Question

A sports news station wanted to know whether people who live in the North or the South are bigger sports fans. For its study, 175 randomly selected Southerners were surveyed and found to watch a mean of 3.8 hours of sports per week. In the North, 152 randomly selected people were surveyed and found to watch a mean of 4.8 hours of sports per week. Find a 95% confidence interval for the true difference between the mean numbers of hours of sports watched per week for the two regions if the South has a population standard deviation of 1.7 hours per week and the North has a population standard deviation of 1.2 hours per week. Let Population 1 be people who live in the South and Population 2 be people who live in the North. Round the endpoints of the interval to one decimal place, if necessary.

Answer 1

Using the z-distribution, it is found that the 95% confidence interval for the difference is (-1.3, -0.7).

What are the mean and the standard error for each sample?

Considering the data given:

$\mu_S = 3.8, n = 175, s_S = \frac{1.7}{\sqrt{175}} = 0.1285$

$\mu_N = 4.8, n = 152, s_N = \frac{1.2}{\sqrt{152}} = 0.0973$

What is the mean and the standard error for the distribution of differences?

The mean is the subtraction of the means, hence:

$\overline{x} = \mu_S - \mu_N = 3.8 - 4.8 = -1$

The standard error is the square root of the sum of the variances of each sample, hence:

$s = \sqrt{s_S^2 + s_N^2} = \sqrt{0.1285^2 + 0.0973^2} = 0.1612$

What is the confidence interval?

It is given by:

$\overline{x} \pm zs$

We have a 95% confidence interval, hence the critical value is of z = 1.96.

Then, the bounds of the interval are given as follows:

• $\overline{x} - zs = -1 - 1.96(0.1612) = -1.3$

• $\overline{x} + zs = -1 + 1.96(0.1612) = -0.7$

More can be learned about the z-distribution at brainly.com/question/25890103

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