Business Stats Ken Black Case Answers

Case Notes 1

ANSWERS TO CASES Chapter 1 DiGiorno Pizza: Introducing a Frozen Pizza to Compete with Carry-Out In conducting research for the launching of a new product it is imperative that the target population be identified. In this case, who are the people most likely to be interested in purchasing and consuming frozen pizzas in lieu of carry-out pizzas? How are these people to be identified for sampling (Chapter 7 refers to this group as the “frame”)? Should a test market city or area be used? Why or why not? What mode of survey such as telephone, mail, or personal interview should be used? When should these people be surveyed? Does time of day, day of the week, or season of the year make any difference? What types of measurements should be taken? Some possible measurements might include dollar amounts spent per week of pizza per family, number of pizzas purchased per month, percentage of family pizza consumption that is frozen pizza, and total amount spent per month on take out food. 1. One population that was identified was “pizza lovers”. These people may have been previously identified by market researchers based on number of pizzas purchased per month, use of coupons, or from previous surveys. Another population mentioned in the case was women ages 25 to 54. The advertisements shown on national TV were likely aimed at the general population because Kraft was attempting to achieve broader goals such as brand name recognition and a dissemination of the “fresh-baked taste” message. In each of the research efforts, the market research company selected only a sample of the population. SMI-Alcott sent out 1,000 surveys to pizza lovers, the Loran Marketing Group conducted focus groups (which usually have no more than 15 people per group) with women 25 to 54, and Product Dynamics used focus groups to conduct blind taste tests. The market research companies (SMI-Alcott, the Loran Marketing Group, and Product Dynamics), took various measurements on sample members and from these measurements likely computed statistics. Some of these measurements may have included the ranking of various frozen pizza brands based on taste or status, numerical ratings of various types of pizza in terms of taste (perhaps, for example, on a scale from 1 to 7), amount of time sample members are willing to spend cooking a pizza, amount of money spent per month on pizzas, and percentage of sample who recognize the DiGiorno name. Using these measurements, sample statistics such as average amount of money spent per

Case Notes 2

month on pizzas per family or proportion of the sample that recognized the DiGiorno name can be computed. From these sample statistics, population parameters can be estimated such as the percentage of all adults in the country who recognize the DiGiorno name; and the average amount a family spends on take-out pizza per month? This is a good place to introduce the estimation concepts of Chapter 8 intuitively. One can discuss point estimates (sample statistics) and the notion of sampling error. Kraft likely used known descriptive market statistics in their product decision making such as total annual amount of dollars spent in the U.S. on frozen pizza; population demographics of the U.S. including age, number and size of households, average household income; and number of competitors in the frozen pizza market. 2. a. number of pizzas per week b. age of purchaser c. zip code d. dollars spent per month e. time between purchases f. rating of taste g. ranking of four brands h. geographic location i. quality rating j. identification number k. gender

ratio level ratio level nominal level ratio level ratio level ordinal level (but some researchers treat as interval) ordinal level nominal level ordinal level nominal level nominal level

Case Notes 3

Chapter 2 Soap Companies Do Battle The pie chart is useful in displaying the market shares in one device adjacent to each other. Many decision makers are used to viewing pie charts in connection which budgets and therefore might feel more at ease with a pie chart. On the other hand, as mentioned in the text, when percentages are close such as with Dial and “Others” in 1983, it can be difficult to discern the difference using the pie chart slices. In this case, the bar chart shown above is more desirable. 1. Shown below are pie charts for the 1983, the 1991, and the 1999 market shares.

Case Notes 4

An examination of the pie charts from 1983 through 1999 reveals that the slice sizes of Unilever have grown and the sizes of the Procter & Gamble slices have shrunk. Shown below are the actual percentage figures for the three time periods so that you have the option of displaying the data in different ways: Company Procter & Gamble Unilever Dial Colgate-Palmolive Others

1983 Share

1991 Share

1999 Share

37.1 24.0 15.0 6.5 17.4

30.5 31.5 19.0 8.0 11.0

28.4 38.5 14.8 9.3 9.1

2. Shown below is a histogram of the weekly sales of bars of soaps over the year. The histogram was constructed using 10 classes. In MINITAB, the student has the option of trying several different values for the number of intervals. In Excel, students can explore various bin options. The shape of the histogram will somewhat change according to the number of class intervals. Note the shape of this histogram is mound shaped with some skewness to the right. The center of the distribution appears to be near to 20 million as would be expected since Procter & Gamble sells about 20 million bars per week. Note, however that some weeks actually average as much as 39 million bars per week and others only 12 million bars. What inventory, production, and human resource implications might this have? How does a company “cope” with such fluctuations?

