g nowIn an effort to counteract student cheating, the professor of a large class created four versions of a midterm exam, distributing the four versions among the 304students in the class. After the exam, some students from the class got together and petitioned to nullify the results on the grounds that the four versions werenot equal in difficulty.To investigate the students' assertion, the professor examined the means and variances of the scores on the different versions of the exam, obtaining thefollowing information (the exam had 200 possible points).GroupVersion AVersion BVersion CVersion DSample Sample Samplesizemeanvariance76426.476428.076541.476416.2Send data to calculator V157.1147.2151.5159.3Taking the 76 scores for each version of the exam as a sample of scores for that version, the professor performed a one-way, independent-samples ANOVA testof the equality of the population mean scores for the four versions.Answer the following, carrying your intermediate computations to at least three decimal places and rounding your responses to at least one decimal place.ExplanationSend data to Excel(a) What is the value of the mean square for treatments (the"between groups" mean square) that would be reported in theANOVA test?(b) What is the value of the mean square for error (the "withingroups" mean square) that would be reported in the ANOVAtest?CheckO Search0W GroupVersion AVersion BVersion CVersion D3°Faining nowSample SampleSamplesizemean76767676ExplanationSend data to calculator V157.1147.2151.5159.3Samplevariance426.4428.0541.4416.2Taking the 76 scores for each version of the exam as a sample of scores for that version, the professor performed a one-way, independent-samples ANOVA testof the equality of the population mean scores for the four versions.Answer the following, carrying your intermediate computations to at least three decimal places and rounding your responses to at least one decimal place.CheckSend data to Excel(a) What is the value of the mean square for treatments (the"between groups" mean square) that would be reported in theANOVA test?(b) What is the value of the mean square for error (the "withingroups" mean square) that would be reported in the ANOVAtest?XO Search0S0HWEspañol?0

Given Information: Group Sample size Sample mean Standard variance Version A 76 157.1 426.4 Version B 76 147.2 428.0 Version C 76 151.5 541.4 Version D 76 159.3 416.2First, find the grand mean: X¯¯=X¯A+X¯B+X¯C+X¯D4=157.1+147.2+151.5+159.34=153.775 The sum of squares between groups is computed as follows: SSbetween=nAX¯A-X¯¯2+nBX¯B-X¯¯2+nCX¯C-X¯¯2+nDX¯D-X¯¯2=76157.1-153.7752+76147.2-153.7752+76151.5-153.7752+76159.3-153.7752=840.2275+3285.5275+393.3475+2319.9475=6839.05 The sum of squares within groups is computed as follows: SSwithin=∑XAi-X¯A2+∑XBi-X¯B2+∑XCi-X¯C2+∑XDi-X¯D2 where, ∑XAi-X¯A2=SA2nA-1=426.476-1=31980 ∑XBi-X¯B2=SB2nB-1=428.076-1=32100 ∑XCi-X¯C2=SC2nC-1=541.476-1=40605 ∑XDi-X¯D2=SD2nD-1=416.276-1=31215 This implies: SSwithin=∑XAi-X¯A2+∑XBi-X¯B2+∑XCi-X¯C2+∑XDi-X¯D2=31980+32100+40605+31215=135900 Total sum of squares is computed as: SStotal=SSbetween+SSwithin=6839.05+135900=142739.05 Degrees of freedom are computed as follows: dfbetween=K-1=4-1=3 dfwithin=N-K=476-4=300 dftotal=N-1=476-1=303 (a) The value of mean square for treatments (between groups) is: MSbetween=SSbetweendfbetween=6839.053=2279.683333≈2279.683 (b) The value of mean square for error (within groups) is: MSwithin=SSwithindfwithin=135900300=453