Question

In a recent year, the Federal Communications Commission reported that the mean wait for repairs for AT&T customers was at least 23.9 hours. In an effort to improve this service, suppose that a new repair service process was developed. This new process, used for a sample of 100 repairs, resulted in a sample mean of 22.1 hours and a sample standard deviation of 6.3 hours. Is there evidence that the population mean amount is less than 23.9 hours? (use a 0.10 level of significance)

Step 1: Specify the population value of interest

Step 2: Formulate the appropriate null and alternative hypotheses and indicate which category it belongs to (two tailed, lower tail, or upper tail test).

Step 3: Specify the desired level of significance and the sample size; determine the type of hypothesis test (two-tailed, upper tail, or lower tail test).

Step 4: Find the critical value and determine the rejection region. With an unknown σ, we will use t-value as the critical value. Critical Value(s) is(are): Rejection Region:

Step 5: Obtain sample evidence (sample statistic x-bar or ) and compute the test statistic z*(Standardized value of x-bar or ).

Step 6: Reach a decision and interpret the result.

Answer 1

1) population value of interest is whether mean wait for repairs for AT&T customers has improved by using new repair service

2)

Ho :µ =23.9

Ha :µ <23.9

it is left/lower tailed test

3)

Level of Significance , α = 0.05

sample std dev ,  s = 6.300

Sample Size ,n = 100

Sample Mean,  x̅ =22.10

degree of freedom=DF=n-1=99

t test will be used, because population std dev is unknown

4)

critical t value, t*=-1.6604 [Excel formula =t.inv(α,df) ]

Rejection Region:

reject Ho, when t-stat< t-critical value

5)

Standard Error , SE =s/√n =0.6300

t-test statistic=(x̅ - µ )/SE = -2.86

6)

since, t-stat = -2.86<-1.6604 , so, reject Ho

hence, there is enough evidence to conclude that mean wait for repairs for AT&T customers has improved by using new repair service at α=0.05