Respuesta :
Testing the hypothesis using the z-distribution, it is found that since the p-value of the test is of 0.0262 < 0.05, it can be concluded that the new booklet is effective.
At the null hypothesis, we test if there has been no improvement, that is, the time is still of 3 hours. Hence:
[tex]H_0: \mu = 3[/tex]
At the alternative hypothesis, we test if there has been improvement, that is, the time is of less than 3 hours. Thus:
[tex]H_1: \mu < 3[/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 population.
- n is the sample size.
In this problem, the values of the parameters are: [tex]\mu = 3, \sigma = 0.2, \overline{x} = 2.9, n = 15[/tex].
Then, the value of the test statistic is:
[tex]z = \frac{2.9 - 3}{\frac{0.2}{\sqrt{15}}}[/tex]
[tex]z = -1.94[/tex]
The p-value of the test is the probability of finding a sample mean of 2.9 hours or less, which is the p-value of z = -1.94.
Looking at the z-table, z = -1.94 has a p-value of 0.0262.
Since the p-value of the test is of 0.0262 < 0.05, it can be concluded that the new booklet is effective.
A similar problem is given at https://brainly.com/question/25413422
