Question

A marketing research company desires to know the mean consumption of milk per week among people over age 32. A sample of 440 people over age 32 was drawn and the mean milk consumption was 3.4 liters. Assume that the population standard deviation is known to be 0.8 liters. Construct the 98% confidence interval for the mean consumption of milk among people over age 32. Round your answers to one decimal place.

Answer 1

Answer:

The 98% confidence interval for the mean consumption of milk among people over age 32 is between 3.3 and 3.5 liters.

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:

$\alpha = \frac{1 - 0.98}{2} = 0.01$

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 $1 - 0.01 = 0.99$, so Z = 2.327.

Now, find the margin of error 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.

$M = 2.327\frac{0.8}{\sqrt{440}} = 0.1$

The lower end of the interval is the sample mean subtracted by M. So it is 3.4 - 0.1 = 3.3 liters

The upper end of the interval is the sample mean added to M. So it is 3.4 + 0.1 = 3.5 liters

The 98% confidence interval for the mean consumption of milk among people over age 32 is between 3.3 and 3.5 liters.