Testing the hypothesis in this problem which is a two-tailed test, we can conclude that there is not sufficient evidence to conclude that the mean time of installation has changed, since the p-value of the test is 0.0364 > 0.01,

At the null hypothesis, we test if the mean is of 25 minutes, that is:

$H_0: \mu = 25$

At the alternative hypothesis, we test if the mean has changed, that is, if it is different than 25 minutes.

$H_1: \mu \neq 25$

We have the standard deviation for the sample, thus, the t-distribution is used. The value of the test statistic is:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

In which:

X is the sample mean.
$\mu$ is the value tested at the null hypothesis.
s is the standard deviation of the sample.
n is the sample size.

In this problem:

25 is tested at the null hypothesis, thus $\mu = 25$ .
Sample mean of 26.2 minutes, thus $X = 26.2$ .
Sample of 51, thus $n = 51$ .
Variance of 16, thus $s = \sqrt{16} = 4$ .

The value of the test statistic is:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{26.2 - 25}{\frac{4}{\sqrt{51}}}$

$t = 2.14$

The p-value of the test is found using a two-tailed test(test if the mean is different of a value), with t = 2.14 and 50 degrees of freedom.

Using a t-distribution calculator, this p-value is of 0.0364.

Since the p-value of the test is 0.0364 > 0.01, there is not sufficient evidence to conclude that the mean time of installation has changed.

A similar problem is given at brainly.com/question/23777908

3 0

3 years ago

A. 4.9 b. 3.7 c. 10.9

stepan [7]

B because the equation

5 0

3 years ago

What is the surface area of the figure shown?

Oduvanchick [21]

It has to be 740 ft2

3 0

3 years ago

At 3:15 p.m a train from New York to Washington D.C.is traveling at 60 miles per hour (mph). You must be in Washington D.C. by 5

xenn [34]

Answer:

they would arrive at 4:15 so they will not make it in time 60 mins is 1 hour so that is the answer

3 0

3 years ago