Respuesta :
Answer:
[tex]t=\frac{24-25}{\frac{2.2}{\sqrt{14}}}=-1.70[/tex]
The degrees of freedom are given by:
[tex]df=n-1=14-1=13[/tex]
The p value for this case would be given by:
[tex]p_v =P(t_{(13)}<-1.70)=0.0565[/tex]
Step-by-step explanation:
Information given
[tex]\bar X=24[/tex] represent the mean height for the sample
[tex]s=2.2[/tex] represent the sample standard deviation
[tex]n=14[/tex] sample size
[tex]\mu_o =25[/tex] represent the value that we want to test
t would represent the statistic
[tex]p_v[/tex] represent the p value for the test
Hypothesis to verify
We want to cehck if the true mean is lees than 25 mph, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 25[/tex]
Alternative hypothesis:[tex]\mu < 25[/tex]
The statistic would be given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Replacing the info given we got:
[tex]t=\frac{24-25}{\frac{2.2}{\sqrt{14}}}=-1.70[/tex]
The degrees of freedom are given by:
[tex]df=n-1=14-1=13[/tex]
The p value for this case would be given by:
[tex]p_v =P(t_{(13)}<-1.70)=0.0565[/tex]
