Respuesta :
Answer:
[tex]t=\frac{180-175}{\frac{4}{\sqrt{15}}}=4.84[/tex]
[tex]df=n-1=15-1=14[/tex]
[tex]p_v =P(t_{(14)}>4.84)=0.000131[/tex]
We got a very low value for the p value so then we have enough evidence to reject the null hypothesis at any significance level commonly used. And the best conclusion based on the possible options are:
None of the above are true
Step-by-step explanation:
Data given
[tex]\bar X=180[/tex] represent the sample mean
[tex]s=4[/tex] represent the sample standard deviation
[tex]n=15[/tex] sample size
[tex]\mu_o =175[/tex] represent the value that we want to test
[tex]\alpha=0.1[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
System of hypothesis
We need to conduct a hypothesis in order to check if the true mean is above the limit of 175 degrees, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 175[/tex]
Alternative hypothesis:[tex]\mu > 175[/tex]
The statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
And replacing we got:
We can replace in formula (1) the info given like this:
[tex]t=\frac{180-175}{\frac{4}{\sqrt{15}}}=4.84[/tex]
P-value
The degrees of freedom are given by:
[tex]df=n-1=15-1=14[/tex]
Since is a one sided test the p value would be:
[tex]p_v =P(t_{(14)}>4.84)=0.000131[/tex]
We got a very low value for the p value so then we have enough evidence to reject the null hypothesis at any significance level commonly used. And the best conclusion based on the possible options are:
None of the above are true