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Introduction to Statistics and Econometrics
Case Problem
SPECIALTY TOYS
Specialty Toys
Specialty faces the decision of how many Weather Teddy units to order for the coming holiday season. of the management team suggested order quantities of 15000, 18000, 24000 or 28000 units. The wide range of order quantities suggested indicate considerable disagreement concerning the market potential. The product management team asks you for an analysis of the stock-out probabilities for various order quantities,
an estimate of the profit potential, and to help make an order quantity recommendation. Specialty expects to sell Weather Teddy for $24 based on a cost of $16 per unit. If inventory remains after the holiday season, Specialty will sell all surplus inventory for $5 per unit.
After reviewing the sales history of similar products, Specialty’s senior sales forecaster predicted an expected demand of 20,000 units with a 0.95 probability that demand would be between 10,000 units and 30,000 units.
Questions ◦ Approximate the demand distribution using Normal distribution and sketch the distribution. ◦ Compute the probability of a stock-out for the order quantities suggested by of the management team. ◦ Compute the projected profit for the order quantities suggested by the management team under three scenarios: worst case in which sales is 10,000 units, most likely case in which sales is 20,000 units and best case in which sales is 30,000 units.
Questions ◦ One of Specialty’s managers felt that the profit potential was so great that the order quantity should have a 70% chance of meeting demand and only a 30% chance of any stock-outs. What quantity would be ordered under this policy, and what is the projected profit under the three sales scenarios? ◦ Provide your own recommendation for an order quantity and note the associated profit projections.
Normal Distribution
.025
10,000
.95
20,000
.025
30,000
At x = 30, 000, x − µ 30000 − 20000 = = 1.96 z =
σ
σ
∴σ = 5102 Mean and Standard deviation are µ 20000, σ 5102 = =
Stock out situation The management team suggested order quantities of 15000, 18000, 24000 or 28000 units. If order quantity is 15000,
x−µ
15000 − 20000 z= = = −0.98 σ 5102 P[ X > 15000] = P[ Z > −0.98] = 0.3365 + 0.5 = 0.8865
Stock out situation
For order quantity being 18000, the probability of stock out is 0.6517.
At 24,000, the probability of stock out is 0.2177.
At 28,000, the probability of stock out is 0.0582.
Profit Projection
For order quantity being 15,000, profit projection is Sales
Unit Sales
Total Cost
at $24
at $5
Profit
10,000
240,000
240,000
25,000
25,000
20,000
240,000
360,000
0
120,000
30,000
240,000
360,000
0
120,000
At 18,000, Sales Unit Sales
Total Cost
at $24
at $5
Profit
10,000
288,000
240,000
40,000
-8,000
20,000
288,000
432,000
0
144,000
30,000
288,000
432,000
0
144,000
At 24,000, Sales Unit Sales
Total Cost
at $24
at $5
Profit
10,000
384,000
240,000
70,000
-74,000
20,000
384,000
480,000
20,000
116,000
30,000
384,000
576,000
0
192,000
At 28,000, Sales Unit Sales
Total Cost
at $24
at $5
Profit
10,000
448,000
240,000
90,000
-118,000
20,000
448,000
480,000
40,000
72,000
30,000
448,000
672,000
0
224,000
Profit potential being high
Order quantity should have a 70% chance of meeting demand and only a 30% chance of any stock-outs.
Q − 20000 = z = 0.52 ∴= Q 22, 653 5102
For Q=22,653, the projected profits under the three scenarios are Sales
Unit Sales
Total Cost
at $24
at $5
Profit
10,000
362,488
240,000
63,265
-59,183
20,000
362,488
480,000
13,265
130,817
30,000
362,488
543,672
0
181,224
Assignment Problem 3
Are male college students more easily bored than their female counterparts? This question was examined in the article “Boredom in Young Adults – Gender and Cultural Comparisons” (Journal of Cross-Cultural Psychology, 1991). The authors istered a scale called the Boredom Proneness Scale to 97 male and 148 female U.S. college students. Does the accompanying data the research hypothesis that the mean Boredom Proneness Rating is higher for men than for women? Test the appropriate hypothesis using a 0.05 significance level.
