Respuesta :
Using the t-distribution, it is found that since the test statistic is greater than the critical value for the right-tailed test, it is found that there is enough evidence to conclude that the students' claim that they work more than 19 hours is correct.
At the null hypothesis, it is tested if they spend on average 19 hours a week working, that is:
[tex]H_0: \mu = 19[/tex]
At the alternative hypothesis, it is tested if they spend more than 19 hours a week working, that is:
[tex]H_1: \mu > 19[/tex]
We have the standard deviation for the sample, hence, the t-distribution is used.
The test statistic 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.
Searching the problem on the internet, it is found that the values of the parameters are:
[tex]\overline{x} = 24.2, \mu = 19, s = 12.59, n = 25[/tex]
Hence, the value of the test statistic is:
[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{24.2 - 19}{\frac{12.59}{\sqrt{25}}}[/tex]
[tex]t = 2.07[/tex]
The critical value for a right-tailed test, as we are testing if the mean is greater than a value, with a significance level of 0.05 and 25 - 1 = 24 df is of [tex]t^{\ast} = 1.71[/tex]
Since the test statistic is greater than the critical value for the right-tailed test, it is found that there is enough evidence to conclude that the students' claim that they work more than 19 hours is correct.
To learn more about the t-distribution, you can take a look at https://brainly.com/question/13873630
