Can you write it in text please so I can copy and paste it!! Thank you 7. For the preceding problem you should find that thereare significant differences among the three treatments.The primary reason for the significance is that the meanfor treatment III is substantially larger than the meansfor the other two treatments. To create the followingdata, we started with the values from problem 6 andsubtracted 3 points to each score in treatment III. Noticethat subtracting a constant causes the mean to changebut has no influence on the variability of the sample.In the resulting data, the mean differences are muchsmaller than those in problem 6.TreatmentII33N = 1257G = 48317EX = 236%3D13M = 3M = 4М — 5T = 12T = 16T = 20SS = 8SS = 12SS = 16%3Dy b. Use an ANOVA with a = .05 to determinewhether there are any significant differencesamong the three treatment means. (Does youranswer agree with your prediction in part a?)

a) Applying ANOVA test for a single factor, Null and alternative hypothesis: Ho: µ1 = µ2 = µ3 (All means are equal) Ha : Not all means are equal. Calculations for SS(between) and SS(within):SS(between) = 8 SS(within) = 36 k = no of treatments = 3 N = total sum = 4+4+4 = 12 Degree of freedom (df) between = df(between) = k-1 = 3-1 = 2 Degree of freedom (df) within = df(within) = N-k = 12-3 = 9 Degree of freedom (df) total = df( total ) = N-1 = 12-1 = 11 MS(between) = SS(between)/df(between) = 8/2 = 4 MS(within) = SS(within/df(within) = 36/9 = 4 Test statistic (F) = MS(between)/MS(within) = 4/4 = 1df(numerator) = df(between) = 2df(denominator) = df(within) = 9For 0.05 significance level,The test statistic follows the F-distribution with (2, 9) degrees of freedom.F critical =F(df(numerator) ,df(denominator)) = 4.2565 (from F distribution table) b) Decision criteria : Reject Ho,if Test statistic (F) < F critical.Since test statistic (F)< F critical we fail to reject the null hypothesis.The result is not significant at the 0.05 significance level.