Download Problems chapter 2

April 10, 2007
Problems chapter 2
Problem 2.1
During 25 days 50 bills issued by a private hospital were picked at random daily and
checked for errors. Table 2.1 shows the number of bills with one or more errors found on
each of the 25 days.
Is the process ”issuing bills” in statistical control? How is the quality of the products
produced by this process?
Table 2.1
Day
Number of bills with one or more errors
1
19
2
13
3
14
4
13
5
18
6
12
7
16
8
16
9
20
10
18
11
15
12
17
13
16
14
21
15
18
16
17
17
19
18
18
19
21
20
15
21
19
22
21
23
13
24
17
25
20
_____________
426
Problem 2.2
Indicate for each of the following sample variables what type of control chart you want to
use and why.
1. The relative frequency per month of personnel who leave their position.
2. The monthly deviation between account and budget.
3. The monthly relative frequency of Cesarian.
4. The number of employed who gets a needle lesion per month.
5. The result of an investigation based on questionnaires. The questionnaires are sent
every month to 25 randomly selected employers. Each employer scores his job
satisfaction using a scale from 1 to 5.
6. The average time it takes to clean a room used for clinical examination after a patient
has left it. The daily average is calculated from five randomly selected patient visits.
7. The frequency of errors per medication per month at a medical department.
Problem 2.3
The bills mentioned in problems 2.1 were examined as follows; the date and code of
treatment, the ID of the physician, the amount paid, and a number of other items were
checked and the number of errors thus found noted for each bill. Table 2.2 shows the
number of errors found per day (per 50 bills picked at random).
Table 2.2
Day #
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
Sum
Number of errors found
26
19
18
15
19
16
21
25
28
29
22
24
25
29
25
22
27
27
30
20
27
28
16
24
26
588
Is the process in statistical control?
Problem 2.4
Table 2.3 shows the total number of patients per month who were catheterised and the
number of these who were infected as a result of the catheterisation.
What type of control chart do you want to use and why?
Control if the value found during month # 1 was within the control limits. Perform the
same control for month # 24, and month # 29.
Table 2.3
Month
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
In all
Patients
213
212
182
286
248
228
258
340
68
201
290
229
271
226
213
268
311
251
211
200
232
273
246
321
167
190
255
302
268
369
240
267
246
255
199
362
8986
Number of patients infected
15
12
15
21
23
19
15
13
9
12
20
19
22
13
17
19
25
29
20
31
28
30
26
39
9
7
11
16
4
14
13
18
11
21
18
25
659
Problem 2.5
Table 2.4 shows the number of patients operated per month and the number of surgical
complications observed in these patients.
What type of control chart do you want to use?
Is the value of month # 7 within the control limits?
Table 2.4
Month
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
Ialt
Patients
21
15
17
17
15
30
26
18
17
30
31
22
25
28
28
22
16
32
18
26
22
23
21
14
534
Number of surgical complications
6
6
10
5
6
13
20
6
3
7
10
11
11
9
11
8
7
10
9
7
7
13
9
2
206
Problem 2.6
Table 2.5 column 1 shows the peak expiratory flow rate/l/min (PEFR) measured daily in a
patient suffering from asthma. Column 2 shows the value of the moving range. The mean
is 167.6 and the moving range is 72.79.
Calculate an X chart. Which values (if any) are outside the three standard deviation
 ( X i  X )2 is used to estimate the estimate the
control limits? When the equation
15  1
standard deviation, it becomes 73.21. Calculate an X chart, using this estimate.
How many values are outside the control limits? Discuss the result.
Table 2.5
PEFR
Moving range
121
140
99
300
150
150
100
122
152
367
200
138
175
150
150
19
41
201
150
0
50
22
30
215
167
62
37
25
0
Solutions
2.1 A p chart is used to assess the process. The mean is 0.341. The standard deviation is
0.067. The three standard deviations UCL is 0.542 and the corresponding LCL is 0.14.
21
12
The largest fraction of errors is
= 0.42 and the smallest is
= 0.24. So all fractions
50
50
are within the control limits. Therefore, in all likelihood the process is in statistical control.
The quality is poor with 34.1% errors.
2.2 1: a p chart is used; 2: an X chart is used; 3: a p chart is used; 4: If the number of
employees is reasonable constant a c chart may be used. Otherwise a u chart should be
used. 5 and 6: an X chart combined with an S chart should be used. 7: A u chart is used.
A medication is an inspection unit; u is calculated as the number of medication errors
divided by the number of medications inspected per month.
2.3 The number of inspection units is constant. Therefore, a c chart may be used.
UCL = 38.1 error; centreline = 23.5 error; LCL = 8.97 error. The process is in statistical
control.
659
= 0.0733.
8986
For sample # 1 the UCL1 is 0.127, and the LCL1 is 0.020. The sample value is 0.07.
Thus the value is within the control limits.
2.4 A p chart for variable sample size is used. p̂ is
For sample # 24 the UCL24 is 0.117, and the LCL24 is 0.030. The sample value is
0.121. Thus the value is above the UCL24.
For sample # 29 the UCL29 is 0.121, and the LCL29 is 0.026. The sample value is
0.015. Thus the value is below the LCL29.
2.5 A u chart should be used. u is
206
=0.3858. The standard deviation of sample # 7 is
534
0.3858
= 0.122. The UCL7 is 0.752, and the LCL7 is 0.020. The observed value, which
26
20
is
= 0.769, lies above the UCL.
26
2.6 The mean is 167.6, the mean of the moving range is 72.79, and the standard deviation
72.79
is
= 64.53. The UCL is 361.2 and the LCL is – 25.98.
1.128
Value # 10 (367) is outside the control limits. (When this value is removed value # 4
(300) gets outside the control limits; and when this value is also removed, and the chart
recalculated using the remaining 13 values, the mean is 142.1, the standard deviation is
28.44, UCL = 227.4, and LCL = 56.8).
When the standard deviation is estimated as 73.21, the UCL is 387.2, and the LCL is –
52.0. Now all values are within the control limits. The reason is that the standard deviation
includes the variation between the sample means. Since the process is not in statistical
control the standard deviation becomes larger than that calculated using the average of
the within sample standard deviations. Thus the lack of statistical control is masked.