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,

Required:
a. 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?
b. State the null and alternative hypotheses for the test.

Answer 1

a) The lower bound of the confidence interval is below 4.50 minutes, which means that there is not sufficient evidence to say that the mean wait time for customers is longer than 4.50 minutes with a 98% level of confidence.

b) The null hypothesis is
`H_0: \mu \leq 4.50`, and the alternate hypothesis is
`H_a: \mu > 4.50`

Explanation:

Finding the confidence interval:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1 - 0.98)/(2) = 0.01`

Now, we have to find z in the Ztable as such z has a pvalue of
`1 - \alpha`.

That is z with a pvalue of
`1 - 0.01 = 0.98`, so Z = 2.327.

Now, find the margin of error M as such

`M = z(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

`M = 2.327(2.1)/(√(53)) = 0.67`

The lower end of the interval is the sample mean subtracted by M. So it is 4.94 - 0.67 = 4.27 minutes

The upper end of the interval is the sample mean added to M. So it is 4.94 + 0.67 = 5.61 minutes.

a. 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?

The lower bound of the confidence interval is below 4.50 minutes, which means that there is not sufficient evidence to say that the mean wait time for customers is longer than 4.50 minutes with a 98% level of confidence.

b. State the null and alternative hypotheses for the test.

Testing if there is evidence to say that the mean wait time for customers is longer than 4.50 minutes, which is the alternate hypothesis.

So the null hypothesis is
`H_0: \mu \leq 4.50`, and the alternate hypothesis is
`H_a: \mu > 4.50`