Case Notes 5

The stem and leaf plot for these data is shown below. The advantage of the stem and leaf over histograms, pie charts, bar charts, and others is that the stem and leaf retains the original data in case the researcher wants to calculate other statistics on the numbers. Production people would likely find the histogram the most interesting because it displays to them where the bulk of production occurs and the magnitude of the unusual size runs. Stem 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32

Leaf 2 6 7 0, 4, 4, 5 8 0, 1, 1, 4, 5 2, 3, 4, 5, 7 1, 3, 6, 9 0, 3, 3, 4, 4, 6, 7, 9 3, 4, 4 5, 8 1, 4, 8, 9 0, 3 2, 2 2, 2, 3, 3, 6, 9

6 8

Case Notes 6 33 34 35 36 37 38 39

8

Case Notes 7

Chapter 3 Coca-Cola Goes Small in Russia Shown below is MINITAB output describing the sample of 50 bottle fills. Note that the mean fill is 200.12 oz. with a standard deviation of 0.42. The minimum fill is 199.20 oz. and the maximum is 201.10 oz. The median fill is 200.15 oz. The measure of skewness (.0367) demonstrates a very slight positive skewness. However the histogram and the high p-value associated with the normality test indicate that the data are approximately normally distributed. The mean fill of 200.12 oz. indicates that, on average, the sample fills are very near to 12 oz. and are, if anything, giving a slight amount of free product away to the consumer. Under the empirical rule, using µ = 200.12 and σ = .42, 95% of the fill should be within 200.12 + .42 or between 199.70 and 200.54 oz. Descriptive Statistics: Bottle Fills Variable Bottle F Variable Bottle F

N 50

Mean 200.12

Minimum 199.20

Median 200.15

Maximum 201.10

TrMean 200.12 Q1 199.80

StDev 0.42 Q3 200.40

Descriptive Statistics Variable: Bottle Fills Anderson-Darling Normality Test A-Squared: P-Value:

199.2

199.6

200.0

200.4

200.8

201.2

Mean StDev Variance Skewness Kurtosis N Minimum 1st Quartile Median 3rd Quartile Maximum

95% Confidence Interval for Mu

0.217 0.833 200.118 0.419 0.175384 3.67E-02 -1.4E-01 50 199.200 199.800 200.150 200.400 201.100

95% Confidence Interval for Mu 199.999 199.95

200.05

200.15

200.25

0.350

95% Confidence Interval for Median

200.237

95% Confidence Interval for Sigma 0.522

95% Confidence Interval for Median 199.967

200.233

SE Mean 0.06

Case Notes 8 2. The bottles have a label that claims there are 20 oz. of fluid therein. This sample of 150 bottles has an average of 20.003 oz. with a median of 20.005 and a mode of 20.004. All three of these statistics indicate that, if anything, the bottles are slightly overfilled overall. The standard deviation of fills is .027 oz. The MINITAB normality statistics and histogram overlaid with the normal curve indicate that the data are approximately normally distributed. We can apply the empirical rule. Approximately 68% of the fills are within 20.003 + 1(.027), from 19.976 to 20.03 oz.; and 95% within 20.003 + 2(.027), from 19.949 to 20.057. The measure of skewness (-.085) indicates a slight negative skewness. The box plot indicates that there is an outlier on both the lower and the upper end (extreme underfilled bottle and extreme overfilled bottle). An examination of the minimum value shows that there is a bottle with 19.92 oz. which has a Z score of –3.07 or over three standard deviations below the mean. The maximum value is 20.09 which has a Z score of 3.22 or almost three-and-a-quarter standard deviations above the mean. Production and quality management people can better interpret these extreme fills in light of company goals and specifications. Overall, the fills are averaging more than 20 oz., the fills are approximately normally distributed, and the standard deviation of fills is about .03 of an ounce.

