Question

The US Census reported that it takes workers an average of 28 minutes to get home from work with standard deviation 5 minutes. A random sample of 10 workers in a large metropolitan area showed an average time to get home from work to be 32 minutes. Is this evidence that workers in the large city take longer than 28 minutes to get home from work.

Answer 1

Using the z-distribution, as we have the standard deviation for the population, to test the hypothesis, it is found that this is evidence that workers in the large city take longer than 28 minutes to get home from work.

What are the hypothesis tested?

At the null hypothesis, it is tested if the mean time is of 28 minutes, that is:

$H_0: \mu = 28$

At the alternative hypothesis, it is tested if the mean time is greater than 28 minutes, that is:

$H_1: \mu > 28$.

What is the test statistic?

The test statistic is given by:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

The parameters are:

• $\overline{x}$ is the sample mean.

• $\mu$ is the value tested at the null hypothesis.

• $\sigma$ is the standard deviation of the sample.

• n is the sample size.

In this problem, the values of the parameters are given by:

$\overline{x} = 32, \mu = 28, \sigma = 5, n = 10$

Hence:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{32 - 28}{\frac{5}{\sqrt{10}}}$

$z = 2.53$

What is the decision?

Considering that we have a right-tailed test, as we are testing if the mean is greater than a value, with a standard significance level of 0.05, the critical value is of $z^\ast = 1.645$.

Since the test statistic is greater than the critical value, it is found that this is evidence that workers in the large city take longer than 28 minutes to get home from work.

To learn more about the z-distribution, you can check brainly.com/question/26454209