Master of Business Administration in Management
STAT 500 Statistics for Managers
1. Applying Six Sigma total quality management technique learned in the course to your workplace, identify a product or service whose quality needs to be monitored and controlled. (20 %)
a) Set up control chart(s) for the product or service studied.
b) On the basis of your results, indicate whether the process is in control.
c) What is your recommended policy if the process is in control or NOT in control?
Answers:
From the data gathered in my work place, the high voltage output of a power supply is specified as 350 ± 5 V dc at 20 milliamps (mA). Subgroups of four power supply units are drawn from the process and inspected approximately every half-hour. The data from 25 subgroups are show in the following table along with the average and range for each subgroup.
DC-voltage output at 20 mA
Subgroup
Samples
Calculations
number
a
b
c
d
TOTAL
MEAN
RANGE
1
348.5
350.2
348.3
350.3
1397.30
349.325
2
2
351.3
351.2
347.1
349.7
1399.30
349.83
4.2
3
348.5
350.5
348.5
349
1396.50
349.13
2
4
351.4
350.4
348.6
353.2
1403.60
350.90
4.6
5
349.4
348
349.6
351.1
1398.10
349.53
3.1
6
351.1
348.1
349.2
350.1
1398.50
349.63
3
7
348.3
349.9
350.7
348.5
1397.40
349.35
2.4
8
349.9
349.1
349
349.6
1397.60
349.40
0.9
9
349.2
348.7
348.8
350.3
1397.00
349.25
1.6
10
349.2
351.6
351.9
349.2
1401.90
350.48
2.7
11
350.1
350.5
351.2
347.9
1399.70
349.93
3.3
12
350.4
350.8
350.3
352.6
1404.10
351.03
2.3
13
347.7
349.6
348.6
349.3
1395.20
348.80
1.9
14
349
351.1
350.2
348
1398.30
349.58
3.1
15
350.7
349.3
349.3
350.2
1399.50
349.88
1.4
16
350
351.8
352.3
349.8
1403.90
350.98
2.5
17
350.1
349.8
349.6
349.2
1398.70
349.68
0.9
18
351.1
350.6
346.9
349.8
1398.40
349.60
4.2
19
351.4
349.3
349.7
349.6
1400.00
350.00
2.1
20
348.8
349.6
351.3
349.2
1398.90
349.73
2.1
21
349.4
350.2
350.2
351.8
1401.60
350.40
2.4
22
351.7
351.6
349.9
347.1
1400.30
350.08
4.6
23
350.4
349
349.2
349.6
1398.20
349.55
1.4
24
349.4
348.7
350.3
348.8
1397.20
349.30
1.6
25
349.6
349.1
349.6
351.2
1399.50
349.88
2.1
349.81
3.7
SPC: X – bar chart
SPC: R chart
From the given information, the computer Mean is 282.45 with Std. Dev. of1.5927. Assuming that the distribution of voltage output is approximately normal, the proportion of the units are expected to meet specifications of 347.25 and 352.75 V to be in control. To reach a 347.25 proportion of units, specifications must range from 345, 344, 349.1 and 350.9. To reach 352.75 however, specifications must be: 366, 345, 350.9 and 349.1. However, the charts indicate that none of the proportion of the units tested actually met these specifications which mean that distribution is normal.
SPC charts were developed by industrial quality control engineers to ensure the consistent delivery of quality products. Specifically, SPC charts are used to detect systematic changes in manufacturing processes through the identification of unusual product samples (for example, a sample of a part that is distinctly larger or smaller than other samples of this part). When unusual product samples are detected, attempts are made to discover their cause in the production process (for example, an improperly calibrated machine). Finally, this causal information is used to improve the production process (for example, recalibrate the machine) (Doty, 1996; Pitt, 1994; Wheeler & Chambers, 1992).
In an SSD chart time is plotted on the horizontal axis, and the outcome variable is plotted on the vertical axis. This provides a running record of outcome variability. This same chart is known in the SPC literature as a "run chart." SPC charts are constructed by adding three elements to a run chart, which provides a more objective basis for the visual analysis of SSD data (Pfadt & Wheeler, 1995). A central value (for example, mean outcome) is computed and plotted as a solid line; this is referred to as the center line (the CL). Upper and lower control limits corresponding to plus and minus three standard deviations from this midpoint are computed and plotted as dashed lines parallel to the CL; these are referred to as the upper control limit (UCL) and the lower control limit (LCL), respectively.
