Question

A bank with a branch located in a commercial district of a city has developed an improved process for serving customers during the noon to 1 P.M. lunch period. The waiting time (operationally defined as the time elapsed from when the customer enters the line until he or she reaches the teller window) of all customers during this hour is recorded over a period of one week. A random sample of 15 customers is selected, and the results (in minutes) are as follows: 4, 5, 3, 5.2, 3.6, 4.8, 5, 3.7, 6, 3.2, 4.5, 4.7, 5.14, 6.14, 3.86 Suppose that another branch located in a residential area is also concerned with the noon to 1 P.M. lunch period. Another set of random sample of 15 customers is also selected, and the results (in minutes) are as follows:

Answer 1

Answer:

There is evidence of a significant difference in the mean waiting time between the two branches (P-value=0.0003).

Step-by-step explanation:

The question is incomplete.

"A bank with a branch (Branch 1) located in a commercial district of a city has developed an improved process for serving customers during the noon-to-1 p.m. lunch period. The waiting time (operationally defined as the time elapsed from when the customer enters the line until he or she reaches the teller window) needs to be shortened to increase customer satisfaction. A random sample of 15 customers is selected and the results (in minutes) are shown in the table on the left. In addition, suppose that another branch (Branch 2), located in a residential area, is also concerned with the noon-to-1 p.m. lunch period. A random sample of 15 customers is selected and the results are shown in the table on the left. Assuming that the population variances from both branches are equal, is there evidence of a significant difference in the mean waiting time between the two branches? Use alpha = 0.01."

Branch 1 data: [4.21; 5.55; 3.02; 5.13; 4.77; 2.34; 3.54; 3.2; 4.5; 6.1; 0.38; 5.12; 6.46; 6.19; 3.79]

Branch 2 data: [9.66; 5.9; 8.02; 5.79; 8.73; 3.82; 8.01; 8.35; 10.49; 6.68; 5.64; 4.08; 6.17; 9.91; 5.47]

First, we state the null and alternative hypothesis

$H_0: \mu_1=\mu_2\\\\ H_1: \mu_1\neq\mu_2$

The significance level is 0.01.

The standard deviation is estimated as:

$s_M=\sqrt{\frac{s_1^2+s_2^2}{n} } =\sqrt{\frac{2.683+4.336}{15} }=0.684$

The mean values of the samples for the two branches are:

$M_1=4.287\\\\M_2=7.115$

The test statistic t can be calculated as

$t=\frac{(M_1-M_2)-0}{s_M} =\frac{4.287-7.115-0}{0.684}=\frac{-2.828}{0.684}= -4.135$

The degrees of freedom are:

$df=n_1+n_2-2=15+15-2=28$

The p-value for t=-4.135 and df=28 is P=0.0003.

The p-value (0.0003) is much smaller than the significance level, so the effect is significant. There is evidence to reject the null hypothesis.