Question

The director of the IRS has been flooded with complaints that people must wait more than 40 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 which they must wait before seeing an IRS representative. The average waiting time for the sample is 60 minutes with a standard deviation of 22 minutes.

(a) Is there overwhelming evidence to support the claim that the wait time to see an IRS representative is more than 40 minutes at a 0.100 significance level?

Answer 1

`t=(60-40)/((22)/(√(400)))=18.18`

The degrees of freedom are given by:

`df=n-1=400-1=399`

The p value for this case would be given by:

`p_v =P(t_((399))>18.18) \approx 0`

For this case since the p value is a very low value we can conclude that we have enough evidence to reject the null hypothesis and we can conclude that the true mean for this case is significantly higher than 40 minutes

Explanation:

Information provided

`\bar X=60` represent the sample mean

`s=22` represent the sample standard deviation

`n=400` sample size

`\mu_o =40` represent the value to verify

`\alpha=0.1` represent the significance level

t would represent the statistic

`p_v` represent the p value

Hypothesis to test

We want to verify if the true mean for this case is higher than 40 minutes, the system of hypothesis would be:

Null hypothesis:
`\mu \leq 40`

Alternative hypothesis:
`\mu > 40`

The statistic would be given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Replacing the info given we got:

`t=(60-40)/((22)/(√(400)))=18.18`

The degrees of freedom are given by:

`df=n-1=400-1=399`

The p value for this case would be given by:

`p_v =P(t_((399))>18.18) \approx 0`

For this case since the p value is a very low value we can conclude that we have enough evidence to reject the null hypothesis and we can conclude that the true mean for this case is significantly higher than 40 minutes