Question

The director of the IRS has been flooded with complaints that people must wait more than 45 minutes before seeing an IRS representative. To determine the validity of these complaints, the IRS randomly selects 400 people entering IRS offices across the country and records the times that they must wait before seeing an IRS representative. The average waiting time for the sample is 55 minutes with a standard deviation of 15 minutes.
Are the complaints substantiated with alpha = .05?

Answer 1

Answer:

$t=\frac{55-45}{\frac{15}{\sqrt{400}}}=13.33$    

The degrees of freedom are given by:

$df=n-1=400-1=399$  

The p value for this case would be:

$p_v =P(t_{(399)}>13.33) \approx 0$  

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

$\bar X=5$ represent the mean for the average waiting

$s=15$ represent the sample standard deviation

$n=400$ sample size  

$\mu_o =45$ represent the value to verify

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ 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:$\mu \leq 45$  

Alternative hypothesis:$\mu > 45$  

The statistic for this case would be given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

Replacing the info we got:

$t=\frac{55-45}{\frac{15}{\sqrt{400}}}=13.33$    

The degrees of freedom are given by:

$df=n-1=400-1=399$  

The p value for this case would be:

$p_v =P(t_{(399)}>13.33) \approx 0$  

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