$FL2@(#) IBM SPSS STATISTICS DATA FILE MS Windows 20.0.0 ##########################Y@18 Jul 1215:43:51 ###########################PRETEST ########################POSTTEST########################GROUP ########################IQ ##############?#No treatment control #######@#reading intervention ########################################################################## ############ ####################################################### #######3###PRETEST=Pretest POSTTEST=Posttest GROUP=group IQ=IQ############################################M###Pretest:$@Role('0' )/Posttest:$@Role('0' )/group:$@Role('0' )/IQ:$@Role('0' )############ ###windows-1252#######deeeeeeee eeeeee ? f ff fff f ? fff#####d@ff#####`c@ \f\f\f$FL2@(#) IBM SPSS STATISTICS MS Windows 22.0.0.0 ################$#########Y@26 Jun 1613:52:10 ###########################PRETEST ########################POSTTEST########################GROUP ##############? #a1 #######@#a2 #######@#a3 ######################################################################### ############ ##################################################-###PRETEST=Pretest POSTTEST=Posttest GROUP=Group########################$###################=###Pretest: $@Role('0' )/Posttest:$@Role('0' )/Group:$@Role('0' )################UTF8#######eeeee}eeeewqenexzeffffff{lfofofxifqjfne fggggg}ggg|g{tg}gxsg#### #4 Data Drug A Low a2 Control a1 Mean Count Variance Std Dev Total SS Calc Sum 10 13 17 20 15 4 19.3333 4.3970 Drug B Low a4 High a3 11 17 15 21 18 26 22 28 Grand Mean 16.5 23 15.25 4 4 20 21.6667 24.6667 4.6547 4.9666 a1 a2 a3 a4 a5 27.5625 52.5625 150.0625 18.0625 3.0625 5.0625 10.5625 68.0625 0.0625 33.0625 3.0625 0.5625 27.5625 7.5625 115.5625 22.5625 14.0625 10.5625 45.5625 162.5625 SS total 58.25 77.75 256.25 71.25 314.25 777.75 Group SS Calc a1 Source df 8 12 16 19 13.75 4 22.9167 4.7871 High a5 a2 a3 0.25 9 SS Treatment 3 7 10 12 8 4 15.3333 3.9158 a4 210.25 MS a5 6.25 SS treatment 240.25 466 SS error 311.75 F 4 466 116.5000 Error 15 311.75 20.7833 Total 19 777.75 5.605 Anova: Single Factor (Check with Excel) SUMMARY Groups a1 a2 a3 a4 a5 Count Sum 4 4 4 4 4 60 55 32 66 92 Average 15 13.75 8 16.5 23 Variance 19.33333 22.91667 15.33333 21.66667 24.66667 ANOVA Source of Variation Between Groups Within Groups 466 311.75 MS F P-value F crit 4 116.5 5.605453 0.0057825 3.055568 15 20.783333 Total 777.75 19 SS df 8.1 Rested Deprived 11.8 13.8 5.3 12.3 8.8 11.5 14.8 12.8 11.8 17 16.3 15 10 15.3 14 11.5 Grand Mean 11.6 13.65 12.625 8 8 16 12.608571 3.94 Mean Count Variance Rested Deprived 0.680625 1.380625 53.655625 0.105625 14.630625 1.265625 4.730625 0.030625 0.680625 19.140625 13.505625 5.640625 6.890625 7.155625 SS Total 1.890625 1.265625 132.65 Group SS Calc Rested Deprived SS treatment 8.405 8.405 16.81 SS error 115.84 Source df Treatment MS F 1 16.81 16.8100 Error 14 115.84 8.2743 Total 15 132.65 8.1b) Relative Difference related to variability d -0.713 2.8765 R2 0.127 2 0.06 Sum Deprived 0.04 0.0225 39.69 1.8225 7.84 4.6225 10.24 0.7225 0.04 11.2225 22.09 1.8225 2.56 2.7225 5.76 4.6225 88.26 27.58 S pooled 2.8765058 d -0.71267 Total SS Calc SS Rested 2.03160 8.5 2 Could not get book answer Could not get book answer Power Could not get book answer 8.6 2 -0.0072 Book answer = -.007 =(127.8 - (3*59.7))/(7053 + 5 #4e) i) Control vs. all 4 test groups Coefficients a1 a2 a3 a4 a5 1 -0.25 -0.25 -0.25 -0.25 L = control mean * 1 + (a2 mean * -1/4) + ... L -0.3125 ai^2/ni 0.3125 sL 2.54849 t -0.1226 SSB 0.3125 Fcritical for dfnumerator = 1 and df denominator = 15 at =.05 F 0.02 4.54 NOT critical Coefficients ii) Drug A vs. Drug B. a1 a2 a3 0 0.5 L -8.875 ai^2/ni 0.25 sL 2.27944 t -3.8935 SSB 315.063 F 15.16 Coefficients Coefficients 0.5 a5 -0.5 -0.5 N = 16 a = 4 so N - a = 12 Fcritical for dfnumerator = 1 and df denominator = 12 at =.05 4.75 Critical! iii) Low vs. High for Drug A a1 a2 a3 0 0.5 L 2.875 ai^2/ni 0.125 sL 1.61181 t 1.78371 SSB 66.125 F 3.18 a4 a4 -0.5 a5 0 0 N = 8 a = 2 so N - a = 6 Fcritical for dfnumerator = 1 and df denominator = 6 at =.05 5.99 Not Critical iv) Low vs. High for Drug b a1 a2 a3 0 0 a4 0 a5 0.5 -0.5 L -3.25 ai^2/ni 0.125 sL 1.61181 t -2.0164 SSB 84.5 F 4.07 4e) i N = 8 a = 2 so N - a = 6 Fcritical for dfnumerator = 1 and df denominator = 6 at =.05 5.99 Not Critical ii iii iv 0.3125 315.0625 66.125 Sum 84.5 466 = SSA =(127.8 - (3*59.7))/(7053 + 59.7) = 15 at =.05 = 12 at =.05 = 6 at =.05 = 6 at =.05 #6.1 a) Bonferroni Pooled Variance Variance a1 14.500 a2 17.188 a3 11.500 a4 16.250 15.5875 Pooled s /2k 3.9481 0.01 t with 16 degrees of freedom and .01 significance (one-tailed) 3.2520 F 9.0786823438 Scheffe F at .05. 4, 16 3.01 6.2 Harmonic mean 8 MSE 20.7833333333 tDunnett for 16 df and p .05 2.73 D Dunnett 6.2228631875 k-1 MSE (1/n1 + 1/n2) 4 20.783333 0.5 F' 11.185511 a5 18.500 4.1-a,b,c,d,e 6.1 - a & c 6.2 - a & b 6.3 - a & c 8.1 8.5 8.6 See included Excel for calculations. 1) Determine the Grand Mean (mean of ALL the data) = 15.25 2) Find each \"treatment mean\" (including the control) a1 = 15, a2 = 13.75, a3 = 8, a4 = 16.5, a5 = 23 3) Find the total variability in the response variability. = 777.75 = SStotal This means the sum of (each element in the table - the grand mean)2 4) Determine how much of the variability is account for by our treatment. Find SS treatment This means the Sum of: the count of each group (4 in this problem) * (that treatment's mean - the grand mean)2 SS1 = 0.25 SS2 = 9 SS3 = 210.25 SS4 = 6.25 SS5 = 240.25 SStreatment = 465.75 5) Finally, we need to determine how much variability is left over. SS error = SStotal - SStreatment = 312 6) Determine the degrees of freedom (df): Total = N (total number of all observations) - 1 = 19 Treatment: Number of \"treatment\" groups = 5 - 1 = 4 Error: Total - Treatment = 19 - 4 = 15 7) Fill in the ANOVA table: Source df SS MS F Treatment k-1=4 466 116.5 5.605 Error 19 - 4 = 15 311.75 20.7833 Total N - 1 = 19 777.75 8) Find the Mean Squares = SS/Df MStreatment = 116.5 MSerror = 20.7833 9) Finally we calculate the F stat for our overall case: = 116.5 / 20.7833 = 5.605 (rounds to 5.61) 10) Use an F table to determine if this is critical: F(0.05, 4,15) = 3.06 (see below) Since our F is greater than the Fcritical, this is significant with = 0.05. 