Ss error ss within ss subject 13275 6787 6488 source

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SS error = SS within – SS subject 132.75 – 67.87 = 64.88 Source SS Df Ms F ratio Between 154.12 3 51.37 16.62 Within 132.75 28 Subject 67.87 7 Error 64.88 21 3.09 total 286.87 31 (b) Whenever appropriate, estimate effect sizes with □ 2 p and with d, and conduct Tukey’s HSD test. N 2 p = 154.12 154.12 + 64.88 = 154.12 2.19 = 0.70 Tukey’s HSD Test: HSD = q √ MS error n K = 4 Df error = 20 Q = 3.96 HSD = 3.96 √ 3.09 8 = 3.96 [0.62] = 2.46 X0=1.63 X1=3.13 X2=5 X3=7.5 X0=1.63 1.50 3.37 5.87 X1=3.13 1.87 4.37 X2=5 2.50 X3=7.5
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TOPIC 6 EXERCISES D = x 1 x 2 MS error D [x3, x0] = 5.87 3.09 = 5.87 1.76 = 3.34 D [x3, x1] = 4.37 3.09 = 4.37 1.76 = 2.48 D [x3, x2] = 2.50 3.09 = 2.50 1.76 = 1.42 D [x2, x0] 3.37 3.09 = 3.37 1.76 = 1.91 Chapter 17 question 17.7 Recall the experiment described in Review Question 16.11 on page 314, where errors on a driving simulator were obtained for subjects whose orange juice had been laced with controlled amounts of vodka. Now assume that repeated measures are taken across all five conditions for each of five subjects. (Assume that no lingering effects occur because sufficient time elapses between successive tests, and no order bias appears because the orders of the five conditions are equalized across the five subjects.) (a) Summarize the results in an ANOVA table. If you did Review Question 16.11 and saved your results, you can use the known values for SS between, SS within, and SS total to short-circuit computations. SS between = 523.36 SS within = 388.40 SS total = 911.76 SS subject = ∑ T 2 subject - G 2 n
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TOPIC 6 EXERCISES [ ( 46 ) 2 5 + ( 36 ) 2 5 + ( 25 ) 2 5 + ( 50 ) 2 5 + ( 34 ) 2 5 ] - ( 191 ) 2 25 [ 2116 5 + 1296 5 + 625 5 + 2500 5 + 1156 5 ¿ 36481 25 1538.6 – 1459.24 = 79.36 SS error = SS within – SS subject 388.40 – 79.36 = 309.04 df between = 4 df within = 20 df total = 24 df subject = n-1 = 5-1 = 4 df error = df within – df subject = 20 -4 =16 Source SS Df Ms F ratio Between 523.36 4 103.84 6.77 Within 388.40 20 Subject 79.36 4 Error 309.04 16 19.32 total 911.76 24 (b) If appropriate, estimate the effect sizes and use Tukey’s HSD test. HSD = q √ MS error n K = 5 df error = 16 q = 4.33 Ms error = 19.32 N = 5
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TOPIC 6 EXERCISES HSD = 4.33 19.32 5 = 4.33 (1.77) = 8.53 X0 = 3 X1 = 4 X2 = 5.20 X4 = 11.20 X6 = 14.80 X0 = 3 1 2.20 8.20 11.80 X1 = 4 1.20 7.20 10.80 X2 = 5.20 6.20 9.60 X4 = 11.20 3.60 X6 = 14.80 d= x 1 x 2 MS error d (x6, x3) = 11.80 19.32 = 11.80 4.40 = 2.68 d (x6, x4) = 10.80 19.32 = 10.80 4.40 = 2.46 d (x6,x2) = 9.60 19.32 = 9.60 4.40 = 2.18 All values of d are > than 0.80, so all of these are significantly different pairs of means. Chapter 17 question 17.8 While analyzing data, an investigator treats each score as if it were contributed by a different subject even though, in fact, scores were repeated measures. What effect, if any, would this mistake probably have on the F test if the null hypothesis were: (a) True? When the researcher notes the null hypothesis is true, both estimates of between and within variability would reflect only random error.
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  • Fall '14
  • dargon
  • Null hypothesis, Cohen, researcher, Statistical significance, Statistical power

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