Respuesta :
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:
[tex]H_0: \mu = 25[/tex]
At the alternative hypothesis, we test if the mean has changed, that is, if it is different than 25 minutes.
[tex]H_1: \mu \neq 25[/tex]
We have the standard deviation for the sample, thus, the t-distribution is used. The value of the test statistic is:
[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]
In which:
- X 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.
In this problem:
- 25 is tested at the null hypothesis, thus [tex]\mu = 25[/tex].
- Sample mean of 26.2 minutes, thus [tex]X = 26.2[/tex].
- Sample of 51, thus [tex]n = 51[/tex].
- Variance of 16, thus [tex]s = \sqrt{16} = 4[/tex].
The value of the test statistic is:
[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{26.2 - 25}{\frac{4}{\sqrt{51}}}[/tex]
[tex]t = 2.14[/tex]
- 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 https://brainly.com/question/23777908