Gender
Sample Size
Sample Mean
Male Female
97 148
10.40 9.26
Sample Standard deviation 4.83 4.68
Solution H 0 : µ1 = µ2 H1 : µ1 > µ2 n1 97 n2 148 = = X 1 10.40 X 2 9.26 = = s1 4.83 s2 4.68 = = Z =
X1 − X 2 = s12 s22 + n1 n2
Z 0.05 = 1.645
1.14 = 1.83 0.3885
As Z > Zα , we reject the null hypothesis at 5% level of significance. Hence it can be concluded that the mean Boredom Proneness Rating is higher for men than for women at 5% level of significance.
Assignment Problem 2
A manufacturer has just marketed a new appliance with a one year warranty. The product development plan anticipates that the cost of meeting the warranty will be Rs 5000 per appliance, on the average, with a standard deviation of Rs 4000. As the warranty period expires for the first appliances sold, the manufacturer will note the actual costs of warranty servicing (X). Two testing approaches have been suggested for deciding whether or not the average warranty costs are exceeding the Rs 5000 target. (1) Wait for warranty data on the first 100 appliances sold and conclude that the average cost will exceed the target if > Rs 5800. (2) Wait for warranty data on the first 400 appliances sold and conclude that the average cost will exceed the target if > Rs 5400.
A timely conclusion about actual warranty costs is desirable because the manufacturer wishes to reduce the warranty if these are proving too generous to be profitable. On the other hand, an unnecessary reduction of warranty would adversely affect future sales of the industrial appliance. Assume that the warranty data on the initial sales will constitute random sample observations.
Questions Define the parameter and the alternatives H0 and H1 which are of interest in this situation. Sketch the power curves for the two test approaches on the same graph, assuming that the anticipated standard deviation of warranty costs is accurate. How do the risks of the two approaches compare if the mean warranty cost is on target? If it is Rs1000 higher than the target? The test approach with n=400 was eventually chosen and the sample results were sample mean =Rs 5340 and s= Rs 3840. Using the specified decision rule for this approach, what conclusion should be drawn about average warranty costs? Estimate the alpha risk of your test when =Rs 5000. Estimate the beta risk of your test when the mean warranty cost is Rs1000 higher than the target.
Hypotheses H 0 : µ = 5000 H1 : µ > 5000
Type I error P[Rejecting H0| H0 is true]
Type I error Rule 1 Reject H 0 if
X > 5800. Here n = 100 .
P X > 5800 | µ = 5000 = 0.0228 α=
Rule 2 Reject
H0
if X > 5400. Here n = 400 .
α= P X > 5400 | µ = 5000 = 0.0228
Power of a test The power of a statistical hypothesis test is the probability of rejecting the null hypothesis when the null hypothesis is false. Power = (1 - β)
Table showing Power µ
Rule 1
Rule 2
5100
0.0401
0.0668
5400
0.1587
0.5
5700
0.4013
0.9332
6000
0.6915
0.9999
6300
0.9878
1
Power Curve 1.2
1
0.8
0.6
Rule 1 Rule 2
0.4
0.2
0 µ5000
µ5100
µ5400
µ5700
µ6000
µ6300
= n 400 = X Rs 5340 = s Rs 3840
As X = 5340 , Rule 2 will not reject the null hypothesis. Therefore, we conclude that the warranty cost does not exceed Rs 5000.
Risks If σ is known, α = 0.0228 If σ is unknown, α = 0.0188
Beta risk when the mean warranty cost is Rs 1000 higher than the target. If σ is known, β is negligible. If σ is unknown, β is negligible.
