Specialty Toys Problem.pdf

7/22/2019 Specialty Toys Problem.pdf

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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


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


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


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


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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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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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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