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.

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

That is z with a pvalue of , so Z = 2.575.

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 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.

a) y = -3x +2

b) y = 1/2x -6

these 2 can be combined into one equation.

-3x +2 = .5x -6

8 = 3.5x

x = 8 / 3.5 = 2.2857142857

y = -3 * (8 / 3.5) +2

y = (-24 / 3.5) + 2

y = -4.8571428571

step-by-step explanation:

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