Question

The board of a major credit card company requires that the mean wait time for customers when they call customer service is at most 4.50 minutes. To make sure that the mean wait time is not exceeding the requirement, an assistant manager tracks the wait times of 53 randomly selected calls. The mean wait time was calculated to be 4.94 minutes. Assuming the population standard deviation is 2.10 minutes, is there sufficient evidence to say that the mean wait time for customers is longer than 4.50 minutes with a 98% level of confidence?
1. State the null and alternative hypotheses for the test.
2. Compute the value of the test statistic.
3. Draw a conclusion and interpret the decision.

Answer 1

Answer:

H0 : μ = 4.50

H1 : μ > 4.50

Test statistic = 1.525

we conclude that there is no sufficient evidence to conclude that the mean wait time for customers is longer than 4.50.

Step-by-step explanation:

H0 : μ = 4.50

H1 : μ > 4.50

Test statistic :

(xbar - μ) ÷ s/sqrt(n)

(4.94 - 4.50) ÷ 2.10/ sqrt(53)

0.44 / 2.10/ sqrt(53)

= 1.525

α = 1 - 98% = 0.02

Decision region :

Reject H0 ; if Pvalue < α

Pvalue = p(Z < 1.525) = 0.936

Pvalue = 0.936

Since Pvalue > α ; We fail to reject H0

Hence, we conclude that there is no sufficient evidence to conclude that the mean wait time for customers is longer than 4.50.