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:

$H_0: \mu = 3$

At the alternative hypothesis, we test if there has been improvement, that is, the time is of less than 3 hours. Thus:

$H_1: \mu < 3$

We have the standard deviation for the population, thus, the z-distribution is used. The test statistic is given by:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

The parameters are:

$\overline{x}$ is the sample mean.
$\mu$ is the value tested at the null hypothesis.
$\sigma$ is the standard deviation of the population.
n is the sample size.

In this problem, the values of the parameters are: $\mu = 3, \sigma = 0.2, \overline{x} = 2.9, n = 15$ .

Then, the value of the test statistic is:

$z = \frac{2.9 - 3}{\frac{0.2}{\sqrt{15}}}$

$z = -1.94$

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 brainly.com/question/25413422

8 0

3 years ago

Three sisters each purchase two dresses and a purse the purse costs 26.00 if the grand total was 205.50 how much did each dress

Oliga [24]

3 sisters bought 2 x 3 = 6 dress and and 1 x 3 = 3 purses.
3 sisters bought 6 dresses and and 3 purses.

Find 3 purses:
1 purse = $26
3 purses = $26 x 3 = $78

Find 6 dresses:
Total = $205.50
6 dresses = $205.5 - $78
6 dresses = $127.50

Find 1 dress:
6 dresses = $127.50
6 dresses = $127.50 ÷ 6
1 dress = $21.25

Answer: $21.25

6 0

3 years ago