Question

A research company desires to know the mean consumption of meat per week among males over age 43. A sample of 1384 males over age 43 was drawn and the mean meat consumption was 3 pounds. Assume that the population standard deviation is known to be 1.3 pounds. Construct the 99% confidence interval for the mean consumption of meat among males over age 43. Round your answers to one decimal place.

Answer 1

The 99% confidence interval for the mean consumption of meat among males over age 43 is between 2.9 pounds and 3.1 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`.

That 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(1.3)/(√(1384)) = 0.1`

The lower end of the interval is the sample mean subtracted by M. So it is 3 - 0.1 = 2.9 pounds

The upper end of the interval is the sample mean added to M. So it is 3 + 0.01 = 3.1 pounds

The 99% confidence interval for the mean consumption of meat among males over age 43 is between 2.9 pounds and 3.1 pounds.