Question

A marketing research company desires to know the mean consumption of meat per week among people over age 3131. A sample of 369369 people over age 3131 was drawn and the mean meat consumption was 2.42.4 pounds. Assume that the population standard deviation is known to be 0.70.7 pounds. Construct the 85%85% confidence interval for the mean consumption of meat among people over age 3131. Round your answers to one decimal place.

Answer 1

Answer:

The 85% confidence interval for the mean consumption of meat among people over age 31 is between 2.3 and 2.5 pounds

Step-by-step explanation:

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

$\alpha = \frac{1-0.85}{2} = 0.075$

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 $1-0.075 = 0.925$, so $z = 1.44$

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

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

$M = 1.44\frac{0.7}{\sqrt{369}} = 0.1$

The lower end of the interval is the sample mean subtracted by M. So it is 2.4 - 0.1 = 2.3 pounds

The upper end of the interval is the sample mean added to M. So it is 2.4 + 0.1 =  2.5 pounds

The 85% confidence interval for the mean consumption of meat among people over age 31 is between 2.3 and 2.5 pounds