Case Notes 9

Chapter 4 Colgate-Palmolive Makes a “Total” Effort 1. Two probabilities are given in this case. The first is the marginal probability U.S. household had purchased Total for the first time:

that a

P(T1) = .21 The second is the conditional probability that a household purchased Total (for a second time) given that it had purchased it before: P(T2  T1) = .43 The percentage of U.S. household that purchased Total at least twice can be computed as: P(T1  T2) = P(T1) · P(T2T1) = (.21)(.43) = .0903 A little over nine percent (9.03%) of U.S. households purchased Total at least twice during this time period. 2. If age is independent of the willingness to try Total, then the marginal probability of an age category should equal the conditional probability that someone is in that age category given that they are willing to try Total. Here, for example, the probability that someone is in the 45-64 age category is: P(45-64) = .20 The conditional probability that someone is in the 45-64 age category given their willingness to try Total is: P(45-64T) = .24 If age is independent of willingness to try Total, then the following must be true: P(45-64) = P(45-64T) but .20 ≠ .24. Therefore, age is not independent of the willingness to try Total.

Case Notes 10

3. The probability that a person is either in the 45-64 age category or purchased Total is a union probability and is computed as: P(45-64  T) = P(45-64) + P(T) - P(45-64  T) = .20 + .21 - .0903 = .3197 The probability that a person purchased Total given that they are in the 45-64 age category can be computed as: P(T45-64) =

P (T ∩ 45 − 64) .0903 = = .4515 P ( 45 − 64) .20

4. Let S = saw the commercial, N = didn’t see the commercial, Pi = prior probability P(S) = .32

P(N) = .68

P(TS) = .40

P(TNS) = .1206

Event

Prior

P(TPi)

S

.32

.40

N

.68

.1206

P(T  Pi)

P(Pi T)

(.32)(.40) = .128

.61

(.68)(.1206) = .082

.39

P(T) = .210

Case Notes 11

Chapter 5 Fuji Film Introduces APS 1. n = 30, p = .40. The expected number is:

µ = n·p = 30(.40) = 12 The probability that six or fewer purchase an APS camera is: Prob(x < 6  n = 30 and p = .40) = 30

C6 (.40) 6 (.60) 24 +30 C5 (.40)5 (.60) 25 + ⋅ ⋅ ⋅ +30 C0 (.40)0 (.60)30 =

.0115 + .0041 + .0012 + .0003 + .0000 + .0000 + .0000 = .0171 If the market share actually is 40% (p = .40), the expected number of purchasers from a sample of 30 is 12. Six or fewer are considerably less than the expected number (12). How often would one get six or fewer out of thirty when twelve is expect – less than 2% (.0171) of the time. Therefore, if six or fewer out of thirty actually purchase, it is likely that the market share is not 40%. 2.

Let λ = 2.4 complaints/100,000 rolls. Shown below are some of the values for the Poisson distribution with λ = 2.4 from MINITAB: x 0 1 2 3 4 5 6 7 8 9 10 11 12

Probability .0907 .2177 .2613 .2090 .1254 .0602 .0241 .0083 .0025 .0007 .0002 .0000 .0000

Case Notes 12

With a Poisson distributed average rate of 2.4 complaints/100,000 rolls, the probability of getting 7 complaints/100, 000 rolls by chance is .0083 or less than 1%. Often researchers like to use the cumulative probability of x > 7 on a problem like this. The probability of randomly getting 7 or more complaints/100,000 rolls when the average is only 2.4 is . 0117 (.0083 + .0025 + .0007 + .0002). This is still quite unlikely to happen by chance. If seven complaints are actually logged, then managers would have to strongly consider the possibility that the average rate of 2.4 complaints has increased. This is, as has been stated, unacceptable to management. 3. This is a hypergeometric problem with: N = 52, n = 10, A = 18 (customer satisfaction), and x = 7 The probability is computed as: 18

C7 ⋅34 C3 = .0120 52 C10

The probability that seven out of the ten successful products were created for customer service when only eighteen of the fifty-two original products were created for customer service is about 1% (.012). Since this is unlikely to happen by chance, it is likely that there is something inherently more successful about creating a product for customer service reasons than for revenue growth.

