Instruction 11-10
the CEO of a Large Australian Health Fund

Multiple Choice

Instruction 11-10
The CEO of a large Australian health fund is attempting to show the benefits of managed health care to an insurance underwriting company. The CEO believes that certain types of doctors are more cost-effective than others. One theory states that Primary Specialty is an important factor in measuring the cost-effectiveness of physicians. To investigate this, the CEO obtained independent random samples of 20 physicians from each of four primary specialties - General Practice (GP) , Internal Medicine (IM) , Paediatrics (PED) , and Family Physicians (FP) - and recorded the total charges per member per month for each. A second factor which the CEO believes influences total charges per member per month is whether the doctor is a foreign or Australian medical school graduate. The CEO theorises that foreign graduates will have higher mean charges than Australian graduates. To investigate this, the CEO also collected data on 20 foreign medical school graduates in each of the four primary specialty types described above. So information on charges for 40 doctors (20 foreign and 20 Australian medical school graduates) was obtained for each of the four specialties. The results for the ANOVA are summarised in the following table.
$$\begin{array} { | l | l | l | l | l | l | } \hline \text { Source } &{ d f } & { \text { S5 } } &{ \text { MS } } & { F } & { \text { PR } > \mathrm { F } } \\\hline \text { Specialty } & 3 & 22.855 & 7.618 & 60.94 & 0.0001 \\\hline \text { Med school } & 1 & 105 & 105 & 0.84 & 0.6744 \\\hline \text { Interaction } & 3 & 850 & 297 & 2.38 & 0.1348 \\\hline \text { Error } & 152 & 18,950 & & & \\\hline \text { Total } & 159 & 42,800 & & & \\\hline\end{array}$$
-Referring to Instruction 11-10,what degrees of freedom should be used to determine the critical value of the F ratio against which to test for differences in the mean charges for doctors among the four primary specialty areas?

A) numerator df = 1, denominator df = 152
B) numerator df = 1, denominator df = 159
C) numerator df = 3, denominator df = 152
D) numerator df = 3, denominator df = 159