One-way Analysis of VarianceQuiz 2 Name ANSWER KEYStat 217SHOW YOUR WORK! No work, no credit. Here’s some handy dandy formulas:SST=SSG+SSE MS=SS/DF F=MSG/MSEA pair of sociologists are interested in the effects of child abuse on the child’s verbal IQ . Three groups of children take part in a study: “abuse”, “neglect”, or “control.” Being in the “abuse” group is confirmed by an investigator; being in the “neglect” group is based on legal findings with regard to lack of adequate care; and the “control” group is selected from out-patients at a clinic. The MINITAB output is:One-way Analysis of VarianceAnalysis of Variance for IQ Source DF SS MS F PGroup 2 1090.9 545.5 ***** 0.000Error 61 ****** ****Total 63 3084.1 Individual 95% CIs For Mean Based on Pooled StDevLevel N Mean StDev ----------+---------+---------+------Abuse 32 79.091 6.767 (---*---) Neglect 16 78.087 4.280 (-----*-----) Control 16 88.244 4.463 (----*-----) ----------+---------+---------+------Pooled StDev = ***** 80.0 85.0 90.0(a) Give an estimate of 3 244.883x is our best guess for the mean average verbal IQ score of non-abused, non-neglected children.(b) Give the value of SSE, MSE and the F statistic. SSE = SST – SSG = 3084.1 – 1090.9 = 1993.2MSE = SSE/DFE = 1993.2/61 = 32.68F = MSG/MSE = 545.5/32.68 = 16.69(c) Give an estimate of . is estimated by 72.568.32 MSE(d) Check the assumption that the population standard deviations are the same (SHOW YOUR WORK!). Using the rule of thumb from our textbook, since none of the sample standard deviations s1 = 6.767, s2 = 4.28 and s3 = 6.463 are more than twice the others, then we may assume that the population standard deviations are equal, 321.-10 0 10-2.5-2.0-1.5-1.0-0.50.00.51.01.52.02.5Normal ScoreResidualNormal Probability Plot of the Residuals(response is IQ)(e) Check the assumption that the residuals are normal. Explain what you’re checking.Since the residuals in the normal probability plot follow a straight line, then we will assume that the residuals are normally distributed,  ~ N(0,).(f) Conduct the overall test: State hypotheses to be tested using valid statistical notation. H0: 1=2=3Ha: not all of the i ‘s are equal Circle One: Which of the following is evidence against Ho (and for Ha) in this test?A. When the variability among the group means is large compared to the variability within the groups.B. When the variability among the group means is similar to the variability within the groups.C. When the variability among the group means is small compared to the variability within the groups.D. When the variability among the group means is equal to the variability within the groups. What is the distribution of the test statistic (given that HO is true),? F ~ F(DFG,DFE) = F(2,61) What is the p-value?The p-value is so small that MINITAB reports it as 0.000. Make a decision at the 0.05 significance level. Since the p-value = 0.000 < .05, then we REJECT H0 in favor of the alternative, Ha Make a conclusion in terms of the problem. There is an effect of the level of abuse on children’s mean verbal IQ scoreORThere is a significant difference between children’s mean verbal IQ scoredepending on the level of