Answer to Question #199875 in Statistics and Probability for meaca

Question

A bank manager wants to implement a new system for customer waiting in the branch office. Because of cost consideration, he wants to implement the new system only if it reduces the mean waiting time by more than 3 minutes. A random sample of n₁ = 100 waiting times observed under the current system gives a sample mean x₁ = 8.79 and a sample variance s₁2 = 4.8237. A random sample of n₂ = 100 waiting times observed during the trial run of the new system yield a sample mean X₂ = 5.14 and a sample variance s₂² 1.7927. Will the manager implement the new system or not. (Use a level of significance of 5%). Let:

u1 = the mean customer waiting time under the current system u2 the mean customer waiting time under the new system

Answer 1

$H_0:$ $\Delta>3$

$H_{\alpha}:\Delta\le3$

$z=\frac{\mu_1-\mu_2-\Delta}{\sqrt{s_1/n_1+s_2/n_2}}$

where $\mu_1,\mu_2$ are means of current and new systems,

$\Delta$ is difference of means,

$s_1,s_2$ are variances of samples,

$n_1,n_2$ are sizes of samples.

$z=\frac{8.79-5.14-3}{\sqrt{4.8237+1.7927}/10}=2.53$

p-value:

$P(\Delta>3)=1-P(z<2.53)=1-0.9943=0.0057$

Since p-value$=0.57\ \%<\alpha=5\ \%$ , we reject null hypothesis. So, the manager will not implement the new system.