252review 11 14 06 Open this document in Outline view The feedback that I get is that there are people who are absolutely sure that there is material we never covered on the final exam I also remember one very indignant woman who was absolutely sure that you couldn t compare two means unless one was a population mean This document may help M REVIEW Assume that 05 in all the following In each case be sure to clearly state the null hypothesis and the location of the do not reject or reject region Note that other methods that we have studied but which are not mentioned in this review are noted in boldface 1 Means The following data is used in the sections on means medians variances and distributions It relates to different methods of training people to pass an exam Method 1 74 88 82 93 55 70 Method 2 78 80 65 57 89 Method 3 68 83 50 91 84 77 94 Note x 3 x 1 720 462 x 2 3 x 2 1 81 92 36518 x 2 369 x 2 2 27879 59140 x1 77 00 s1 13 74 n1 6 x 2 73 80 s 2 12 72 n 2 5 x 3 80 00 s 3 13 87 n 3 9 These are only given for convenience you should be able to compute every one of these sums a Are the means for these three methods the same A multiple test of means is usually an Analysis of Variance For this to be valid we must assume that the samples come from populations with an underlying Normal distribution and common variance Note that data is not cross classified so that a 2 way ANOVA is not applicable H 0 1 2 3 H 1 Not all means equal Because there are many examples available the computation of SSB and SST is not shown here You should be able to do that computation When you are finished you should have an ANOVA table with the numbers below F Source SS DF MS F 05 H0 Between 126 2 63 0 34 F 2 17 3 59 ns Column means equal Within 3131 17 184 Total 3257 19 Explanation This is one way ANOVA because the columns are considered three independent random samples with no cross classification Since the Sum of Squares SS column must add up 3131 is found by subtracting 126 from 3257 Since n 20 the total degrees of freedom are n 1 19 Since there are 3 random samples or columns the degrees of freedom for Between is 3 1 2 Since the Degrees of Freedom DF column must add up 17 29 2 The Mean Square MS column is MSB found by dividing the SS column by the DF column 63 is MSB and 17 is MSW F and MSW is compared with F 05 from the F table df 1 2 df 2 17 We accept the null hypothesis if our computed F is less than or equal to 3 59 Because our computed F is less than the table F do not reject H 0 b Is the mean for method 1 above the mean for method 2 The data that we can use is repeated here x1 77 00 s1 13 74 s12 188 79 n1 6 x 2 73 80 s 2 12 72 s 22 161 80 n 2 5 Because the question does not include an equality it must be an alternative hypothesis Neither 77 nor 73 8 is a population mean so these numbers cannot be in the null hypothesis since a null hypothesis cannot include sample statistics We thus conclude that our alternative hypothesis is H 1 1 2 so that our null hypothesis must be the opposite H 0 1 2 The method taught in class assumes that the data comes from a Normal Distribution and that these are two samples from populations with equal variances We could do this problem using the test ratio method the critical value method or a one sided confidence interval Only the test ratio will be done here For the general formulas used see the formula table x x1 x 2 77 00 73 80 3 20 DF n1 n 2 2 6 5 2 9 s 2p n1 1 n2 1 n1 n2 2 s12 s22 5 188 79 4 161 80 943 95 647 20 176 79 9 9 176 69 1 1 6 5 Our hypotheses are 05 176 69 0 36667 9 t 05 1 833 s x s p 1 1 n1 n2 64 786 8 049 H 0 1 2 H 0 0 or Since this is a 1 sided alternate hypothesis you H 1 1 2 H 1 0 cannot use 2 sided confidence intervals or tests Note that 77 and 74 80 cannot appear in the hypotheses because they are sample means not population means Test Ratio t n is below t 1 x 0 3 20 0 9 0 398 The do not reject region is t t 05 Since our t s x 8 049 n2 2 9 t 05 1 833 do not reject the null hypothesis 2 c Is the mean for method 3 above 78 The data that we can use is repeated here x 3 80 00 s 3 13 87 n 3 9 H 0 3 78 Note that because the question involves an inequality it is an alternative hypothesis H 1 3 78 13 87 4 623333 9 n We use only a Test Ratio here but remember that all such problems can be done using a critical value or a sx s one sided confidence interval t x 0 80 78 0 4326 The do not reject region is sx 4 62333 n 1 8 8 t t 05 Since our t is below t t 05 1 860 do not reject the null hypothesis Remember for comparing means or medians of 2 samples if the parent distribution is Normal use methods above for means If the parent distribution is not Normal use the Wilcoxon Signed Rank test or the Wilcoxon Mann Whitney test If the samples are independent use Wilcoxon MannWhitney test or a method that is appropriate for comparing means of independent samples If the data is cross classified use a method for means of paired data or the Wilcoxon Signed Rank test 2 Medians a Are the medians for these three methods the same Since this refers to medians instead of means and if we assume that the underlying distribution is not Normal we use the nonparametric rank test analogue to ANOVA the Kruskal Wallis Test Note that data is not cross classified so that the Friedman Test is not applicable The null hypothesis is H 0 Columns come from same distribution or medians are equal The data are repeated in order The second number in each column is the rank of the number among the 20 numbers in the three groups Method 1 Method 2 Method 3 55 2 57 3 50 1 70 6 65 4 68 5 74 7 78 …
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