4b) i) I = 1 - (2 + 3 + 4 + 5)/4 ii) (2 + 3)/2 - (4 + 5)/2 iii) 2 - 3 iv) 4 - 5 => a1 = 1 a2 = a3 = a4 = a5 = -1/4 => a1 = 0 a2 = a3 = a4 = - a5 = - => a1 = 0 a2 = 1 a3 = -1 a4 = 0 a5 = 0 => a1 = 0 a2 = 0 a3 = 0 a4 = 1 a5 = -1 4c) Mutually Orthogonal => #1 Coefficients sum to zero i) 1 ---- =0 ii) 0++--=0 iii) 0+1-1+0+0=0 iv) 0+0+0+1-1=0 #2 Sum of products of the coefficients in pairwise comparisons = 0. i) vs. ii) (1*0) + (- * ) + (- * ) + (- * -) + (- * -) = 0 i) vs. iii) (1*0) + (- * 0) + (- * 1) + (- * -1) + (- * 0) = 0 i) vs. iv) (1*0) + (- * 0) + (- * 0) + (- * 1) + (- * -1) = 0 ii) vs. iii) (0*0) + ( * 0) + ( * 1) + (- * 1) + (- * 0) = 0 ii) vs. iv) (0*0) + ( * 0) + ( * 0) + (- * 1) + (- * -1) = 0 iii) vs. iv) (0*0) + (0 * 0) + (0 * 0) + (1 * 0) + (-1 * 0) = 0 d. Find the sum of squares associated with these comparisons and test their significance. See Excel for calcs and critical comparisons 4e) i 0.312 5 ii 315.062 5 iii 66.125 iv 84.5 Sum 466 = SSA 6.1 a) (Bonferroni) Not getting book answer 6.2 Not getting book answer 8.1 a) 1) Determine the Grand Mean (mean of ALL the data) = 12.65 2) Find each \"treatment mean\" Rested = 11.6 Deprived = 13.65 3) Find the total variability in the response variability. = 132.65 = SStotal This means the sum of (each element in the table - the grand mean)2 4) Determine how much of the variability is account for by our treatment. Find SS treatment This means the Sum of: the count of each group (4 in this problem) * (that treatment's mean - the grand mean)2 SS1 = 8.405 SS2 = 8.405 SStreatment = 16.81 5) Finally, we need to determine how much variability is left over. SS error = SStotal - SStreatment = 115.84 6) Determine the degrees of freedom (df): Total = N (total number of all observations) - 1 = 15 Treatment: Number of \"treatment\" groups = 2 - 1 = 1 Error: Total - Treatment = 15 - 1 = 14 7) Fill in the ANOVA table: Source df SS MS F Treatment k-1=1 16.81 16.8100 Error 15 - 1 = 14 115.84 8.2743 Total N - 1 = 15 132.65 2.0316 8) Find the Mean Squares = SS/Df MStreatment = 16.81 MSerror = 8.2743 9) Finally we calculate the F stat for our overall case: = 16.81/8.2743= 2.0316 (rounds to 2.03) 10) Use an F table to determine if this is critical: F(0.05, 1,15) = 4.60 (as before) Since our F is less than the Fcritical, this is NOT significant with = 0.05. 8b) Relative difference related to variability (Cohen's d) (M1 and M2 below indicate (means)) See Excel for Calcs d = -0.713 Total observed variability attributed to the treatment => R2 See Excel for Calcs R2 = 0.127 => 12.7% of the variation is attributed to the treatment Estimate of this value in the population: 2 = 0.06 Note: SSEffect => SStreatment and dfEffect = dftreatment 8c) We can increase the sample size or perform a one-tailed type of test rather than the two-tailed test that is inherent in ANOVA. Both of these will increase the power (1 - ) of the test without lowering the . 8.5 2 = could not get book answer = could not get book answer Power = could not get book answer 8.6 2 = -0.0072 Since 2 is negative, we round it to zero. Since there is no apparent variation, expanding the study will not, generally, yield statistically significant results. $FL2@(#) IBM SPSS STATISTICS MS Windows 22.0.0.0 ################$#########Y@26 Jun 1613:52:10 ###########################PRETEST ########################POSTTEST########################GROUP ##############? #a1 #######@#a2 #######@#a3 ######################################################################### ############ ##################################################-###PRETEST=Pretest POSTTEST=Posttest GROUP=Group########################$###################=###Pretest: $@Role('0' )/Posttest:$@Role('0' )/Group:$@Role('0' )################UTF8#######eeeee}eeeewqenexzeffffff{lfofofxifqjfne fggggg}ggg|g{tg}gxsg####