BUS 308 Statistics for Managers | Standardized Guidance

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Week 3 – Standardized Guidance – BUS 308: Statistics for ManagersAfter mastering this week’s material, you shouldUnderstand:oWhen the paired t-test is used.oWhen the Analysis of Variance (ANOVA) test is used.oHow to calculate the effect size measure for each test.oThe differences between the one factor ANOVA, the two factor ANOVA withoutreplication, and the two factor ANOVA with replication.oHow to interpret the F statistic.Be able to:oDevelop the null and alternate hypothesis statement for both the paired t-test and theANVOA tests from a research question.oUse Excel to perform a paired t-test.oInterpret the results of a paired t-test.oUse Excel to perform ANOVA tests.oInterpret the results of ANOVA tests.oCalculate appropriate effect size measures for each test.Continuing with Mean DifferencesFor Week Three, we will continue to look at mean differences.First we will look at dependent groups –these exist when we take more than one measure from the subject (Tanner & Youssef-Morgan, 2013).Comparing pre and post-test scores would be an example of dependent measures.This would be anexample of a 2 sample situation, we can also expand this analysis to 3 or more measures on the samesubjects.Then we will move on to testing means from 3 or more independent groups, where – as withthe t-tests we examined last week – we measure different subjects in each group (Lind, Marchel, &Wathen, 2008).Dependent/Paired t-testChapter 7 (Tanner & Youssef-Morgan, 2013) introduces the idea of a dependent variable – “repeatedmeasures of the same variable within the same group of subjects over time” (p. 164).This issue ofmeasuring the same subject over time or on two similar measures introduces some complexity into ouranalysis.Due to differences in initial measures, differences in related measures are not due solely to thefactor being measured, but have a relationship to the initial results.For example, someone 6 feet tall cangenerally jump higher than someone 5 feet tall; if we train both individuals in techniques to jump higher,the taller person will still have an advantage – so treating the second measure as unrelated to the initialstarting point is not reasonable.There are several techniques available to us to take related measuressuch as this into account.This week we will look at the paired t-test, used when we have 2 measures on each subject.Chapter 7 inthe text (Tanner & Youssef-Morgan, 2013) also discusses the within-subjects F (AKA within-subjectsANOVA).We will this later with the other forms of ANOVA introduced this week.We start with another version of the t-test.Recall that last week we looked at the single sample t-test, the2 sample t-test with equal variance and the 2 sample t-test with unequal variance (and noted how we canuse this version to perform a one sample t-test with no variance in the Ho sample).We now look at thepaired t-test.The uniqueness of this test is that for each subject in the sample we have two measures.For example, determining if the difference in scores between a pre- and post-test for a course in statisticsis significant would be an example of a paired t-test situation.Another example would be how people didin different subjects, for example is there a significant difference between scores in an English and
Statistics class?Note that we cannot just pair up measures – for example, attempting to pair a male andfemale test score is NOT an example of a paired situation (Lind, Marchel, & Wathen, 2008).The paired t-test t equals the mean difference between scores divided by the standard error of the meanof these differences.The important information we are interested in for paired situations lies in thedifference between scores.By focusing on the differences rather than the actual scores, we arecontrolling for the differences shown between subjects in the initial measure – how good or poor the initialscore was is not important, we only care about how much difference exists (Lind, Marchel, & Wathen,2008).Review: Choosing among the different versions of the t-test is fairly simple.If we have:One group with only 1 measure: use the one-sample t-test (or trick excel by using the two sampleunequal variance t-test with one variable equaling only the hypothesized value)Two groups with the same measure for each: use either the equal or unequal variance 2 samplet-testOne group that has 2 measures on each subject: use the paired t-test.Excel ExampleIn our equal pay case, we do have two measures for each employee that we could compare for meanequality – the salary and the midpoint.Each is measured in dollar units, so they are comparable.Thereare a couple of tests we could make.We could compare everyone’s salary and midpoint to see if theaverage were equal or not.We could also look at each gender individually – that is, are the male andfemale salary averages