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Answer:
There is evidence of a significant difference in the mean waiting time between the two branches (P-value=0.0003).
Step-by-step explanation:
The question is incomplete.
"A bank with a branch (Branch 1) located in a commercial district of a city has developed an improved process for serving customers during the noon-to-1 p.m. lunch period. The waiting time (operationally defined as the time elapsed from when the customer enters the line until he or she reaches the teller window) needs to be shortened to increase customer satisfaction. A random sample of 15 customers is selected and the results (in minutes) are shown in the table on the left. In addition, suppose that another branch (Branch 2), located in a residential area, is also concerned with the noon-to-1 p.m. lunch period. A random sample of 15 customers is selected and the results are shown in the table on the left. Assuming that the population variances from both branches are equal, is there evidence of a significant difference in the mean waiting time between the two branches? Use alpha = 0.01."
Branch 1 data: [4.21; 5.55; 3.02; 5.13; 4.77; 2.34; 3.54; 3.2; 4.5; 6.1; 0.38; 5.12; 6.46; 6.19; 3.79]
Branch 2 data: [9.66; 5.9; 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]
First, we state the null and alternative hypothesis
[tex]H_0: \mu_1=\mu_2\\\\ H_1: \mu_1\neq\mu_2[/tex]
The significance level is 0.01.
The standard deviation is estimated as:
[tex]s_M=\sqrt{\frac{s_1^2+s_2^2}{n} } =\sqrt{\frac{2.683+4.336}{15} }=0.684[/tex]
The mean values of the samples for the two branches are:
[tex]M_1=4.287\\\\M_2=7.115[/tex]
The test statistic t can be calculated as
[tex]t=\frac{(M_1-M_2)-0}{s_M} =\frac{4.287-7.115-0}{0.684}=\frac{-2.828}{0.684}= -4.135[/tex]
The degrees of freedom are:
[tex]df=n_1+n_2-2=15+15-2=28[/tex]
The p-value for t=-4.135 and df=28 is P=0.0003.
The p-value (0.0003) is much smaller than the significance level, so the effect is significant. There is evidence to reject the null hypothesis.
The standard deviation and the degree of freedom from the information about the bank branches are 0.6840 and 28 respectively.
How to compute the standard deviation?
From the information given, the standard deviation will be calculated thus:
= (✓2.683 + ✓4.336) / ✓15
= 0.684
The degree of freedom is calculated thus:
= (n1 + n2) - 2
= (15 + 15) - 2
= 30 - 2
= 28
Lastly, the test statistic will be:
= (4.287 - 7.115)/0.684
= -4.135
In conclusion, the null hypothesis will be rejected since the p value is smaller than the level of significance.
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