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nydimaria [60]
2 years ago
7

A sports news station wanted to know whether people who live in the North or the South are bigger sports fans. For its study, 17

5 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.
Mathematics
1 answer:
Free_Kalibri [48]2 years ago
4 0

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

<h3>What are the mean and the standard error for each sample?</h3>

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

<h3>What is the mean and the standard error for the distribution of differences?</h3>

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

<h3>What is the confidence interval?</h3>

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

#SPJ1

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