Question

Question:

Answer 1

Using the t-distribution, as we have the standard deviation for the sample, it is found that since the absolute value of the test statistic is greater than the critical value for the two-tailed test, there is enough evidence to conclude that the sample does not come from a population whose mean is 1,600 hours.

What are the hypothesis tested?

At the null hypothesis, it is tested if the mean is of 1600 hours, that is:

$H_0: \mu = 1600$

At the alternative hypothesis, it is tested if the mean is different of 1600 hours, that is:

$H_1: \mu \neq 1600$

What is the test statistic?

It is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

• $\overline{x}$ is the sample mean.

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

• s is the standard deviation of the sample.

• n is the sample size.

In this problem, the values of the parameters are:

$\overline{x} = 1500, \mu = 1600, s = 120, n = 100$

Hence, the value of the test statistic is:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{1500 - 1600}{\frac{120}{\sqrt{100}}}$

$t = -8.33$

What is the decision rule?

Considering a two-tailed test, as we are testing if the mean is different of a value, with 100 - 1 = 99 df and a significance level of 0.01, the critical value is of $t^{\ast} = 1.66$.

Since the absolute value of the test statistic is greater than the critical value for the two-tailed test, there is enough evidence to conclude that the sample does not come from a population whose mean is 1,600 hours.

More can be learned about the t-distribution at brainly.com/question/13873630