Respuesta :
Answer:
[tex]t=\frac{15.5-18.7}{\frac{5.7}{\sqrt{11}}}=-1.862[/tex]
The degrees of freedom are given by:
[tex]df=n-1=11-1=10[/tex]
The p value would be given by:
[tex]p_v =2*P(t_{(10)}<-1.862)=0.0922[/tex]
We see that the p value is higher than the significance level so we don't have enough evidence to ocnclude that the true mean is different from 18.7 months in jail at 5% of significance.
Step-by-step explanation:
Information given
[tex]\bar X=15.5[/tex] represent the sample mean for the jail time
[tex]s=5.7[/tex] represent the sample standard deviation
[tex]n=11[/tex] sample size
[tex]\mu_o =18.7[/tex] represent the value that we want to compare
[tex]\alpha=0.05[/tex] represent the significance level
t would represent the statistic
[tex]p_v[/tex] represent the p value for the test
System of hypothesis
We want to check the hypothesis if the true mean for the jail time is equal to 18.7 or no, the system of hypothesis are:
Null hypothesis:[tex]\mu = 18.7[/tex]
Alternative hypothesis:[tex]\mu \neq 18.7[/tex]
Since we don't know the population deviation the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Rreplacing we got:
[tex]t=\frac{15.5-18.7}{\frac{5.7}{\sqrt{11}}}=-1.862[/tex]
The degrees of freedom are given by:
[tex]df=n-1=11-1=10[/tex]
The p value would be given by:
[tex]p_v =2*P(t_{(10)}<-1.862)=0.0922[/tex]
We see that the p value is higher than the significance level so we don't have enough evidence to ocnclude that the true mean is different from 18.7 months in jail at 5% of significance.