Question

A research company desires to know the mean consumption of milk per week among males over age 25. A sample of 715 males over age 25 was drawn and the mean milk consumption was 2.9 liters. Assume that the population standard deviation is known to be 1.2 liters. Construct the 90% confidence interval for the mean consumption of milk among males over age 25. Round your answers to one decimal place.

Answer 1

Answer:

The 90% confidence interval for the mean consumption of milk among males over age 25 is (2.8 lt, 3.0 lt).

Step-by-step explanation:

The (1 - α) % confidence interval for population mean is:

 $CI=\bar x\pm z_{\alpha /2}\ \frac{\sigma}{\sqrt{n}}$

The information provided is:

 $n=715\\\bar x=2.9\ \text{lt}\\\sigma=1.2\ \text{lt}$

Confidence level = 90%

Then, α = 10%

Compute the critical value of z for α = 10% as follows:

 $z_{\alpha/2}=z_{0.10/2}=z_{0.05}=1.645$

*Use a z-table.

Compute the 90% confidence interval for population mean as follows:

$CI=\bar x\pm z_{\alpha /2}\ \frac{\sigma}{\sqrt{n}}$

     $=2.9\pm 1.645\times \frac{1.2}{\sqrt{715}}\\\\=2.9\pm 0.074\\\\=(2.826, 2.974)\\\\\approx (2.8\ \text{lt}, 3.0\ \text{lt})$

Thus, the 90% confidence interval for the mean consumption of milk among males over age 25 is (2.8 lt, 3.0 lt).