equal to their respective midpoint averages?We can also note that since none ofthe other variables have the same measurement scale, there are no other paired comparisons that wecould make.Question:Is the female mean salary greater than the female mean midpoint?Step 1: Ho: Female mean salary <= female mean midpoint(Note: Remember that the null always has tohave the = sign in it, since our question did not imply an equality, the question represents the alternativeand we back into the null as the opposite statement.)Ha: Female mean salary > female mean midpoint (Note: since we have a one way question that does notinclude an equal term in the question, the alternate is generally expresses the question.If the questionhad been mean salary greater than or equal to the mean midpoint, the null would have expressed thequestion: Ho: mean salary => mean midpoint, and Ha: mean salary < mean midpoint.)Step 2: Test to use: Paired t-test.(Note: we have two measures – salary and midpoint – on each person,thus it is paired data.)Step 3: Decision rule: If the p-value is < 0.05 and the t statistic is in the left side of the itsdistribution/mean, then we reject the null hypothesis.(Note: since we have a one tail test, we need toensure that the calculated salary mean statistic is in fact less than the midpoint mean; this translates to anegative t-value – our hint as to which side of the curve to use is given by arrow in the alternate pointingto the left side of the curve.In order for everything to line up correctly, the t-test needs to enter thevariables in the same order as listed in the hypothesis statements.)Step 4: Execution - Excel solution:Copy and paste the female salary and midpoint columns to a newspreadsheet (or to the spreadsheet with the question on the assignment file) – be sure to include thevariable names at the top of each column.In the data, Data Analysis button select the T-test: Paired TwoSample for Means option.In the variable 1 range, enter the salary range; in the variable 2 range, enterthe midpoint range.Click on the labels box (if you have labels in the input ranges).Click on the OutputRange button and then enter the upper left corner cell where you want the output to go.Clicking OK will
give you the following (note: for this example, the actual values came from an earlier version of the dataset – they do not match the data in your assignment file, but shouldbe close).t-Test: Paired Two Sample for MeansSalaryMidpointMean3834.88Variance334.6667223.5267Observations2525Pearson Correlation0.994941Hypothesized Mean Difference0df24t Stat4.177739P(T<=t) one-tail0.000168t Critical one-tail1.710882P(T<=t) two-tail0.000336t Critical two-tail2.063899Step 5: Conclusion - Reading and interpreting the table.The first thing to check is the label names – seethe light blue highlighted names.If we have numbers here, a data entry mistake has been made – theinput ranges did not include the variable names and the labels option was checked.This resulted in thefirst values being considered as the labels and 1 fewer data point in each variable for the analysis (seethe observation numbers).In deciding if we reject the null hypothesis or not, we have two things we can check initially – the p-valueand the comparison of the t stat with the t critical value.Our t stat (4.18 rounded) is more than the criticalvalue (1.71 for the one tail test we have) and the p-value (0.000168) is less than 0.05.So far, so good.Since we have a one tail test, we need one additional step (as discussed last week) – is the t stat in thecorrect tail for our alternate to be true.The arrow in the alternate points to the right tail and our t stat is inthe right tail (a positive value), so our conclusion is: Reject the null hypothesis – there is enough evidenceto say that the average salary is more than the average midpoint.Effect size.The effect size, Cohen’s d, for a paired t-test is D-bar (the average difference)/standard deviation.Alternately, this equals the t-value divided by the square root of n (Lakens, 2013) – a much easier way tocalculate the paired t-test effect size.So, since we rejected the null hypothesis for this test, we have d =4.178/square root of 25 = 4.178/5 = 0.84, a high value meaning the variable interaction was moreresponsible for the rejection of the null than the sample size – an outcome of practical significance.Analysis of Variance (ANOVA)ANOVA – the analysis of variance – is used when we have means for 3 or more groups that we want tocompare for mean equality at the same time.Why do we want another test for mean equality – why can’twe just use multiple t-tests comparing groups 2 at a time?There are two answers to this question.Thefirst is that using multiple t-tests can become somewhat burdensome.To compare 3 groups (a, b, and c)we would need 3 tests (a-b, b-c, a-c), to compare 4 groups (a, b, c, and d) we would need 6 tests (a-b, a-c, a-d, b-c, b-d, c-d), etc. – adding even a single group greatly increases the number of tests we need toperform (Lind, Marchel, & Wathen, 2008).Second, our confidence in the correctness of all the tests decreases the more tests we use.Recall thatwith an alpha of .05, we are 95% confident in our outcome – there is always the chance that the sampledid not accurately reflect the population.So, with 3 tests, our confidence is 95% with the first test, 95%with the second test, and 95% with the third test; when put together the probability of correctly rejectingthe null in all 3 cases is .95*.95*.95 = 0.857 – much below our desired level of 95%.With the 6 tests
