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STAT 3130 Statistical Methods I Session 2 One Way Analysis of Variance (ANOVA) ANOVA In Statistical Methods I, we covered several different forms of t-tests. For example, we used ttests to answer questions such as: Are cars going over the speed limit on this residential street? Did students who took a preparatory course, score better on the standardized exam than students who did not take the course? ANOVA The general setting for all of these questions is that some quantitative variable (speed, test scores) has been measured for one or two categories of subjects. What if we have more than two categories across which we want to compare the value of some quantitative variable? For example, lets say that we wanted to compare the mean weight loss of subjects who were put on one of four diet plans. For ease of discussion, lets call these plans A, B, C and D. ANOVA The following approach would be tempting… H0: Plan A = Plan B H0: Plan B = Plan C H1: plan A Plan B H1: plan B Plan C H0: Plan C = Plan D H0: Plan A = Plan C H1: plan C Plan D H1: plan A Plan C H0: Plan A = Plan D H0: Plan B = Plan D H1: plan A Plan D H1: plan B Plan D ANOVA …but wrong. Apart from being very cumbersome, there is a critical problem – we are inflating our probability of making a type 1 error. Think about that – lets use alpha = .05. If we ran 6 separate tests, that would generate a cumulative probability of a type 1 error of .3. We could lower the alpha value to .05/6 – I hear you saying. But this has its own problems – what happens if the number of tests increase to 8 or 10? Our alpha value would become so low, we would almost never reject the null (recall Power). ANOVA What we need is a single test which will allow us to evaluate all of the relationships simultaneously while using a reasonable alpha level. The test needs to be able to provide information regarding differences in the mean values of multiple subject groupings. To accomplish this, we use the Analysis of Variance or ANOVA procedure. ANOVA Lets discuss how to use ANOVA to test a hypothesis by returning to our dieters… In this instance there are four levels (diet plans) to a single factor (weight loss). The hypothesis statements would look like this: H0: All level means are equal. In other words, all four of the diet plans generate approximately the same amount of weight loss. H1: Not all of the level means are equal. In other words, at least one of the plans’ weight loss mean is statistically significant different from the other plans’ means. ANOVA Prior to executing the test, we must check for three important assumptions about our data: 1. All the groups are normally distributed. 2. All the populations sampled have approximately equal variance (you can check this by generating side-by-side boxplots). The rule of thumb is that the largest std is <2x the smallest std. 3. The samples of the groups are independent of each other and subjects within the groups were randomly selected. As with most, but not all, statistical tests, if our samples are large, we can relax our assumptions and work around non normal data. ANOVA Lets examine the hypothesis statements in more detail: H0: µa = µb = µc = µd H1: µa ≠ µb ≠ µc ≠ µd Consider – what would the hypothesized distributions look like under H0 and H1? ANOVA Ok. We understand the concept, we have the hypotheses, we have the assumptions – we need a test statistic. In ANOVA, we use the F-distribution. In the science of statistics, whenever you need to evaluate a ratio of variances you will be using an F-statistic. The ratio in question here is: The variation BETWEEN the groups The variation WITHIN the groups Question – what kind of value would indicate difference versus no difference? ANOVA The result of this ratio is the F-statistic. You can see the FDistribution on page 675 of your book. As the number of groups and observations increases, the distribution will start to appear normal. Lets start working with an example… ANOVA Returning to the diet plans… PLAN Mean PLAN A 14 14 20 22 26 27 20.50 PLAN B 15 18 23 25 28 30 23.17 PLAN C 32 36 40 42 45 45 40.00 PLAN D 33 38 42 44 46 47 41.67 OVERALL MEAN 31.33 ANOVA Our hypotheses statements would be: H0: The four diets plans have the same results (the mean weight loss is the same) H1: At least one of the diet plans has a different result (the mean weight loss is different) We will now calculate our test statistic: The variation BETWEEN the groups The variation WITHIN the groups ANOVA To calculate the F-Statistic, we use the following table (refer to page 682 in your book): SOURCE SUM OF SQUARES DEGREES OF FREEDOM MEAN SQUARE F-stat BETWEEN SSB # levels – 1 SSB/(# levels – 1) {SSB/(# levels – 1)} {SSW(n- # levels)} WITHIN SSW n- # levels SSW/(n- # levels) TOTAL SST (SSB + SSW) n-1 ANOVA For those who are interested: SST _ = SSW _ + SSB _ ij(Xij-X)2 = ij(Xij-Xj)2 + nj(Xj-X)2 ANOVA For the present problem: SOURCE 1 SUM OF SQUARES DEGREES OF FREEDOM MEAN SQUARE BETWEEN1 2195.67 3 731.89 WITHIN2 601.67 20 30.08 TOTAL 2797.34 23 SSB = 6(10.832 + 8.172 + 8.672 +10.332) 2SSW = (159.50 + 166.83 + 134 + 141.33) = 601.67 F-stat 24.33 ANOVA Now…what to do with an F-statistic of 24.33? This is a fairly strong statistic – recall that as the variance ratio approaches 1, the null is true. As the variance ratio grows larger than 1, we can more confidently reject the null. As with all test statistics, this result will translate into a p-value. The p-value associated with this statistic is less than .001. Based upon this result, we can confidently reject the null hypothesis and conclude that at least one of the results is different. Lets execute this same problem in SAS…