Respuesta :
Answer:
[tex]t=\frac{13.88-13.2}{\frac{3.14}{\sqrt{44}}}=1.436[/tex]
[tex]df=n-1=44-1=43[/tex]
[tex]p_v =P(t_{(43)}>1.436)=0.079[/tex]
We see that the p value i higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 13.2 *the value of 2008 ).
Step-by-step explanation:
Information given
[tex]\bar X=13.88[/tex] represent the sample mean
[tex]s=3.14[/tex] represent the sample standard deviation
[tex]n=44[/tex] sample size
[tex]\mu_o =13.2[/tex] represent the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test
Hypothesis to test
We want to conduct a hypothesis in order to check if the true mean has increased from 2008 , and the system of hypothesi are:
Null hypothesis:[tex]\mu \leq 13.2[/tex]
Alternative hypothesis:[tex]\mu > 13.2[/tex]
The statistic for this case is:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Calculating the statistic
Replacing the info given we got:
[tex]t=\frac{13.88-13.2}{\frac{3.14}{\sqrt{44}}}=1.436[/tex]
P-value
The degrees of freedom are:
[tex]df=n-1=44-1=43[/tex]
Since is a right tailed test the p value is:
[tex]p_v =P(t_{(43)}>1.436)=0.079[/tex]
Decision
We see that the p value i higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 13.2 *the value of 2008 ).