Assignment Problem 1 An aluminum company is experimenting with a new design for electrolytic cells in smelter pot rooms. A major design objective is to maximize a cell’s expected service life. Thirty cells of the new design were started and operated under similar conditions, and failed at the following ages (in days): Two items of concern to management are: (1) the mean service life of this design, and (2) the comparative performance of this design with the standard industry design, which is known to have a mean service life of 1,300 days. Management does not want to conclude that the new design is superior to the standard one unless the evidence is fairly strong.
Assuming the distribution of service life of the cell is normal; calculate an appropriate 95% confidence interval for the mean service life of cells of the new design. Justify your choice of one or two-sided confidence interval. Should management conclude that the new design is superior to the standard one with respect to mean service life? Comment. It has been suggested that the logarithms of service life are more normally distributed than the original observations. Take logarithms (to base 10) of the service-life data and calculate the same type of confidence interval as in (a) for the mean log-service-life of cells of the new design. Cells of the standard design are known to have mean log-service-life of 3.095 (to base 10). Can management claim with confidence that the mean log-service-life is greater for cells of the new design than for cells of the standard design? Graph histograms of the original data and the log-data. Does the distribution of log-service-lives appear to be more normal than the distribution of service-lives, as suggested? Comment A management objective is to obtain a large total service life for the cells. Is the mean service life or the mean log-service-life the more relevant measure here? Explain.
Confidence interval s ,∞ X − tn −1 (α ) n
= X 1379.133 = s 399.0149 = t0.05 (29) 1.699
The confidence interval is
(1255.361, ∞ )
This does not that the new design is superior to the standard one.
Logarithms of service times
= Y 3.120485 = sY 0.134747 = t0.05 (29) 1.699
The confidence interval is ( 3.078, ∞ )
This does not that the new design is superior to the standard one.
Histogram 10 9 8
Frequency
7 6 5 4 3 2 1 0 500
800
1100
1400
1700 Bin
2000
2300
More
Histogram 10 9 8
Frequency
7 6 5 4 3 2 1 0 2.79
2.89
2.99
3.09
3.19 Bin
3.29
3.39
More
Case Problem
SPECIALTY TOYS
Specialty Toys
Specialty faces the decision of how many Weather Teddy units to order for the coming holiday season. of the management team suggested order quantities of 15000, 18000, 24000 or 28000 units. The wide range of order quantities suggested indicate considerable disagreement concerning the market potential. The product management team asks you for an analysis of the stock-out probabilities for various order quantities,
an estimate of the profit potential, and to help make an order quantity recommendation. Specialty expects to sell Weather Teddy for $24 based on a cost of $16 per unit. If inventory remains after the holiday season, Specialty will sell all surplus inventory for $5 per unit.
After reviewing the sales history of similar products, Specialty’s senior sales forecaster predicted an expected demand of 20,000 units with a 0.95 probability that demand would be between 10,000 units and 30,000 units.
Questions ◦ Approximate the demand distribution using Normal distribution and sketch the distribution. ◦ Compute the probability of a stock-out for the order quantities suggested by of the management team. ◦ Compute the projected profit for the order quantities suggested by the management team under three scenarios: worst case in which sales is 10,000 units, most likely case in which sales is 20,000 units and best case in which sales is 30,000 units.
Questions ◦ One of Specialty’s managers felt that the profit potential was so great that the order quantity should have a 70% chance of meeting demand and only a 30% chance of any stock-outs. What quantity would be ordered under this policy, and what is the projected profit under the three sales scenarios? ◦ Provide your own recommendation for an order quantity and note the associated profit projections.
Normal Distribution
.025
10,000
.95
20,000
.025
30,000
At x = 30, 000, x − µ 30000 − 20000 = = 1.96 z =
σ
σ
∴σ = 5102 Mean and Standard deviation are µ 20000, σ 5102 = =
Stock out situation The management team suggested order quantities of 15000, 18000, 24000 or 28000 units. If order quantity is 15000,
x−µ
15000 − 20000 z= = = −0.98 σ 5102 P[ X > 15000] = P[ Z > −0.98] = 0.3365 + 0.5 = 0.8865
Stock out situation
For order quantity being 18000, the probability of stock out is 0.6517.
