{\bar X-\mu}` <
`-1.96 * {(\sigma)/(√(n) ) }` ) = 0.95

P(
`\bar X-1.96 * {(\sigma)/(√(n) ) }` <
`\mu` <
`\bar X+1.96 * {(\sigma)/(√(n) ) }` ) = 0.95

95% confidence interval for
`\mu` = [
`\bar X-1.96 * {(\sigma)/(√(n) ) }` ,
`\bar X+1.96 * {(\sigma)/(√(n) ) }` ]

= [
`3-1.96 * {(1.4)/(√(2092) ) }` ,
`3+1.96 * {(1.4)/(√(2092) ) }` ]

= [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].