Question

A research company desires to know the mean consumption of meat per week among people over age 27. A sample of 1179 people over age 27 was drawn and the mean meat consumption was 1.5 pounds. Assume that the population standard deviation is known to be 1.2 pounds. Construct the 99% confidence interval for the mean consumption of meat among people over age 27. Round your answers to one decimal place.

Answer 1

The 99% confidence interval for the mean consumption of meat among people over age 27 is between 1.4 pounds and 1.6 pounds.

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1-0.99)/(2) = 0.005`

Now, we have to find z in the Ztable as such z has a pvalue of
`1-\alpha`.

So it is z with a pvalue of
`1-0.005 = 0.995`, so
`z = 2.575`

Now, find the margin of error M as such

`M = z*(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

`M = 2.575*(1.2)/(√(1179)) = 0.1`

The lower end of the interval is the sample mean subtracted by M. So it is 1.5 - 0.1 = 1.4 pounds

The upper end of the interval is the sample mean added to M. So it is 1.5 + 0.1 = 1.6 pounds

The 99% confidence interval for the mean consumption of meat among people over age 27 is between 1.4 pounds and 1.6 pounds.