Case Notes 13

Chapter 6 Mercedes Goes after Younger Buyers 1. Prob(x > 42,000µ = 43,215 and σ = 2981): z=

x −µ

σ

=

42,000 − 43,215 = -0.41 2981

From Table A.5, the area for z = -0.41 is .1591. Prob(x > 42,000) = .5000 + .1591 = .6591 = 65.91% Almost 66% of Mercedes dealers would be priced out of competition with this BMW model. Prob (x > 43,215µ = 34,990 and σ = 2367): z=

x −µ

σ

=

43,215 − 34,990 = 3.47 2367

From Table A.5, the area for z = 3.47 is .4997. Prob(x > 43,215) = .5000 - .4997 = .0003 = 0.03% Virtually none of the BMW dealers are pricing the 328 ci more than the average price of the Mercedes CLK320. Prob(x < 34,990µ = 43,215 and σ = 2981): z=

x −µ

σ

=

34,990 − 43,215 = -2.76 2981

From Table A.5, the area for z = -2.76 is .4971. Prob(x < 34,990) = .5000 - .4971 = .0029 = 0.29% About .3% of the Mercedes dealers are pricing CLK320 less than the average price of the BMW 320 ci. Prob(x < 37,059µ = 43,215 and σ = 2981):

Case Notes 14 z=

x −µ

σ

=

37,059 − 43,215 = -2.07 2981

From Table A.5, the area for z = -2.07 is .4808. Prob(X < 37,059) = .5000 - .4808 = .0192 = 1.92% Less than 2% of the Mercedes dealers price the CLK320 less than $37,059. Conclusion: There is little overlap in the prices of the two cars and it could be concluded that they are not really competing with each other pricewise.

2. CLK: Prob. =

a = 24

x1 = 26

x2 = 30

30 − 26 = .4 = 40% 34 − 24

328is: a = 25 Prob. =

b = 34

b = 35

x1 = 26

x2 = 30

30 − 26 = .4 = 40% 35 − 25

The same proportion of 328is cars fall into this category (26-30 mpg). However, an examination of the end points of the mileage distributions of each car reveals that the upper end for 328’s is 1 mpg. higher than for the 328is’s and the lower end for CLK’s is 1 mpg. lower. Both cars have very close gas mileage figures. Each car more than 30 mpg.: For CLK: Prob. =

34 − 30 = .4 = 40% 34 − 24

For 328is: Prob. =

35 − 30 = .5 = 50% 35 − 25

A higher proportion of 328’s are in the more than 30 mpg. category than CLK’s.

Case Notes 15

3.

λ = 1.37 cars/3 hours, µ = 1/1.37 = .73 of 3 hours = 2.19 hours For 1 hour: 1 hour = .333 of 3 hours. x0 = 0.333. The cumulative probability of this time interval is .3663. This means that there is a 36.63% chance that there will be less than one hour between sales. For 12 hours: 12 hours = 4 times 3 hours. x0 = 4. The cumulative probability for this time interval is . 9958 meaning that there is a 99.58% chance that there will be less than 12 hours between sales. The complement of this is that there is a 1 - .9958 = .0042 = 0.42% chance that there will be more than 12 hours between sales. Managers know that there is an almost 37% chance of a sale within every hour. They need to determine how much staffing it takes to sell a car every hour or less. Given that it takes several potential buyers and often multiple visits to the dealership to sell one car and that it is relatively likely (probability almost 75%) that they will close a sale every 3 hours (x0 = 1), the dealership should never go without having salespeople around and may have to have several employees around all the time. By having good exponential and Poisson distribution data, one can, to some extent, track the impact of advertising on sales by testing values of λ using random arrival data in time periods following advertising to determine if λ has increased. For example, if λ = 1.37 every 3 hours but following a advertising campaign, there is a randomly selected 3 hours period and 5 cars are sold, then management might be able to statistically justify that the λ has increased and then conclude that the advertising campaign was the cause. In many businesses, the value of lambda changes with time of day, day of the week, and season of the year. In the car business, there may be an increase in sales on the weekend, in the evening, or perhaps in the fall when new models arrive. Students should always be cautioned about using the same value of lambda for all time periods. Many students know intuitively that lambda varies over time.

