A bank with branches located in a commercial district of a city and in a residential area has the business objective of developing an improved process for serving customers during the noon-to-1 P.M. lunch period. Management decides to first study the waiting time in the current process. The waiting time is defined asthe number of minutes that elapses from when the customer enters the line until he or she reaches the teller window. Data are collected from a random sample of 15 customers at each branch. Complete parts (a) through (d) below.Click the icon to view the bank branch data......a. Assuming that the population variances from both banks are equal, is there evidence of a difference in the mean waiting time between the two branches? (Use a = 0.1.)Letbe the mean waiting time of the commercial district branch and u,be the mean waiting time of the residential area branch. Determine the hypotheses. Choose the correct answer below.Waiting time dataA. Ho: H1 # H2H1: H1 =H2B. Ho: H1 = H2Residential9.535.938.15Commercialc. Ho: H12 H2H1: H1 H25.473.155.435.54Find the test statistic.4.868.81tSTAT =(Round to two decimal places as needed.)2.273.77%3D3.658.193.498.5810.48Find the critical value(s).4.256.076.870.215.375.164.03(Use a comma to separate answers as needed. Round to two decimal places as needed.)6.546.46Choose the correct conclusion below.6.029.783.735.43A. Do not reject Ho. There is sufficient evidence that the mean waiting time between the two branches differ.B. Reject Ho: There is sufficient evidence that the mean waiting time between the two branches differ.C. Do not reject Ho. There is insufficient evidence that the mean waiting time between the two branches differ.PrintDoneD. Reject Ho. There is insufficient evidence that the mean waiting time between the two branches differ.b. Determine the p-value in (a) and interpret its meaning.p-value =(Round to three decimal places as needed.)

Given : Let u1, be the mean waiting time of the commercial district branch and u₂ be the mean waiting time of the residential area branch.solution : For commercial x̄1 = 4.2946667 s1 = 1.6565319 n1 = 15 For Residential x̄2 = 7.128 s2 = 2.1123048 n2 = 15 Significance level, α= 0.1 Hypothesis test : The null and alternative hypothesis is Ho : u1 = u2 H1 : u1≠ u2 The hypothesis test is a two tailed test pooled variance : =(( 15 -1) 1.6565319 ^2+( 15 -1) 2.1123048 ^2)/ 15 + 15 -2 = 3.603 pooled variance is 3.603 Test statistic test statistic for equal variance t test is (( 4.2946667 - 7.128 )-( 0.0 ))/√( 3.603 (1/ 15 +1/ 15 )) = -4.09 The test statistic is -4.09 Degree of freedom : df=n1+n2-2= 15 + 15 -2= 28 critical value : critical value at α = 0.1 tα/2=±1.7 critical values are (-1.7 ,1.7) As t <- 1.7 we reject the null hypothesis therefore option b) is correct answerb) p-value : p-value with test statistic t= -4.09 and df= 28 is P-value =2P(t>|tobs|) = 2 x P(t >| -4.09 |) = 0 ...... from t distribution table ∴ P-value is 0