ONE-WAY ANOVA AND TWO-SAMPLE COMPARISON1) The basic procedure.ANOVA is an acronym for Analysis Of Variance. It is a methodology for testing hypotheses regarding population means in various contexts. One-way ANOVA refers to the methodology for testing the equality of two or more population means, based on independent samples from each of the populations. Thus, if mu1, mu2, … ,muk denote the means of k populations, one-way ANOVA is a methodology for testing H_0 : mu1= mu2= … =muk,against the alternative that not all are equal. Here we will demonstrate the use of Minitab for carrying out this test procedure using the data from www.stat.psu.edu/~mga/401/labs/05/lab6/anova.fe.data.txt. The data are about total Fe for four types of iron formation (1= carbonate, 2= silicate, 3= magnetite, 4= hematite).If the data from the different populations (also called factor levels in the ANOVA jargon) are given in different columns, then use the sequence of commandsStat >ANOVA>One-way (Unstacked)>Enter C1-C4 for Response, 95 for confidence level>OK.[If the data from all factor levels are stored in one column, there must also be a second column which indicates the group membership of each observation in the first column. Call this second column “formation”. The sequence of commands in this case are: Stat>ANOVA>One-way>Enter Fe as response and formation as Factor>OK ]The output that Minitab produces (other software packages produce similar outputs) is:One-way ANOVA: C1, C2, C3, C4 Source DF SS MS F PFactor 3 509.1 169.7 10.85 0.000Error 36 563.1 15.6Total 39 1072.3S = 3.955 R-Sq = 47.48% R-Sq(adj) = 43.10% Individual 95% CIs For Mean Based on Pooled StDevLevel N Mean StDev -----+---------+---------+---------+----C1 10 26.080 3.391 (-----*------)C2 10 24.690 4.425 (------*-----)C3 10 29.950 2.854 (-----*-----)C4 10 33.840 4.831 (------*-----) -----+---------+---------+---------+---- 24.0 28.0 32.0 36.0Pooled StDev = 3.955The ANOVA table gives the decomposition of the total sum of squares into asum of squares due to the population differences (Factor) and a sum of squares due to the intrinsic error. Thus, 509.1 + 563.1 = 1072.3 (not really, due to rounding). MS = SS/DF, and the F statistic is the ratio of the MS for Factor over MS for error. Typically, statistics books also have F-tables, where the value of the F statistic can be looked up. We do willnot learn how to do that because Minitab produces the p-value, and does so in much greater accuracy than what we could do from the F-tables. [The F-distribution is characterized by two degrees of freedom (here the degrees of freedom of the F-statistic are 3 and 36) and thus contain only selected percentiles.] Because the p-value is small, the hypothesis of equality of the population means is rejected. Following the ANOVA table there is information about the estimate of the standard deviation, which is assumed to be the same in all populations (here the estimate is S=3.955, and it is also given in the last line of the output), and the R-Sq, which has the same significance as explained in the activity for regression. The individual sample means, estimated standard errors, and 95% CI for each population mean are also given. 2) Multiple Comparisons for One-Way ANOVA.When the null hypothesis is rejected it means that, the data strongly suggest that, at least one of the population means is different from the others. When k>2, additional testing needs to be done to identify which means appear to be different. This additional testing is called multiple comparisons. It involves performing all pair-wise comparisons (i.e. testing the null hypothesis of equality of each possible pairs of means) in such a way that the probability of committing a type I error for any of these pair-wise test procedures does not exceed the designated level of significance alpha. One of the ways of doing multiple comparisons, is to perform the aforementioned ANOVA test for each pair-wise comparison, at an adjusted level of significance. The adjusted level equals the designatedalpha divided by the total number of pair-wise comparisons. This is called the Bonferroni method. Here we will demonstrate a different method, called the Tukey method, for doing pairwise comparisons, in such a way that the overall level of significance is alpha. . Stat >ANOVA>One-way (Unstacked)>Enter C1-C4 for Response, 95 for confidence level>Click Comparisons select Tukey’s, enter family error rate (5 for overall level of significance 0.05)>OK>OKThe additional Minitab output, with my comments in brackets, is:Tukey 95% Simultaneous Confidence IntervalsAll Pairwise ComparisonsIndividual confidence level = 98.93%C1 subtracted from: Lower Center Upper +---------+---------+---------+---------C2 -6.155 -1.390 3.375 (------*------)C3 -0.895 3.870 8.635 (------*-----)C4 2.995 7.760 12.525 (------*------) +---------+---------+---------+--------- -14.0 -7.0 0.0 7.0[These are simultaneous CI for the differences mu1-mu2, mu1-mu3, and mu1-mu4. Ifa CI does not contain 0, the two means are declared significantly different. Thus, mu1 is significantly different from mu4, but not significantly different from mu2 or from mu3.]C2 subtracted from: Lower Center Upper +---------+---------+---------+---------C3 0.495 5.260 10.025 (------*-----)C4 4.385 9.150 13.915 (------*------) +---------+---------+---------+--------- -14.0 -7.0 0.0 7.0[These are simultaneous CI for mu2-mu3 and mu2-mu4. None of these CI contains zero, and thus mu2 is significantly different from both mu3 and mu4.]C3 subtracted from: Lower Center Upper +---------+---------+---------+---------C4 -0.875 3.890 8.655 (------*-----) +---------+---------+---------+--------- -14.0 -7.0 0.0 7.0[This is a simultaneous CI for mu3-mu4. The CI contains zero,