Respuesta :
Using the t-distribution, it is found that since the test statistic is above the critical value for the left-tailed test, it is found that this advanced training seminar did not reduce response time, hence option C is correct.
At the null hypothesis, we test if the mean time has not been reduced, that is, it still is of 9 minutes, hence:
[tex]H_0: \mu = 9[/tex]
At the alternative hypothesis, we test if it has been reduced, that is, it is less than 9 minutes, hence:
[tex]H_1: \mu < 9[/tex]
We have the standard deviation for the sample, thus, 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.
For this problem, the values of the parameters are: [tex]\overline{x} = 8, \mu = 9, s = 4.2, n = 36[/tex]
Then, the value of the test statistic is:
[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{8 - 9}{\frac{4.2}{\sqrt{36}}}[/tex]
[tex]t = -1.43[/tex]
The critical value for a left-tailed hypothesis test with 35 df and a significance level of 0.05 is [tex]t^{\ast} = -1.69[/tex].
Since the test statistic is above the critical value for the left-tailed test, it is found that this advanced training seminar did not reduce response time, hence option C is correct.
A similar problem is given at https://brainly.com/question/13873630
