Respuesta :
Testing the hypothesis using the z-distribution, it is found that the statement is False.
At the null hypothesis, we test if the mean is still of 7.4 minutes, that is:
[tex]H_0: \mu = 7.4[/tex]
At the alternative hypothesis, we test if the mean is of less than 7.4 minutes, that is:
[tex]H_1: \mu < 7.4[/tex]
We have the standard deviation for the population, thus, the z-distribution is used. The test statistic is given by:
[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\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.
- [tex]\sigma[/tex] is the standard deviation of the sample.
- n is the sample size.
For this problem, the values of the parameters are as follows: [tex]\overline{x} = 7.1, \mu = 7.4, \sigma = 1.2, n = 36[/tex].
Hence, the value of the test statistic is:
[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{7.1 - 7.4}{\frac{1.2}{\sqrt{36}}}[/tex]
[tex]z = -1.5[/tex]
The p-value of the test is the probability of finding a sample mean of 7.1 or lower, which is the p-value of z = -1.5.
Looking at the z-table, z = -1.5 has a p-value of 0.0668.
Since the p-value of the test is 0.0668 > 0.05, there is not enough evidence to conclude that the time is lower, hence the statement is False.
A similar problem is given at https://brainly.com/question/16695704
