Questions and Answers
Question 3
Checkout time at a supermarket is monitored using a mean and a range chart. Six samples of n=20 observations have been obtained, the sample means, and ranges computed:
| Sample | Mean | Range | Sample | Mean | Range |
| 1 | 3.06 | 0.42 | 4 | 3.13 | 0.46 |
| 2 | 3.15 | 0.50 | 5 | 3.06 | 0.46 |
| 3 | 3.11 | 0.41 | 6 | 3.09 | 0.45 |
Using the Factors in the table 10.3, determine the upper and lower limits for mean and range charts. Is the process in control?
Total mean= 18.6: thus average mean = = 18.6/6= 3.10
Sample (n) = 20
A2 =0.18, D3 = 0.41, and D4 = 1.59.
The formula for calculation of the mean chart = ± A2
3.1 ± 0.18(0.45)
= 3.1 ± .081
The upper control limit = 3.1 + 0.81 = 3.181
The Lower Control Limit = 3.1 – 0.81 = 3.019
3.019 ≤ Mean ≤ 3.181. This indicates that all the six means are within the limits.
Total Ranges = 2.7, implying that the average range = 0.45
Range Control limits is obtained through D4 and D3 for the upper and lower limits respectively.
Upper Control Limit = 1.59(0.45) =0 .7155
Lower Control Limit = 0.41(0.45) = 0.1845
0.1845 ≤ R ≤ 0.7155
Since all the points from the above table are within the limits as illustrated above, the process is in control.
Question 7
The postmaster of a small western town receives a certain number of complaints each day about mail delivery. Determine three-sigma control limits using the following data. Is the process in control?
DAY
1 2 3 4 5 6 7 8 9 10 11 12 13 14
Number of
Complaints 4 10 14 8 9 6 5 12 13 7 6 4 2 10
Mean Complaints = 110/14 = 7.857
Since the sample is large as reflected b more than 10 n, we adopt the formula; Average Complaints ± thrice the square root of the mean complaint
7.857 ± 8.409
From this calculation, the upper chart limit is 16.266. The lower chart limit is assumed to be zero. This is because the sample is large and the limit does not include negatives.
0 ≤ A.C ≤ 16.266
From the calculation, above, all values within the table fall in the limits. This indicates that the process is in control.
Question 9
After a number of complaints about its directory assistance, a telephone company examined samples of calls to determine the frequency of wrong numbers given to callers. Each sample composed of 100 calls. Determine 95 percent limits. Is the process stable (i.e…, in control)? Explain.
SAMPLE
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Number of Errors 5 3 5 7 4 6 8 4 5 9 3 4 5 8 6 7
Average Defectives calculated by
This generates 87/ 16 (100) = 0.54. In order to obtain control limits, we apply the formula
Implying that
= 0.54 ± 0.044, this indicates that
The upper control limit = 0.54 + 0.044 = 0.10 (following conversion into fraction forms)
The lower control limit = 0.54 – 0.044 = 0.01 (following conversion into fraction forms)
From the calculation and table above, it is ideal to conclude that the process is out of control. This is because ¾ of the values appear to be above 0.04 despite all values falling within the limits.
Question 24
Each of the processes listed is non-centered with respect to the specifications for that process. Compute the appropriate capability index for each, and decide if the process is capable.
| Process | Mean | Standard Deviation | Lower Spec | Upper Spec |
| H | 15.0 | 0.32 | 14.1 | 16.0 |
| K | 33.0 | 1.00 | 30.0 | 36.5 |
| T | 18.5 | 0.40 | 16.5 | 20.1 |
In the determination of the capability of the process, we let Upper Specification Limit to be USL, Lower Specification Limit to be LSL, to be the process mean, and s to represent the standard deviation of the process.
For Process H, (Process Mean – LSL) / 3 s = (15 – 14.1) / 3 (32) = 0.93
(USL- Process Mean) / 3 s = (16 – 15) / 96 = 1.04
{0.938, 1.04} = 0.93
Since 0.93 < 1.0, the process is not capable.
For Process K
(Process mean – LSL) / 3 s = (33 – 30)/ 3 = 1.0
USL – (Process Mean) / 3 s = (36.5 – 33) / 3 = 1.17
With assumption that the minimum acceptable process is 1.33, the calculations reveal that the process is not capable because 1.0 < 1.33
For Process T
(Process Mean- LSL) / 3 s = (18.5 – 16.5) / (3)(0.4) = 1.67
(USL – Process Mean ) / 3 s = (20.1-18.5)/ 1.2= 1.33
From the calculation, the computed value = the assumed value, thus 1.33 = 1.33 the process T is capable.
