In: Statistics and Probability
Question B: The following data report length of stay (LOS) for 10 patients of Dr. Jones and 10 patients of Dr. Smith. What is the expected outcome (average outcome) for Dr. Smith? What is the expected outcomes if Dr. Jones if he was seeing Dr. Smith's patients? To answer this question, replace each outcome of Dr. Jones with average outcome of same type of patient seen by Dr. Smith. Is Dr. Smith more efficient than Dr. Jones?
Dr. Smith Patients | Previous MI | CHF | Shock | LOS |
1 | 1 | 1 | 0 | 4 |
2 | 1 | 1 | 0 | 5 |
3 | 1 | 0 | 0 | 4 |
4 | 1 | 0 | 1 | 5 |
5 | 1 | 0 | 1 | 4 |
6 | 1 | 0 | 1 | 4 |
7 | 1 | 0 | 1 | 5 |
8 | 0 | 0 | 0 | 2 |
9 | 0 | 0 | 0 | 2 |
10 | 0 | 0 | 0 | 1 |
Dr. Jones Patients | previous MI | CHF | Shock | LOS |
1 | 1 | 1 | 0 | 5 |
2 | 1 | 1 | 0 | 5 |
3 | 1 | 1 | 0 | 5 |
4 | 1 | 1 | 1 | 5 |
5 | 1 | 0 | 1 | 5 |
6 | 1 | 0 | 1 | 5 |
7 | 1 | 0 | 1 | 5 |
8 | 1 | 0 | 0 | 4 |
9 | 0 | 0 | 0 | 2 |
10 | 0 | 0 | 0 | 2 |
Explanation:-
Given sample size 'n' =10
Mean Or average of Dr smith patients
Mean Or average of Doctor Jones
Drsmithpatients x |
Doctor jones y |
|
||||
4 |
5 |
0.4 | 0.5 | 0.16 | 0.25 | |
5 | 5 | 1.4 | 0.5 | 1.96 | 0.25 | |
4 | 5 | 0.4 | 0.5 | 0.16 | 0.25 | |
5 | 5 | 1.4 | 0.5 | 1.96 | 0.25 | |
4 | 5 | 0.4 | 0.5 | 0.16 | 0.25 | |
4 | 5 | 0.4 | 0.5 | 0.16 | 0.25 | |
5 | 5 | 1.4 | 0.5 | 1.96 | 0.25 | |
2 | 4 | -1.6 | -0.5 | 2.56 | 0.25 | |
2 | 2 | -1.6 | -2.5 | 2.56 | 6.25 | |
1 | 2 | -2.6 | -2.5 | 6.76 | 6.25 | |
Null Hypothesis:H0: There is no significant difference between Dr smith patients and Dr jones patients
Alternative Hypothesis: H1:Dr smith is more efficient than Dr jones
we will use the t - Test statistic
where
Level of significance
Degrees of freedom
The Calculated value at 5% level of significance
Null hypothesis is accepted
Conclusion:-
Dr smith patients is not more than efficient of Dr jones