At 24,000, the probability of stock out is 0.2177.
At 28,000, the probability of stock out is 0.0582.
Profit Projection
For order quantity being 15,000, profit projection is Sales
Unit Sales
Total Cost
at $24
at $5
Profit
10,000
240,000
240,000
25,000
25,000
20,000
240,000
360,000
0
120,000
30,000
240,000
360,000
0
120,000
At 18,000, Sales Unit Sales
Total Cost
at $24
at $5
Profit
10,000
288,000
240,000
40,000
-8,000
20,000
288,000
432,000
0
144,000
30,000
288,000
432,000
0
144,000
At 24,000, Sales Unit Sales
Total Cost
at $24
at $5
Profit
10,000
384,000
240,000
70,000
-74,000
20,000
384,000
480,000
20,000
116,000
30,000
384,000
576,000
0
192,000
At 28,000, Sales Unit Sales
Total Cost
at $24
at $5
Profit
10,000
448,000
240,000
90,000
-118,000
20,000
448,000
480,000
40,000
72,000
30,000
448,000
672,000
0
224,000
Profit potential being high
Order quantity should have a 70% chance of meeting demand and only a 30% chance of any stock-outs.
Q − 20000 = z = 0.52 ∴= Q 22, 653 5102
For Q=22,653, the projected profits under the three scenarios are Sales
Unit Sales
Total Cost
at $24
at $5
Profit
10,000
362,488
240,000
63,265
-59,183
20,000
362,488
480,000
13,265
130,817
30,000
362,488
543,672
0
181,224
Assignment Problem 3
Are male college students more easily bored than their female counterparts? This question was examined in the article “Boredom in Young Adults – Gender and Cultural Comparisons” (Journal of Cross-Cultural Psychology, 1991). The authors istered a scale called the Boredom Proneness Scale to 97 male and 148 female U.S. college students. Does the accompanying data the research hypothesis that the mean Boredom Proneness Rating is higher for men than for women? Test the appropriate hypothesis using a 0.05 significance level.
Gender
Sample Size
Sample Mean
Male Female
97 148
10.40 9.26
Sample Standard deviation 4.83 4.68
Solution H 0 : µ1 = µ2 H1 : µ1 > µ2 n1 97 n2 148 = = X 1 10.40 X 2 9.26 = = s1 4.83 s2 4.68 = = Z =
X1 − X 2 = s12 s22 + n1 n2
Z 0.05 = 1.645
1.14 = 1.83 0.3885
As Z > Zα , we reject the null hypothesis at 5% level of significance. Hence it can be concluded that the mean Boredom Proneness Rating is higher for men than for women at 5% level of significance.
Assignment Problem 2
A manufacturer has just marketed a new appliance with a one year warranty. The product development plan anticipates that the cost of meeting the warranty will be Rs 5000 per appliance, on the average, with a standard deviation of Rs 4000. As the warranty period expires for the first appliances sold, the manufacturer will note the actual costs of warranty servicing (X). Two testing approaches have been suggested for deciding whether or not the average warranty costs are exceeding the Rs 5000 target. (1) Wait for warranty data on the first 100 appliances sold and conclude that the average cost will exceed the target if > Rs 5800. (2) Wait for warranty data on the first 400 appliances sold and conclude that the average cost will exceed the target if > Rs 5400.
A timely conclusion about actual warranty costs is desirable because the manufacturer wishes to reduce the warranty if these are proving too generous to be profitable. On the other hand, an unnecessary reduction of warranty would adversely affect future sales of the industrial appliance. Assume that the warranty data on the initial sales will constitute random sample observations.
