Respuesta :
Answer:
(a) The test statistic value is -4.123.
(b) The critical values of t 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:
H₀: There is no difference in the mean waiting time between the two branches, i.e. μ₁ - μ₂ = 0.
Hₐ: There is a difference in the mean waiting time between the two branches, i.e. μ₁ - μ₂ ≠ 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:
[tex]\bar x_{1}=\frac{1}{n_{1}}\sum X_{1}=\frac{1}{15}[4.21+5.55+...+3.79]=4.29[/tex]
[tex]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[/tex]
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:
[tex]\bar x_{2}=\frac{1}{n_{2}}\sum X_{2}=\frac{1}{15}[9.66+5.90+...+5.47]=7.11[/tex]
[tex]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[/tex]
(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 t-test for two means.
Compute the test statistic value as follows:
[tex]t=\frac{\bar x_{1}-\bar x_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}[/tex]
[tex]=\frac{4.29-7.11}{\sqrt{\frac{1.64^{2}}{15}+\frac{2.08^{2}}{15}}}[/tex]
[tex]=-4.123[/tex]
Thus, the test statistic value is -4.123.
(b)
The degrees of freedom of the test is:
[tex]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}}[/tex]
[tex]=\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}}[/tex]
[tex]=26.55\\\approx 27[/tex]
Compute the critical value for α = 0.05 as follows:
[tex]t_{\alpha/2, m}=t_{0.025, 27}=\pm2.052[/tex]
*Use a t-table for the values.
Thus, the critical values of t are ± 2.052.
