Respuesta :
Answer:
[tex]t=\frac{7.43-8}{\frac{3.6}{\sqrt{64}}}=-1.27[/tex]
The degrees of freedom are given by:
[tex]df=n-1=64-1=63[/tex]
The p value can be calculated like this:
[tex]p_v =P(t_{(63)}<-1.27)=0.104[/tex]
For this case since the p value is higher than the significance level we have enough evidence to conclude that the true mean for the wait time is not significantly less than 8 minutes
Step-by-step explanation:
Information given
[tex]\bar X=7.43[/tex] represent the sampke mean in minutes
[tex]s=3.6[/tex] represent the sample standard deviation
[tex]n=64[/tex] sample size
[tex]\mu_o =8[/tex] represent the value to verify
[tex]\alpha=0.005[/tex] represent the significance level
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value
System of hypothesis
For this case we are trying to proof if inspection station advertises that the wait time for customers is less than 8 minutes, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 8[/tex]
Alternative hypothesis:[tex]\mu < 8[/tex]
Since the population deviation is not known the statistic can be calculated with:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
The statistic for this case is given by:
[tex]t=\frac{7.43-8}{\frac{3.6}{\sqrt{64}}}=-1.27[/tex]
The degrees of freedom are given by:
[tex]df=n-1=64-1=63[/tex]
The p value can be calculated like this:
[tex]p_v =P(t_{(63)}<-1.27)=0.104[/tex]
For this case since the p value is higher than the significance level we have enough evidence to conclude that the true mean for the wait time is not significantly less than 8 minutes