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.
At the null hypothesis, it is tested if the mean is of 1600 hours, that is:
[tex]H_0: \mu = 1600[/tex]
At the alternative hypothesis, it is tested if the mean is different of 1600 hours, that is:
[tex]H_1: \mu \neq 1600[/tex]
It is given by:
[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]
The parameters are:
In this problem, the values of the parameters are:
[tex]\overline{x} = 1500, \mu = 1600, s = 120, n = 100[/tex]
Hence, the value of the test statistic is:
[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{1500 - 1600}{\frac{120}{\sqrt{100}}}[/tex]
[tex]t = -8.33[/tex]
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 [tex]t^{\ast} = 1.66[/tex].
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 https://brainly.com/question/13873630