Answer:
(a) The test statistic value is -4.123.
(b) The critical values of <em>t</em> are ± 2.052.
Step-by-step explanation:
In this case we need to determine whether there is evidence of a difference in the mean waiting time between the two branches.
The hypothesis can be defined as follows:
<em>H₀</em>: There is no difference in the mean waiting time between the two branches, i.e. <em>μ</em>₁ - <em>μ</em>₂ = 0.
<em>Hₐ</em>: There is a difference in the mean waiting time between the two branches, i.e. <em>μ</em>₁ - <em>μ</em>₂ ≠ 0.
The data collected for 15 randomly selected customers, from bank 1 is:
S = {4.21, 5.55, 3.02, 5.13, 4.77, 2.34, 3.54, 3.20, 4.50, 6.10, 0.38, 5.12, 6.46, 6.19, 3.79}
Compute the sample mean and sample standard deviation for Bank 1 as follows:
The data collected for 15 randomly selected customers, from bank 2 is:
S = {9.66
, 5.90
, 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}
Compute the sample mean and sample standard deviation for Bank 2 as follows:
(a)
It is provided that the population variances are not equal. And since the value of population variances are not provided we will use a <em>t</em>-test for two means.
Compute the test statistic value as follows:
Thus, the test statistic value is -4.123.
(b)
The degrees of freedom of the test is:
Compute the critical value for <em>α</em> = 0.05 as follows:
*Use a <em>t</em>-table for the values.
Thus, the critical values of <em>t</em> are ± 2.052.