In: Operations Management
As a hospital administrator of a large hospital, you are concerned with the absenteeism among nurses' aides. The issue has been raised by registered nurses, who feel they often have to perform work normally done by their aides. To get the facts, absenteeism data were gathered for the last three weeks, which is considered a representative period for future conditions. After taking random samples of 60 personnel files each day, the following data were produced
Day |
Aides Absent |
Day |
Aides Absent |
Day |
Aides Absent |
1 |
66 |
6 |
66 |
11 |
33 |
2 |
55 |
7 |
11 |
12 |
88 |
3 |
11 |
8 |
77 |
13 |
1212 |
4 |
33 |
9 |
22 |
14 |
44 |
5 |
77 |
10 |
88 |
15 |
44 |
Because your assessment of absenteeism is likely to come under careful scrutiny, you would like a type I error of only 1 percent. You want to be sure to identify any instances of unusual absences. If some are present, you will have to explore them on behalf of the registered nurses.
a. Design a p-chart. then find the
upper control limit
_________
and the lower control limit
_____
(Enter
your responses rounded to
three decimal
places.
If your answer for the lower control limit is negative, enter this value as
0.)
Since, the number of aides absent have been doubled in the question as defects can't be more than sample size. I AM GONNA GIVE YOU THE REQUIRED SOLUTION FOR SINGLE CORRECT DIGITS.
Total number of aides absent (p) = 6+5+1+3+7+6+1+7+2+8+3+8+12+4+4
Total number of aides absent (p) = 77
Number of samples (m) = 15
Sample size (n) = 60
Fraction defective () = p/(n×m)
Fraction defective () = 77/(60×15)
Fraction defective () = 0.08556
Let Standard Deviation =
Given type 1 error of 1%
Therefore, confidence level = 99%
Value of Z corresponding to 99% confidence = 2.576
UCL = 0.08556 + (2.576 × 0.03611)
UCL = 0.17858
LCL = 0.08556 - (2.576 × 0.03611)
LCL = -0.007459
Since LCL can't be negative, therefore,
LCL = 0