>8:46ChilerellenosamplesMeanStandarddeviationPepe Delgados is looking to redesign its kitchen layout with two goals in mind: (1) reducing theaverage time it takes to produce a chile relleno, and (ii) reducing the variability in the time ittakes to produce a chile relleno. You have two candidate designs (A and B), and you implementeach design for one week, collecting samples for each. Your sample data for production timesare below.Layout LayoutAB252112.33 13.94Expert Q&A1.93 4.39Donea. Test at 5% significance the similarity in variability of the production times for the two layoutsusing the critical value approach.b. Using the result of Question a, appropriately test whether the average production time forLayout A is less than that of Layout B, with 90% confidence using the p-value approach.

The sample data for production time for layouts A and B is given below. layoutALayout Bchile relleno samples2521means12.3313.94standard deviations1.934.39n1 and n2 are the sample sizes for layouts A and Bbe sample means for layouts A and Bs1 and s2 are sample standard deviations for layouts A and BTo test the similarity of variability of production time for layouts A and BThe hypothesis is, Vs It is a two-tailed test. The and are the variance of production times for layouts A and B respectively. Step 3: Calculate test statistic:We use the F-test to test the equality of variance of two populations. The formula for F-statistic is, The F-statistic is calculated as, Thus, the value of F-statistic is 0.1933.Now calculate critical value.using a two-tailed test. level of significance = 0.05for left tail Probability = , Df = n1-1left tail-Fcritical = =F.INV(0.025,20,24) =0.4297for right tail Probability = 1- , Df = n2-1right tail-Fcritical = =F.INV(0.975,20,24) = 2.3272 Decision rule: For a two-tailed test, if the test statistic value lies between critical values, we fail to reject the null hypothesis at the desired level of significance otherwise we reject it. It is observed the value of the f-statistic (0.1933) does not lie between 0.4297 and 2.3272. then reject the null hypothesis at a 0.05 significance level.It is concluded that there is no similarity in the variability of the production times for the two layouts. To test whether the average production time for Layout A is less than that of Layout Buse two sample t-testThe test hypothesis is,H0: Vs H1: population means for layouts A and BThe test statistic for unequal variance is,t t = -1.5588The test statistic value is -1.5588Now calculate p-value p-value = P(T> t)where T follows t-distribution with n1+n2-2P-value =T.DIST(-1.5588,43,TRUE) = 0.0632 ...........by using excel function (=T.DIST(x, df,cumulative)The p-value is 0.0631The level of significance is 0.10Decision: If the p-value is less than then reject H0 otherwise reject to fail H0p-value < Reject H0 at level 0.10There is average production time for Layout A is less than that of Layout Ba) There is no similarity in the variability of the production times for the two layouts. b) There is average production time for Layout A is less than that of Layout B