Question

Exhibit 1

The average wait time on the phone for taxpayers calling the IRS in 2016 was 1,686 seconds. Suppose that the IRS made operational changes in an effort to reduce wait times. To test the effectiveness of these changes, a random sample of 60 phone calls was selected and the wait time of each call was recorded. The average wait time in the sample was calculated to be 1,619 seconds. Assume the population standard deviation for the wait time is 160 seconds. Let the significance level to be 2%.

1. Refer to Exhibit 1. How do you state the hypotheses?

2. Refer to Exhibit 1. What is the test statistics?

3. Refer to Exhibit 1. What is the p-value?

4. Refer to Exhibit 1. What is the critical value?

5. Refer to Exhibit 1. What is your conclusion?

Exhibit 2

According to the 2018 Majoring in Money report by the government-lending institution Sallie Mae, 18- to 20-year-old college students have an average credit card balance of $619. You want to find out if this number is lower now and collect data for a random sample of 36 college students. Their average credit card debt was found to be $519 with a sample standard deviation of $269.26. Using α=0.03, does this sample provide enough evidence to support your hypothesis stated above?

6. Refer to Exhibit 2. What is the p-value of your test?

7. Refer to Exhibit 2. Using the critical value approach, how do you assess the null hypothesis?

8. Refer to Exhibit 2. What is your conclusion?

9. Refer to Exhibit 2. Explain in your own words how Type I and Type II errors can occur in this hypothesis test.

10. Refer to Exhibit 2.  Does changing the value of α from 0.03 to 0.01 affect your conclusion? Why or why not?

Answer 1

Exhibit 1

The average wait time on the phone for taxpayers calling the IRS in 2016 was 1,686 seconds. We are testing whether now the operation has been able to reduce it or not. So we are testing for a left sided one tailed test.

random sample of 60 phone calls

Sample Average = 1,619 seconds

population standard deviation = 160 seconds. since we have population SD we will use z-test

Let the significance level to be =2%.

1. Refer to Exhibit 1. How do you state the hypotheses?

The population avg. waiting time is 1686.

The population avg. waiting time is less than 1686.

2. Refer to Exhibit 1. What is the test statistics?

Test Stat =

where thenull mean = 1686

test Stat =

3. Refer to Exhibit 1. What is the p-value?

p-value = P( Z > |Test Stat |)

= P( Z > 3.24)

= 1 - P( Z<3.24)

= 1 - 0.99941 ...............using normal distribution tables

p-value =

4. Refer to Exhibit 1. What is the critical value?

C.V. at 2% =

=

C.v. = .......using normal percentage tables with p = 2%

5. Refer to Exhibit 1. What is your conclusion?

Since |test stat | > C.V. or as well p -value < 0.02

We reject the null hypothesis and concldue that operation was significantly effective in reducingthe avg. waiting time.

Exhibit 2

average credit card balance of $619. You want to find out if this number is lower now So this is our null mean 619 and the test is again left sided one tailed.

random sample of n = 36 college students.

average credit card debt = $519

sample standard deviation of Sx = $269.26. we will use t-test since pop SD is unknwon

Using α=0.03,

The population avg. card debt of clg students is 619.

The population avg. card debt of clg students is lower than 619.

6. Refer to Exhibit 2. What is the p-value of your test?

Test Stat =

Where the null mean = 619

Tst Stat = -2.2283

p-value = P( >|Test Stat|)

= P( > 2.23)

p- value =

7. Refer to Exhibit 2. Using the critical value approach, how do you assess the null hypothesis?

C.V. at 3% =

=

C.V. =

8. Refer to Exhibit 2. What is your conclusion?

since p-vlue < 0.03 and also |test Stat| > C.V.

we reject the null hypothesis and concldue that significantly the population avg. card debt of clg students is lower than 619.

9. Refer to Exhibit 2. Explain in your own words how Type I and Type II errors can occur in this hypothesis test.

Type 1 error : Rejecting a true null hypothesis

In context: We concldue that the population avg. card debt of clg students is lower than 619 when in fact it is not.

Type 2 : Accepting a false null hypothesis

In context: We concldue that the population avg. card debt of clg students is not lower than 619 when in fact it is.

But here since we are rejecting the null hypothesis there is a chance to make type 1 error.

10. Refer to Exhibit 2.  Does changing the value of α from 0.03 to 0.01 affect your conclusion? Why or why not?

We reject the null hypothesis when either p-value < level of signifcance or |test Stat| > C.V. we need not check both. Since p -value = 0.016 < 0.03 but > 0.01, so we would not reject the null hpyothesis. This means we need more evidence to reject it since acceptance of error as been reduced to 1% that is more strict.