12.56 What test can be used to assess whether there areoverall differences in mean HDL cholesterol among thethree treatment groups?12.57 State the hypotheses to be tested.12.58 What is your best estimate of the variance ofHDL cholesterol within a treatment group?12.59 What is the test statistic to perform the test inProblem 12.56 based on the data in Table 12.40? Table 12.40 Mean HDL cholesterol by treatment groupGroupMean(mg/dl) sd nA 53.9 17.2 31B 52.8 14.8 31C 59.0 14.4 30Overall 55.2 15.6 9212.60 What is the two-sided p-value associated with thetest statistic in Problem 12.59?I need detailed answer for 12.60 please!!!! HypertensionRefer to Table 12.15 (on page 593). A similar two-wayANOVA was run using PROC GLM of SAS comparingmean diastolic blood pressure (DBP) by study group andsex. The results are given in Table 12.41.12.62 Summarize the findings in a few sentences. Table 12.41 is attached in pictures.THANK YOU! TABLE 12.41 SAS GLM procedure output illustrating the effects of study group and sex on DBP using thedata set in Example 12.23 (on page 589)DEPENDENT VARIABLE: MDIASSOURCEMODELERROR745CORRECTED TOTAL 748SOURCESTUDYSEXPARAMETERINTERCEPTSTUDYSEXDF SUM OF SQUARES3DF21SVLVNORMALEFEMALE48186.9927009489199.44205496137386.43475590SASGENERAL LINEAR MODELS PROCEDURETYPE I SS45269.885091532917.10760942STUDYSVLVNORSEXMALEFEMALEESTIMATE76.47708914-17.30065001-10.653023920.000000003.983995820.00000000MEAN SQUARE F VALUE134.1516062.33090031119.73079470F VALUE189.0524.36PR > F0.00010.00010.00010.0001SVMALE0.0001PROB> |T|PROB> |T|LV0.0001T FOR HO:PARAMETER=0115.61-19.40-7.23DF210.00014.94FEMALE0.0001PR > F0.0001ROOT ME10.94215677TYPE III SS46573.928189032917.10760942NOR0.00010.0001PR > T0.00010.00010.0001R-SQUARE0.3507410.0001C.V.15.0906MDIAS MEAN72.50972853F VALUE PR > F194.49 0.000124.36 0.0001STD ERROR OFESTIMATE0.661529120.891747161.472589210.80713389

[Note: when we have multiple questions and we are solving only 1st question as per our guidelines]The sample size and summary statistics mean and standard deviation of HDL cholesterol are given below,GroupMean (mg/dl)SdnA53.917.231B52.814.831C5914.430Overall55.215.692Grant means =55.2and total sample size n =n1+n2+n3 = 31+31+30=92no of treatment = p=3a) We want to test whether there are overall differences in mean HDL cholesterol among the three treatment groups.The appropriate test for the above scenario is the one-way ANOVA test Anova test is used to compare the means of more than two variables.b) The test statistics are,H0: There are no significant differences in mean HDL cholesterol among the three treatment groups.i.e. H0: Vs H1: There are significant differences in mean HDL cholesterol among the three treatment groups.i.e. H1: at least one is different.Now calculate the Anova table degree of freedom is For treatment = p-1 =2For error = n-p = 92-3 =89Total df = n-1 = 92-1 = 91The sum of square for the treatment = where ni is the sample size of ith treatment. is the sample meant of ith treatment.Now SStreat = 31(53.9-55.2)2+31(52.8-55.2)2+31(59-55.2)2 = 52.39+178.56+433.2 = 664.15SStreat = 664.15For errorSSE = where xij is the jth observation of ith treatment.SSE = ............................Sd = = (31-1)(14.4)2+(31-1)(14.8)2+(30-1)(17.2)2 = 8875.2+6571.2+6013.44 = 21459.84SSE = 21459.84SSTotal = SStreat+SSE = 664.15+21459.84 = 22123.99Now mean squareMSteat= MSE= F ratio is,F=The anova table is,source of variationDfSSMSFTreatment2664.15332.0751.3772Error8921459.84241.1218Total9122123.99c) The MSE is the best estimate of the variance of HDL cholesterol within a treatment group.The variance within the group is a measure of how much individual values within that group deviated from their group means.MSE = 241.1218d) The test statistic is,F= 1.3772e) the p-value is,P-value = 2P(F>Fstatistic)Where F follows F-distribution with df1 is df of treatment and df2 is df of errorP-value = 0.257607 0.2576 ............... using Excel function =F.DIST.RT(x,df1,df2)f) Decision :consider level of significance = 0.05If the p-value is less than or equal to >α then reject H0 otherwise reject to fail H0P-value > αReject to fail H0 at level 0.05There are no significant differences in mean HDL cholesterol among the three treatment groups.a) The Anova test is appropriate to test whether there are overall differences in mean HDL cholesterol among the three treatment groups. b) The hypothesis is,H0: There are no significant differences in mean HDL cholesterol among the three treatment groups.H1: There are significant differences in mean HDL cholesterol among the three treatment groups.c) The MSE is the best estimate of the variance of HDL cholesterol within a treatment group.MSE = 241.1218d) The test statistics are 1.3372e) The P-value is 0.2576f) Decision: There are no significant differences in mean HDL cholesterol among the three treatment groups.The Anova table is,source of variationDfSSMSFTreatment2664.15332.0751.3772Error8921459.84241.1218Total9122123.99