Respuesta :
Answer:
[tex]t=\frac{55-45}{\frac{15}{\sqrt{400}}}=13.33[/tex]
The degrees of freedom are given by:
[tex]df=n-1=400-1=399[/tex]
The p value for this case would be:
[tex]p_v =P(t_{(399)}>13.33) \approx 0[/tex]
Since the p value is very low and less than the significance level we have enough evidence to conclude that the true mean is higher than 45
Step-by-step explanation:
Information provided
[tex]\bar X=5[/tex] represent the mean for the average waiting
[tex]s=15[/tex] represent the sample standard deviation
[tex]n=400[/tex] sample size
[tex]\mu_o =45[/tex] represent the value to verify
[tex]\alpha=0.05[/tex] represent the significance level
t would represent the statistic
[tex]p_v[/tex] represent the p value
System of hypothesis
We want to test that complaints that people must wait more than 45 minutes, the system of hypothesis are:
Null hypothesis:[tex]\mu \leq 45[/tex]
Alternative hypothesis:[tex]\mu > 45[/tex]
The statistic for this case would be given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Replacing the info we got:
[tex]t=\frac{55-45}{\frac{15}{\sqrt{400}}}=13.33[/tex]
The degrees of freedom are given by:
[tex]df=n-1=400-1=399[/tex]
The p value for this case would be:
[tex]p_v =P(t_{(399)}>13.33) \approx 0[/tex]
Since the p value is very low and less than the significance level we have enough evidence to conclude that the true mean is higher than 45
