A research company desires to know the mean consumption of meat per week among people over age 29. A sample of 2092 people over age 29 was drawn and the mean meet consumption was 2.9 pounds. Assume that the standard deviation is known to be 1.4 pounds. Construct a 95% confidence interval for the mean consumption of meat among people over age 29. Round your answer to one decimal place.

The 95% confidence interval for the mean consumption of meat among people over age 29 is between 2.8 pounds and 3 pounds.

Step-by-step explanation:

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

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find the margin of error M as such

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

The lower end of the interval is the sample mean subtracted by M. So it is 2.9 - 0.1 = 2.8 pounds

The upper end of the interval is the sample mean added to M. So it is 2.9 + 0.1 = 3 pounds.

The 95% confidence interval for the mean consumption of meat among people over age 29 is between 2.8 pounds and 3 pounds.

30 feet

step-by-step explanation:

that would be 2 1/2 * 12

= 24 + 6

= 30 feet   (answer)