There are a number of SSD situations in which it is important to identify nonnormative events and to monitor the stability of a process (for example, Pfadt et al., 1992; Pfadt & Wheeler, 1995). First, in prevention programs the occurrence of undesirable nonnormative outcomes (for example, excessive weight loss or gain) may signal the need to modify the prevention process. Indeed, the prevention of undesirable outcomes is one of the main goals of SPC. Second, in testing intervention effects nonnormative outcomes that occur when an intervention is implemented may signal an intervention effect. Third, the identification of a nonnormative outcome after successful implementation of an intervention may signal relapse and a need for "booster shots." Fourth, the identification of nonnormative outcomes can stimulate a search for the cause of such outcomes, which then can be incorporated into the process (for example, an increase in caregiver satisfaction associated with increased social support). Fifth, in observation-only designs it is necessary to determine if some nonnormative outcome occurs (for example, postpartum depression), which then may require intervention. Finally, in comparing data across phases, and in deciding when to change phases, it is important to determine if a stable and predictable pattern of data exists within a given phase.
References:
Cox, M.E., & ORme, J.G. (2001). Analyzing Single-Subject Design Data Using Statistical Process Control Charts. Social Work Research. Vol25, No.2, p. 115.
Doty, L. A. (1996). Statistical process control (2nd ed.). New York: Industrial Press.
Pfadt, A., Cohen, I. L., Sudhalter, V., Romanczyk, R. G., & Wheeler, D. J. (1992). Applying statistical process control to clinical data: An illustration. Journal of Applied Behavior Analysis. Vol.25, pp. 551-560.
Pfadt, A., & Wheeler, D. J. (1995). Using statistical process control to make data-based clinical decisions. Journal of Applied Behavior Analysis. Vol.28, pp. 349-370.
Pitt, H. (1994). SPC for the rest of us: A personal path to statistical process control. Reading, MA: Addison Wesley.
Wheeler, D. J., & Chambers, D. S. (1992). Understanding statistical process control (2nd ed.). Knoxville, TN: SPC Press.
2. The PGGW has conducted a statistical analysis of 1,000 long-distance telephone calls made by its customers indicates that the length of these calls is normally distributed with a mean of 240 seconds and standard deviation of 40 seconds. (10 %)
a) What percentage of these calls lasted less than 180 seconds?
Answer:
or 1.5 to the left
Since the area from the mean up to 180s is z=1.5 or 0.4432 then the probability less than 180s is 0.5 - 0.4432 = 6.68%.
b) What is the probability that a particular call lasted between 180 and 300 seconds?
Answers:
The area between 180s and 300s is given by:
or 1.5 to the left or 1.5 to the right
Then the area is 0.4432 + 0.4432 = 0.8864 or 88.64%
c) What is the probability that a particular call lasted for 30 minutes?
Answer:
There is an almost zero (0) probability that a particular call lasted for 30 minutes.
d) What is the length of a particular call if only 1% of the calls are shorter?
Answer:
3. The Tiger Rock Tunnel management knows that vehicles arrive at the rate of 50 per minute during the 6:00-7:00 P.M. rush hour. If a vehicle has just arrived, (5 %)
a) What is the probability that the next auto arrives within 3 seconds?
Answer:
Since the probability of the next auto arrives within 1 second is 83.33%. Then the probability of the next auto arrives within 3 seconds is 2.49%
b) What is the probability that the next auto arrives within 1 second?
Answer:
Then the probability is 83.33/1s = 83.33
4. The manager of Coca Coca wants to estimate the actual amount of soft drink contained in 1-liter bottles. The company has a specification requirement that the standard deviation of the amount of soft drink is equal to 0.02 liter. A random sample of 50 bottles is selected, and the sample mean amount of soft drink per 1-litre bottle is 0.995 liter. (10 %)
a) Set up a 99% confidence interval, estimate the true population mean amount of soft drink included in a 1-liter bottle.
Answer:
To determine the 99% confidence interval estimate of the mean amount of soft drink included in a 1-liter bottle, we have to use the formula of maximum error of estimate i.e.
Using the given data we have,
= 0.00728
where n= 50, σ= 0.02 and zα/2= z0.005= 2.575
Thus, the 99% confidence interval estimate of the population mean net weight of the tea bags produced by the sampled machine is 0.00728.
b) On the basis of your results, do you think the manager has a right to complain to the production manager? Why?
Answer:
No. Since the computed possible error at 99% confidence interval is 0.00728 then it is still allowed to have a mean value of 0.995 liter. Actually 0.995 liter is still in the allowed range i.e. 1- 0.00728 < x < 1+0.00728.
c) Does the population amount of soft drink per bottle have to be normally distributed? Explain.
