Question

A certain HMO is attempting to show the benefits of managed care to an insurance company. The HMO believes that certain types of doctors are more cost-effective than others. One theory is that certfication level is an important factor in measuring the cost-effectiveness of physicians. To investigate this, the HMO obtained independent random samples of 22 physicians from each of the three certification levels --Board certified (C) Uncertified, board eligible (EX and Uncertified, board eligible and recorded the total per member per month chwges for each (a total of 66 physcias)In order to compare the mean charges for the three groups, the data will be subjected to an analysis of variace Give the degree of freedom appropriate for conducting the ANOVA Fest numerator of 64 denominator of 3 numerator of 2 denominator di 63 numerador df = 64, denominator of = 2 numerator d = 3. denominator d = 63

Answer 1

The solution to this problem is given by

Total number of semele units = N = 66 Degree of freedom for denominator = =N-k when R = No. of group 3. di). for denominator = 66-3=63 Degree of freedom for numenarator = = 34-2 So, the ansiller is Numerator df =2, denominator elf = 63 Correcct