Respuesta :
Answer:
[tex]t=\frac{147.3-120}{\frac{50}{\sqrt{15}}}=2.115[/tex]
Now we can calculate the degrees of freedom
[tex]df=n-1=15-1=14[/tex]
If we find a critical value in the t distribution with 14 degrees of freedom who accumulates 0.05 of the area in the right we got [tex]t_{critc}= 1.761[/tex]
Since the calculated value is higher than the critical value we have enough evidence to conclude that the true mean is significantly higher than 120 minutes for the average time of part time jobs
Step-by-step explanation:
Information given
[tex]\bar X=147.3[/tex] represent the sample mean for the amount of time spent at part time jobs
[tex]s=50[/tex] represent the sample standard deviation
[tex]n=15[/tex] sample size
[tex]\mu_o =120[/tex] represent the value to check
[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 analyze if the true mean for the amount of time spent at part time jobs is higher than 120, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 120[/tex]
Alternative hypothesis:[tex]\mu > 120[/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)
Replacing the info given we got:
[tex]t=\frac{147.3-120}{\frac{50}{\sqrt{15}}}=2.115[/tex]
Now we can calculate the degrees of freedom
[tex]df=n-1=15-1=14[/tex]
If we find a critical value in the t distribution with 14 degrees of freedom who accumulates 0.05 of the area in the right we got [tex]t_{critc}= 1.761[/tex]
Since the calculated value is higher than the critical value we have enough evidence to conclude that the true mean is significantly higher than 120 minutes for the average time of part time jobs