Questions Define the parameter and the alternatives H0 and H1 which are of interest in this situation. Sketch the power curves for the two test approaches on the same graph, assuming that the anticipated standard deviation of warranty costs is accurate. How do the risks of the two approaches compare if the mean warranty cost is on target? If it is Rs1000 higher than the target? The test approach with n=400 was eventually chosen and the sample results were sample mean =Rs 5340 and s= Rs 3840. Using the specified decision rule for this approach, what conclusion should be drawn about average warranty costs? Estimate the alpha risk of your test when =Rs 5000. Estimate the beta risk of your test when the mean warranty cost is Rs1000 higher than the target.
Hypotheses H 0 : µ = 5000 H1 : µ > 5000
Type I error P[Rejecting H0| H0 is true]
Type I error Rule 1 Reject H 0 if
X > 5800. Here n = 100 .
P X > 5800 | µ = 5000 = 0.0228 α=
Rule 2 Reject
H0
if X > 5400. Here n = 400 .
α= P X > 5400 | µ = 5000 = 0.0228
Power of a test The power of a statistical hypothesis test is the probability of rejecting the null hypothesis when the null hypothesis is false. Power = (1 - β)
Table showing Power µ
Rule 1
Rule 2
5100
0.0401
0.0668
5400
0.1587
0.5
5700
0.4013
0.9332
6000
0.6915
0.9999
6300
0.9878
1
Power Curve 1.2
1
0.8
0.6
Rule 1 Rule 2
0.4
0.2
0 µ5000
µ5100
µ5400
µ5700
µ6000
µ6300
= n 400 = X Rs 5340 = s Rs 3840
As X = 5340 , Rule 2 will not reject the null hypothesis. Therefore, we conclude that the warranty cost does not exceed Rs 5000.
Risks If σ is known, α = 0.0228 If σ is unknown, α = 0.0188
Beta risk when the mean warranty cost is Rs 1000 higher than the target. If σ is known, β is negligible. If σ is unknown, β is negligible.
Assignment Problem 1 An aluminum company is experimenting with a new design for electrolytic cells in smelter pot rooms. A major design objective is to maximize a cell’s expected service life. Thirty cells of the new design were started and operated under similar conditions, and failed at the following ages (in days): Two items of concern to management are: (1) the mean service life of this design, and (2) the comparative performance of this design with the standard industry design, which is known to have a mean service life of 1,300 days. Management does not want to conclude that the new design is superior to the standard one unless the evidence is fairly strong.
Assuming the distribution of service life of the cell is normal; calculate an appropriate 95% confidence interval for the mean service life of cells of the new design. Justify your choice of one or two-sided confidence interval. Should management conclude that the new design is superior to the standard one with respect to mean service life? Comment. It has been suggested that the logarithms of service life are more normally distributed than the original observations. Take logarithms (to base 10) of the service-life data and calculate the same type of confidence interval as in (a) for the mean log-service-life of cells of the new design. Cells of the standard design are known to have mean log-service-life of 3.095 (to base 10). Can management claim with confidence that the mean log-service-life is greater for cells of the new design than for cells of the standard design? Graph histograms of the original data and the log-data. Does the distribution of log-service-lives appear to be more normal than the distribution of service-lives, as suggested? Comment A management objective is to obtain a large total service life for the cells. Is the mean service life or the mean log-service-life the more relevant measure here? Explain.
Confidence interval s ,∞ X − tn −1 (α ) n
= X 1379.133 = s 399.0149 = t0.05 (29) 1.699
The confidence interval is
(1255.361, ∞ )
This does not that the new design is superior to the standard one.
Logarithms of service times
= Y 3.120485 = sY 0.134747 = t0.05 (29) 1.699
The confidence interval is ( 3.078, ∞ )
This does not that the new design is superior to the standard one.
Histogram 10 9 8
Frequency
7 6 5 4 3 2 1 0 500
800
1100
1400
1700 Bin
2000
2300
More
Histogram 10 9 8
Frequency
7 6 5 4 3 2 1 0 2.79
2.89
2.99
3.09
3.19 Bin
3.29
3.39
More
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