needed for the 4 group comparison, our overall percent becomes 0.735.In order to keep our desiredconfidence in the results we need to perform all of the comparison in one test – the ANOVA (Lind,Marchel, & Wathen, 2008).ANOVA tests mean equality by comparing the variance of the entire data set with the average variance ofthe individual groups.Therefore, it assumes equal population variances for each group.This assumptionis critical, and should always be verified before using the results of an ANOVA analysis.If you have morethan two groups, compare the largest and smallest variances with the F test (discussed in week 2 forverifying equal variances in choosing the appropriate 2 sample t-test to use) outcome – if they are notsignificantly different, none of the others will be (Lind, Marchel, & Wathen, 2008).Just how does testing variance equality tell us that means are equal?The underlying logic says that if (1)the groups come from populations having equal variances AND (2) equal means, then the differentgroups will overlap each other if graphed and the entire data set should have the same mean andvariance as each of the samples (within statistical sampling error).If this is true, then the resulting F willbe small.If however, the groups have equal variances and UNEQUAL means, they will be spread outrather than overlap and the variance of the overall group will be greater than the average variance for theindividual groups and F will be large (Lind, Marchel, & Wathen, 2008).While ANOVA can be used for 2 groups (the results will be identical to those obtained with the twosample equal variance t-test), by convention we use the t for two groups and ANOVA for 3 or moregroups (Lind, Marchel, & Wathen, 2008).In our equal pay example, if we think that age is a reason fordiffering salaries among the grades, we could use ANOVA to see if the average ages for each grade levelare different.ANOVA requires that the different populations being tested have the same variance.This is tested byusing the F-test introduced last week.By testing the sample with the largest variance against the samplewith the smallest variance, we can see if the variances are statistically equal or not.If the extremes arenot significantly different, values between them will also not differ (Lind, Marchel, & Wathen, 2008).Real Life ExampleOne of the course instructors was charged with creating and starting supervisory training classes for acompany, The initial proposed 5-day basic supervisory program in a single week met with some initialresistance, as company had never done that before, and the Vice President of Human Resources did notthink the managers would accept having their supervisors gone for this period.So, a test was offered.The training would be initially scheduled in 3 different formats, and the managers could pick which theywanted their supervisors to attend – once a week, twice a week, or the full week.At the end, the pre- andpost-test scores – and change in scores - were evaluated to see if format made a difference.TheANOVA was used to verify that the pre-test results were the same – that is, the average knowledge levelof supervisors entering the program was not different between the sessions.The paired t-tests showedthat learning occurred in each session.At the end, the ANOVA was again used to see if a differenceexisted between the formats.It did, and this evidence was used to schedule the remainder of the classesin the full 5 day format.Types of ANOVA testsJust as the t-test has 3 versions (one-sample, two-sample, and paired) to choose from, ANOVA also hasseveral choices.The three approaches described in the text match those of the choices that Excel offers– chapter 5’s one way ANVOA is the same as Excel’s ANOVA single factor; chapter 6’s Factorial ANOVAis the same as Excel’s ANOVA: two factor with replication.The authors (Tanner & Youssef-Morgan,2013) make a somewhat confusing comment on p. 184: “Dependent groups ANOVA is not one of theoptions Excel offers.”The Excel option 2 factor ANOVA without replication, when used with eachexample in chapter 7 produce the same results with some additional information – it provides informationto test more than the dependent measures across the same subject.We will discuss this in more detail
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