Question

A bank with branches located in a commercial district of a city and in a residential area has the business objective of developing an improved process for serving customers during the​ noon-to-1 P.M. lunch period. Management decides to first study the waiting time in the current process. The waiting time is defined as the number of minutes that elapses from when the customer enters the line until he or she reaches the teller window. Data are collected from a random sample of 15 customers at each branch. Complete parts​ (a) and​ (b) below.

Commercial   Residential
4.31    9.64
5.51    5.77
3.09    8.14
5.01    5.56
4.94    8.68
2.45    3.57
3.48    8.06
3.05    8.45
4.72 10.65
6.23    6.55
0.23    5.34
5.05 4.24
6.42    6.02
6.28    9.93
3.54 5.69

t-stat:____

p-value: ____

b.)Assuming equal variances between the two populations yields a t-stat test statistic of ____ and a p-value of ____. How do these results compare to the results found in a.)?

Answer 1

Answer- Using R software

Here we have to test Hypothesis

H0: µ₁= µ₂ v/s H1:  µ₁≠ µ₂

where,

µ₁= average weighting time for a customer in current process of bank branches located in commercial district of city.

µ₂= average weighting time for a customer in current process of bank branches located in residential area.

Here we have given,

n₁=n₂=15.

where n1=no. of samples from branches of commercial area.

n₂=no. of samples from branches of residential area.

Answer for a).

Assuming unequal variances between the two populations yields

t_stat = -3.9724, df = 26.454

and

p-value = 0.0004901

Answer for b).

Assuming equal variances between the two populations yields

t_stat= -3.9724, df = 28

and

p-value = 0.0004525

Here by compare these two results by using p-values

we can conclude that

Result of the t test is pretty good with Assuming unequal variances between the two populations yields than Assuming equal variances between the two populations yields

The R soft. code is -

Commercial=c(4.31,5.51,3.09,5.01,4.94,2.45,3.48,3.05,4.72,
6.23,0.23,5.05,6.42,6.28,3.54)

Residential=c(9.64,5.77,8.14,5.56,8.68,3.57,8.06,8.45,10.65,
6.55,5.34,4.24,6.02,9.93,5.69)

t_stat_unequal_variance=t.test(Commercial, Residential,alternative = "two.sided",
var.equal =FALSE )
t_stat_unequal_variance

t_stat_equal_variance=t.test(Commercial, Residential,alternative = "two.sided",
var.equal = TRUE )
t_stat_equal_variance

Output Of R soft. is-

> Commercial=c(4.31,5.51,3.09,5.01,4.94,2.45,3.48,3.05,4.72,
+ 6.23,0.23,5.05,6.42,6.28,3.54)
>
> Residential=c(9.64,5.77,8.14,5.56,8.68,3.57,8.06,8.45,10.65,
+ 6.55,5.34,4.24,6.02,9.93,5.69)
>
> t_stat_unequal_variance=t.test(Commercial, Residential,alternative = "two.sided",
+ var.equal =FALSE )
> t_stat_unequal_variance

   Welch Two Sample t-test

data: Commercial and Residential
t = -3.9724, df = 26.454, p-value = 0.0004901
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-4.245639 -1.351695
sample estimates:
mean of x mean of y
4.287333 7.086000

>
> t_stat_equal_variance=t.test(Commercial, Residential,alternative = "two.sided",
+ var.equal = TRUE )
> t_stat_equal_variance

   Two Sample t-test

data: Commercial and Residential
t = -3.9724, df = 28, p-value = 0.0004525
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-4.241829 -1.355504
sample estimates:
mean of x mean of y
4.287333 7.086000