Respuesta :
Answer:
0.0384 < 0.05, which means that we can conclude that the mean amount of time that college students spend in the shower is significantly different from 5 minutes.
Step-by-step explanation:
A facilities manager at a university reads in a research report that the mean amount of time spent in the shower by an adult is 5 minutes.
This means that the null hypothesis is [tex]H_{0} = 5[/tex]
He decides to collect data to see if the mean amount of time that college students spend in the shower is significantly different from 5 minutes.
This means that the alternate hypothesis is [tex]H_{a} \neq 5[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
Null hypothesis:
Tests [tex]H_{0} = 5[/tex], which means that [tex]\mu = 5[/tex]
In a sample of 8 students, he found the average time was 5.55 minutes and the standard deviation was 0.75 minutes.
This means, respectively, that [tex]n = 8, \mu = 5.55, \sigma = 0.75[/tex]
Test statistic:
The test statistic is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{5.55 - 5}{\frac{0.75}{\sqrt{8}}}[/tex]
[tex]z = 2.07[/tex]
Pvalue:
Since we are testing if the mean is different from a value, and z is positive. The pvalue is two multiplied by 1 subtracted by the pvalue of z = 2.07.
z = 2.07 has a pvalue of 0.9808
2*(1 - 0.9808) = 2*(0.0192) = 0.0384
Decision:
0.0384 < 0.05, which means that we can conclude that the mean amount of time that college students spend in the shower is significantly different from 5 minutes.
