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Introduction to Statistics and
Econometrics
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Case Problem
SPECIALTY TOYS
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Specialty Toys
Specialty faces the decision of how manyWeather Teddy units to order for the comingholiday season. Members of the management
team suggested order quantities of15000, 18000, 24000 or 28000 units. The widerange of order quantities suggested indicateconsiderable disagreement concerning the
market potential. The product managementteam asks you for an analysis of the stock-outprobabilities for various order quantities,
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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.
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After reviewing the sales history of similar
products, Specialtys 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.
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Questions
Approximate the demand distribution usingNormal distribution and sketch the distribution.
Compute the probability of a stock-out for the
order quantities suggested by members 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.
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Questions
One of Specialtys 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.
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Normal Distribution
20,000
.025
10,000 30,000
.025 .95
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At 30,000,
30000 20000
1.96
5102
Mean and Standard deviation are20000, 5102
x
x
z
=
= = = =
= =
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Stock out situation
The management team suggested order
quantities of 15000, 18000, 24000 or 28000
units.
If order quantity is 15000,
15000 200000.98
5102
[ 15000] [ 0.98]
0.3365 0.5 0.8865
xz
P X P Z
= = =
> = >
= + =
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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.
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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
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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
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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
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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
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Profit potential being high
Order quantity should have a 70% chance of meeting demand
and only a 30% chance of any stock-outs.
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For Q=22,653, the projected profits underthe three scenarios are
200000.52 22,653
5102
Qz Q
= = =
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
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Assignment Problem 3
Are male college students more easily bored thantheir female counterparts? This question wasexamined in the article Boredom in Young Adults
Gender and Cultural Comparisons (Journal of
Cross-Cultural Psychology, 1991). The authorsadministered a scale called the BoredomProneness Scale to 97 male and 148 female U.S.college students. Does the accompanying datasupport the research hypothesis that the meanBoredom Proneness Rating is higher for men thanfor women? Test the appropriate hypothesis usinga 0.05 significance level.
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Gender Sample
Size
Sample
Mean
Sample
Standarddeviation
Male
Female
97
148
10.40
9.26
4.83
4.68
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Solution
0 1 2
1 1 2
1 2
1 2
1 2
1 2
2 21 2
1 2
0.05
:
:
97 148
10.40 9.26
4.83 4.68
1.141.83
0.3885
1.645
H
H
n n
X X
s s
X XZ
s s
n n
Z
=
>
= =
= =
= =
= = =
+
=
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As , we reject the null hypothesis at5% level of significance.
Hence it can be concluded that the mean
Boredom Proneness Rating is higher formen than for women at 5% level of
significance.
Z Z>
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Assignment Problem 2
A manufacturer has just marketed a new appliancewith a one year warranty. The product developmentplan anticipates that the cost of meeting the warrantyterms will be Rs 5000 per appliance, on theaverage, with a standard deviation of Rs 4000. As the
warranty period expires for the first appliancessold, the manufacturer will note the actual costs ofwarranty servicing (X). Two testing approaches havebeen suggested for deciding whether or not theaverage warranty costs are exceeding the Rs 5000target. (1) Wait for warranty data on the first 100appliances sold and conclude that the average costwill exceed the target if > Rs 5800. (2) Wait forwarranty data on the first 400 appliances sold andconclude that the average cost will exceed the targetif > Rs 5400.
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A timely conclusion about actual warrantycosts is desirable because the manufacturer
wishes to reduce the warranty terms if
these are proving too generous to beprofitable. On the other hand, an
unnecessary reduction of warranty terms
would adversely affect future sales of the
industrial appliance. Assume that thewarranty data on the initial sales will
constitute random sample observations.
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Questions
Define the parameter and the alternatives H0 and H1 whichare of interest in this situation.
Sketch the power curves for the two test approaches on thesame graph, assuming that the anticipated standard deviationof warranty costs is accurate. How do the risks of the twoapproaches 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, whatconclusion should be drawn about average warranty costs?
Estimate the alpha risk of your test when =Rs 5000. Estimatethe beta risk of your test when the mean warranty cost isRs1000 higher than the target.
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Hypotheses
Type I error
P[Rejecting H0
| H0
is true]
0
1
: 5000
: 5000
H
H
=
>
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Type I error
Rule 1
Reject if . Here .
Rule 2
Reject if . Here .
0H 5800X> 100n =
5800 | 5000 0.0228P X = > = =
0H 5400X> 400n=
5400 | 5000 0.0228P X = > = =
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Power of a test
The power of a statistical hypothesis testis the probability of rejecting the null
hypothesis when the null hypothesis is
false. Power = (1 - )
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Table showing Power
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
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Power Curve
0
0.2
0.4
0.6
0.8
1
1.2
5000 5100 5400 5700 6000 6300
Rule 1
Rule 2
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As , Rule 2 will not reject the null
hypothesis. Therefore, we conclude thatthe warranty cost does not exceed Rs
5000.
400 Rs 5340 Rs 3840n X s= = =
5340X=
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Risks
If is known,
If is unknown,
Beta risk when the mean warranty cost is
Rs 1000 higher than the target.
If is known, is negligible.
If is unknown, is negligible.
0.0228= 0.0188=
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Assignment Problem 1
An aluminum company is experimenting with a newdesign for electrolytic cells in smelter pot rooms. Amajor design objective is to maximize a cellsexpected service life. Thirty cells of the new designwere 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) thecomparative performance of this design with thestandard industry design, which is known to have a
mean service life of 1,300 days. Management does notwant to conclude that the new design is superior tothe standard one unless the evidence is fairly strong.
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Assuming the distribution of service life of the cell is normal; calculate anappropriate 95% confidence interval for the mean service life of cells ofthe new design. Justify your choice of one or two-sided confidence interval.Should management conclude that the new design is superior to thestandard one with respect to mean service life? Comment.
It has been suggested that the logarithms of service life are more normallydistributed than the original observations. Take logarithms (to base 10) of
the service-life data and calculate the same type of confidence interval asin (a) for the mean log-service-life of cells of the new design. Cells of thestandard design are known to have mean log-service-life of 3.095 (to base10). Can management claim with confidence that the mean log-service-lifeis 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 thedistribution 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 relevantmeasure here? Explain.
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Confidence interval
The confidence interval is
This does not support that the new design issuperior to the standard one.
1( ) ,ns
X t
n
( )1255.361,
0.051379.133 399.0149 (29) 1.699X s t= = =
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Logarithms of service times
The confidence interval is
This does not support that the new
design is superior to the standard one.
0.053.120485 0.134747 (29) 1.699YY s t= = =
( )3.078,
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0
1
2
3
4
5
6
7
8
9
10
500 800 1100 1400 1700 2000 2300 More
Frequency
Bin
Histogram
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0
1
2
3
4
5
6
7
8
9
10
2.79 2.89 2.99 3.09 3.19 3.29 3.39 More
Frequency
Bin
Histogram
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