Answer:
Yes, it should be normally distributed since the skewness and kurtosis of distribution affects the central tendency and dispersion of data.
5) A survey is planned to determine the mean annual family medical expenses of employees of a large company. The management of the company wishes to be 95% confident that the sample mean is correct to within +/- $50 of the true population mean annual family medical expenses. A pilot study indicates that the standard deviation can be estimated as $400. (5 %)
a) How large a sample size is necessary?
Answer:
samples
b) If management wants to be correct to within +/- $25, what sample size is necessary?
Answer:
samples
6) The quality control manager at a lightbulb factory needs to determine whether the mean life of a large shipment of lightbulbs is equal to the specified value of 375 hours. The process standard deviation is known to be 100 hours. A random sample of 64 lightbulbs indicates a sample mean life of 350 hours. (15 %)
a) State the null and alternative hypotheses.
Answer:
b) At the 0.05 significance level, is there evidence that the mean life is different from 375 hours?
Answer:
There is no significant evidence that the mean life is different from 375 hours.
c) Set up a 95% confidence interval, estimate the population mean life of the lightbulbs.
Answer:
Critical Region: z< -1.645 and z> 1.645 where
Where = 350, n = 64 and hence
d) What is the relationship between hypothesis testing and confidence interval estimation?
Answer:
In confidence interval estimation, it involves the computation of bounds for which it is reasonable that the parameter in question is in that bound. From the results and in general, for every test of hypothesis there is an equivalent statement about whether the hypothesized parameter value is included in a confidence interval.
7) A large mail-order house believes that there is an association between the weight of the mail it receives and the number of orders to be filled. It would like to investigate the relationship in order to predict the number of orders based on the weight of the mail. From an operational perspective, knowledge of the number of orders will help in the planning of the order-fulfillment process. A sample of 25 mails shipments is selected within a range of 200 to 700 pounds. The results are as follows: (20 %)
Weight of
Orders
Weight of
Orders
Weight of
Orders
Mail (lbs)
(in $000)
Mail (lbs)
(in $000)
Mail (lbs)
(in $000)
216
6.1
384
10.6
528
16.2
283
9.1
404
12.5
501
15.8
237
7.2
426
12.9
628
19
203
7.5
482
14.5
677
19.4
259
6.9
432
13.6
602
19.1
374
11.5
409
12.8
630
18
342
10.3
553
16.5
652
20.2
301
9.5
572
17.1
365
9.2
506
15
a) Assuming a linear relationship, use the least-squares method to find the regression coefficients, the vertical intercept and the slope.
Answer:
Weight of Mails = 5.55 + 32.76 Orders
b) Interpret the meaning of the vertical intercept and the slope.
Answer:
From the given linear regression equation i.e. Weigth of Mails = 5.55 + 32.76 Orders, we can see that there is a positive relationship between the variables. In accordance to the vertical intercept and the slope, the changes are in positive direction. Meaning to say, if the Weigth of Mails increases, there is also an increase in orders or vice versa. However, if the order is zero, it is still expected the weight of mail is at least 5.55.
c) Predict the average number of orders when the weight of the mail is 500 pounds.
Answer:
Weight of Mails = 5.55 + 32.76 Orders
500 = 5.55 + 32.76 Order
Order = (500+5.55)/32.76
Order = 15.43
d) Is your prediction in ( c ) reliable? Explain
Answer:
From the given behavior of data and revealed by their correlation coefficient i.e. r = 0.98643916874012, the computed prediction is reliable.
8) Develop a model to predict the assessed value using the size of the houses and the age of the houses from the following table: (15 %)
House
Value
Size (sq ft)
Age
1
844000
2000
3.42
2
774000
1710
11.5
3
757000
1450
8.33
4
859000
1760
0
5
791000
1930
7.42
6
704000
1200
32
7
758000
1550
16
8
859000
1930
2
9
785000
1590
1.75
10
792000
1500
2.75
11
867000
1900
0
12
793000
1390
0
13
745000
1540
12.58
14
838000
1890
2.75
15
768000
1590
7.17
a) State the multiple regression equation.
Value = 637751.235966945 + 107.251829803543Size -2842.54348013854Age
b) Find the intercept and slopes in this equation.
The intercept is 637751.235966945 and the slopes with respect to value are 107.251829803543 for size and -2842.54348013854 for age.
c) Assess the value of a house that has a size of 1,750 sq ft and is 10 years old.
Value[t] = 637751.235966945 + 107.251829803543(1750) -2842.54348013854 (10)
Value = 797016.5033
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