Case Notes 16

Chapter 7 Shell Attempts to Return to Premier Status 1. The answers to this question will vary. Students should select one of the four types of random sampling or a hybrid (eg. area – stratified) to use in their sampling plan. The target population is that group of people that the researcher want to be able to infer to. For example, if Shell wants to impact all adults in the U.S., then all adults in the U.S. should be their target population. The frame is the list or roster of this target population from which the researchers sample. What is the frame of all adults in the U.S.? This is a difficult question. Many national lists involve some specialty group such as registered Republicans, Visa card users, Internet users, Catholic church members, etc. which really do not capture all U.S. adults. Instead of searching for a national frame, the researcher may want to use some form of area sampling such as selecting a test market city which is thought to be similar in demographics to the U.S. This might make frame identification easier because some test market city directory such as the phone book or voter registration list might be accessible and include most adults. As an example of two-stage sampling, researchers could randomly select a few test market cities and then sample every 100th name from the phone book or interview everyone from a few randomly selected blocks within the test market city. 2.

Shell contracted researchers appear to have stratified on age, ethnicity, household location, occupation, and previous employment with Shell. With regard to opinions about Shell as a “premier” company, some strata that might make sense are age, ethnicity, economic class, education, occupation, gender, and geographic location. It is important to Shell that all segments of the U.S. adult population be reached. In order to test to determine the effectiveness of marketing campaigns, Shell needs to recognize that there will likely be differences in the perceptions of young people and those of old people of what a “premier” company should be because of their life experiences and the types of messages that appeal to them. The same thing is true for different ethic groups (different cultural values may appeal to different groups), economic class (the economics of the household may determine what types of “premier” messages appeal to their needs), education (at what level of education should the messages be targeted), occupation (how does Shell impact various occupations differently? eg. environment, reliability of products, availability of products, pricing), gender (often men and women seek different outcomes from firms), and geographic location (state and regional cultures vary and thus consumer messages should be targeted so as to parallel geographic interests).

Case Notes 17

3. In 1979, p = .12.

ˆ = .25. A new survey of n = 350 resulted in p

ˆ > .25 n = 350 and p = .12): Prob.( p

z=

pˆ − p p ⋅q n

=

.25 − .12 (.25)(.75) = 5.62 350

From Table A.5, the area for z = 5.62 is .5000. ˆ > .25) = .5000 - .5000 = .0000 Prob.( p

It is virtually impossible to randomly select 350 people and have 25% declare that Shell is a “premier” company if in the population only 12% believe that Shell is a “premier” company. This is strong statistical evidence that the 12% figure is no longer true and that the actual population figure is greater. This is a nice segue into section 8.3 of chapter 8 in which the sample data can be used to estimate the actual proportion who now believe that Shell is a “premier” company. It also lays the groundwork for the hypothesis testing to come in chapter 9. 4. Prob.( z=

x

> 2.0  µ = 1.8, σ = .7, and n = 35):

x −µ

σ

n

=

2.0 −1.8 .7 = 1.69 35

From Table A.5, the area for z = 1.69 is .4545 Prob.(

x

> 2.0) = .5000 - .4545 = .0455

There is only a 4.55% probability that the sample mean of 2.0 was obtained by chance. It is likely that the population mean is no longer 1.8 and indeed, is now higher. This is a good place for the instructor to mention .05 as a common standard for low probability and begin the groundwork for chapter 9.

Case Notes 18

Prob.( z=

x

> 2.5  µ = 1.8, σ = .7, and n = 35):

x −µ

σ

n

=

2.5 −1.8 .7 = 5.92 35

From Table A.5, the area for z = 5.92 is .5000. Prob.(

x

> 2.5) = .5000 - .5000 = .0000

It is virtually impossible to randomly obtain a sample mean of 2.5 or more from a sample of 35 with this standard deviation if the population mean is only 1.8. This is, of course, even more conclusive evidence than obtaining a sample mean of 2.0. A discussion of the .0455 probability above and the .0000 value here may result in a preliminary understanding of the notion of p-value introduced in chapter 9.

