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:
![\bar x_{1}=\frac{1}{n_{1}}\sum X_{1}=\frac{1}{15}[4.21+5.55+...+3.79]=4.29](https://tex.z-dn.net/?f=%5Cbar%20x_%7B1%7D%3D%5Cfrac%7B1%7D%7Bn_%7B1%7D%7D%5Csum%20X_%7B1%7D%3D%5Cfrac%7B1%7D%7B15%7D%5B4.21%2B5.55%2B...%2B3.79%5D%3D4.29)
![s_{1}=\sqrt{\frac{1}{n_{1}-1}\sum (X_{1}-\bar x_{1})^{2}}\\=\sqrt{\frac{1}{15-1}[(4.21-4.29)^{2}+(5.55-4.29)^{2}+...+(3.79-4.29)^{2}]}\\=1.64](https://tex.z-dn.net/?f=s_%7B1%7D%3D%5Csqrt%7B%5Cfrac%7B1%7D%7Bn_%7B1%7D-1%7D%5Csum%20%28X_%7B1%7D-%5Cbar%20x_%7B1%7D%29%5E%7B2%7D%7D%5C%5C%3D%5Csqrt%7B%5Cfrac%7B1%7D%7B15-1%7D%5B%284.21-4.29%29%5E%7B2%7D%2B%285.55-4.29%29%5E%7B2%7D%2B...%2B%283.79-4.29%29%5E%7B2%7D%5D%7D%5C%5C%3D1.64)
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:
![\bar x_{2}=\frac{1}{n_{2}}\sum X_{2}=\frac{1}{15}[9.66+5.90+...+5.47]=7.11](https://tex.z-dn.net/?f=%5Cbar%20x_%7B2%7D%3D%5Cfrac%7B1%7D%7Bn_%7B2%7D%7D%5Csum%20X_%7B2%7D%3D%5Cfrac%7B1%7D%7B15%7D%5B9.66%2B5.90%2B...%2B5.47%5D%3D7.11)
![s_{2}=\sqrt{\frac{1}{n_{2}-1}\sum (X_{2}-\bar x_{2})^{2}}\\=\sqrt{\frac{1}{15-1}[(9.66-7.11)^{2}+(5.90-7.11)^{2}+...+(5.47-7.11)^{2}]}\\=2.08](https://tex.z-dn.net/?f=s_%7B2%7D%3D%5Csqrt%7B%5Cfrac%7B1%7D%7Bn_%7B2%7D-1%7D%5Csum%20%28X_%7B2%7D-%5Cbar%20x_%7B2%7D%29%5E%7B2%7D%7D%5C%5C%3D%5Csqrt%7B%5Cfrac%7B1%7D%7B15-1%7D%5B%289.66-7.11%29%5E%7B2%7D%2B%285.90-7.11%29%5E%7B2%7D%2B...%2B%285.47-7.11%29%5E%7B2%7D%5D%7D%5C%5C%3D2.08)
(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:
![m=\frac{[\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}]^{2}}{\frac{(\frac{s_{1}^{2}}{n_{1}})^{2}}{n_{1}-1}+\frac{(\frac{s_{2}^{2}}{n_{2}})^{2}}{n_{2}-1}}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B%5B%5Cfrac%7Bs_%7B1%7D%5E%7B2%7D%7D%7Bn_%7B1%7D%7D%2B%5Cfrac%7Bs_%7B2%7D%5E%7B2%7D%7D%7Bn_%7B2%7D%7D%5D%5E%7B2%7D%7D%7B%5Cfrac%7B%28%5Cfrac%7Bs_%7B1%7D%5E%7B2%7D%7D%7Bn_%7B1%7D%7D%29%5E%7B2%7D%7D%7Bn_%7B1%7D-1%7D%2B%5Cfrac%7B%28%5Cfrac%7Bs_%7B2%7D%5E%7B2%7D%7D%7Bn_%7B2%7D%7D%29%5E%7B2%7D%7D%7Bn_%7B2%7D-1%7D%7D)
![=\frac{[\frac{1.64^{2}}{15}+\frac{2.08^{2}}{15}]^{2}}{\frac{(\frac{1.64^{2}}{15})^{2}}{15-1}+\frac{(\frac{2.08^{2}}{15})^{2}}{15-1}}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%5B%5Cfrac%7B1.64%5E%7B2%7D%7D%7B15%7D%2B%5Cfrac%7B2.08%5E%7B2%7D%7D%7B15%7D%5D%5E%7B2%7D%7D%7B%5Cfrac%7B%28%5Cfrac%7B1.64%5E%7B2%7D%7D%7B15%7D%29%5E%7B2%7D%7D%7B15-1%7D%2B%5Cfrac%7B%28%5Cfrac%7B2.08%5E%7B2%7D%7D%7B15%7D%29%5E%7B2%7D%7D%7B15-1%7D%7D)
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.
