Question

A bank branch located in a commercial district of a city has the business objective of improving the process for serving customers during the noon-to-1:00 P.M. lunch period. The waiting time is defined as the time the customer enters the line until he or she reaches the teller window. The bank wants to determine if the mean waiting time is less than 5 minutes. A random sample of 15 customers is selected. Results found that the sample mean waiting time was 4.287 minutes with a sample standard deviation of 1.638 minutes. Show work if necessary.

Answer 1

Answer:

We conclude that the mean waiting time is more than or equal to 5 minutes at 5% level of significance.

Step-by-step explanation:

We are given that waiting time is defined as the time the customer enters the line until he or she reaches the teller window.  

A random sample of 15 customers is selected. Results found that the sample mean waiting time was 4.287 minutes with a sample standard deviation of 1.638 minutes.

Let $\mu$ = mean waiting time.

SO, Null Hypothesis, $H_0$ : $\mu$ $\geq$ 5 minutes     {means that the mean waiting time is more than or equal to 5 minutes}

Alternate Hypothesis, $H_A$ : $\mu$ < 5 minutes      {means that the mean waiting time is less than 5 minutes}

The test statistics that would be used here One-sample t-test statistics as we don't know about population standard deviation;

                              T.S. =  $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$  ~ $t_n_-_1$

where, $\bar X$ = sample mean waiting time = 4.287 minutes

            s = sample standard deviation = 1.638 minutes

            n = sample of customers = 15

So, the test statistics  =  $\frac{4.287 -5}{\frac{1.638}{\sqrt{15} } }$  ~ $t_1_4$

                                       =  -1.686

The value of t test statistics is -1.686.

Since, in the question we are not given the level of significance so we assume it to be 5%. Now, at 5% significance level the t table gives critical value of -1.761 at 14 degree of freedom for left-tailed test.

Since our test statistic is more than the critical value of t as -1.686 > -1.761, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the mean waiting time is more than or equal to 5 minutes.