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Sagot :
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:
[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]
What is the test statistic?
It is given by:
[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]
The parameters are:
- [tex]\overline{x}[/tex] is the sample mean.
- [tex]\mu[/tex] 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:
[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]
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 [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
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