Question

Three physicians were selected for a study to evaluate the length of stay for patients undergoing a major surgical procedure. All these procedures occurred in the same hospital and were without complications. Eight records were randomly selected from patients treated over the past 12 months.

Physician	Time
A	9
A	12
A	10
A	7
A	11
A	13
A	8
A	13
B	10
B	6
B	7
B	10
B	11
B	9
B	9
B	11
C	8
C	9
C	12
C	10
C	14
C	10
C	8
C	15

1. what is the p value?

1b.

What is the conclusion of your hypothesis test?

a		There are no significant differences between the average times for the physicians
b		All the average times are different between the physicians
c		At least one of the physicians has a mean that is significantly different from the others
d		None of the above

1c.

Which physicians have means that are significantly different? (mark all that apply)

a		Physician A and Physician B
b		Physician A and Physician C
c		Physician B and Physician C
d		None of the above

Answer 1

Solution

Part (a)

Analysis of Variance

ANOVA TABLE		α (assumed ) = 0.05
Source	df	SS	MS	F	Fcrit	p-value
Treatment	2	11.58	5.791666667	1.1235566	3.4668	0.34389652
Error	21	108.25	5.154761905
Total	23	119.83

Answer 1

p-value = 0.3439 Answer 2

Since p-value < α, null hypothesis of No difference in means among the 3 physicians is accepted.

Part (b)

Conclusion

Since null hypothesis of ‘No difference in means among the 3 physicians’ is accepted, we conclude:

Option a Answer 3

Part (c)

Since we have concluded, ‘There are no significant differences between the average times for the physicians’,

question of which physician has a greater mean time does not arise. Hence, Option d Answer 4

Details of ANOVA Calculations

Obsn #	A		B		C
1	9		10		8
2	12		6		9
3	10		7		12
4	7		10		10
5	11		11		14
6	13		9		10
7	8		9		8
8	13		11		15
Total xi.		83		73		86
Grand Total G = ∑xi. = 242
r		3
n		8
N		24
Correction Factor C = G^2/N						2440.166667
∑xij^2		897		689		974
Total SS Raw = Sum∑xij^2						2560
Total SS = Total SS Raw - C						119.83
Treatment SS Raw = ∑xi.^2/n						2452
Treatment SS = Treatment SS Raw - C								11.58
Error SS = Total SS - Treatment SS								108.25
Degrees of freedom df
Total: N - 1
Treatment: r - 1
Error: Total - Treatment = N - r
Mean Sum of squares = SS/df
F = MSTr/MSE
Fcrit = Fr-1, N-r,α
p-value = P(Fr-1, N-r > F)

DONE