Case Notes 19

Chapter 8 Thermatrix 1. n = 115

For 95% confidence, z = 1.96

Use: pˆ ± z

pˆ ⋅ qˆ n

1) Yes: pˆ =

63 = .5475 115

.5475 + 1.96

(.5475)(.4522) = .5478 + .0910 115

.4568 < p < .6388

2) Yes:

pˆ =

86 = .7478 115

.7478 + 1.96

(.7478)(.2522) = .7478 + .0794 115

.6684 < p < .8272

3) Yes:

pˆ =

101 = .8783 115

.8783 + 1.96

(.8783)(.1217) = .8783 + .0598 115

.8185 < p < .9381

Case Notes 20

4) Yes: pˆ =

105 = .9130 115 (.9130)(.0870) = .9130 + .0515 115

.9130 + 1.96

.8615 < p < .9645

2. n = 9 Use: 1)

x

df = 8 x ±t

For 95% confidence, t.025,8 = 2.306

s n

= 3.79

3.79 + 2.306

s = .86 (.86) = 3.79 + .66 9

3.13 < µ < 4.45 2)

x

= 2.74

2.74 + 2.306

s = 1.27 (1.27 ) = 2.74 + .98 9

1.76 < µ < 3.72 3)

x

= 4.18

4.18 + 2.306

s = .63 (.63) = 4.18 + .48 9

3.70 < µ < 4.66

Case Notes 21

4)

x

= 3.34

3.34 + 2.306

s = .81 (.81) = 3.34 + .62 9

2.72 < µ < 3.96 5)

x

= 3.95

3.95 + 2.306

s = .21 (.21) = 3.95 + .16 9

3.79 < µ < 4.11

Case Notes 22

Chapter 9 Frito-Lay Targets the Hispanic Market 1.

a) H0: p = .63 Ha: p ≠ .63 Let α = .05

For a two-tailed test, α/2 = .025

z.025 = + 1.96 (critical value) n = 850 pˆ =

z=

x = 575

575 = .6765 850

pˆ − p p⋅q n

=

.6765 − .63 (.63)(.37) = 2.81 850

Since the observed z = 2.81 > z.025 = 1.96, the decision is to reject the null hypothesis. The proportion of Hispanics that are Mexican Americans is not .63. The sample data indicate that the proportion is higher. b) H0: p = .93 Ha: p < .93 Let α = .05 n = 689 pˆ =

z=

z.05 = - 1.645 (critical value) x = 606

606 = .8795 689

pˆ − p p⋅q n

=

.8795 − .93 (.93)(.07) = -5.20 689

Since the observed z = -5.20 < z.05 = -1.645, the decision is to reject the null hypothesis. The proportion of Hispanic grocery shoppers that are women is less than .93.

Case Notes 23

c) H0: p = .83 Ha: p ≠ .83 Let α = .05

For a two-tailed test, α/2 = .025

n = 438 ˆ p

= .792237

Since the p-value of .042 < α = .05 (the p-value was adjusted by MINITAB for the two-tailed test), the decision is to reject the null hypothesis. The proportion of Hispanics who listen to advertisements in Spanish is not .83. The sample data indicate that the proportion may now be less than .83.

2. a) H0: µ = 31 Ha: µ ≠ 31 x

α = .01

= 28.81

The observed t = -1.52 For a two-tailed test, t23,.005 = -2.07 Since the observed t = -1.52 > t23,.005 = -2.07, the decision is to fail to reject the null hypothesis. There is not enough evidence to say that the average age is different from 31 years. Marketing decision makers must assume that the average age has not changed. b) H0: µ = 45 Ha: µ < 45

α = .05

= 35.67

s = 19.26

x

t=

n = 18

df = 17

t.05, 17 = -1.740

x − µ 35.67 − 45 = s 19.26 = -2.06 n 18

Since the observed t = -2.06 < t.05,17 = -1.740, the decision is to reject the null hypothesis. We conclude that the average expenditure of Hispanic customers on chips per year is less than $ 45.

Case Notes 24

Chapter 10 Seitz Corporation Producing Quality Gear-Driven and Linear-Motion Products 1. Comparing last year’s mean transactions to this year’s: n1 = 20 n2 = 25

x 1 = 2300 x 2 = 2450

s1 = 500 s2 = 540

H0: µ1 - µ2 = 0 Ha: µ1 - µ2 ≠ 0 df = n1 + n2 - 2 = 20 + 25 - 2 = 43

α = .05

α/2 = .025

t.025,43 = + 2.021 ( x1 − x 2 ) − ( µ1 − µ 2 ) t =

2

2

s1 (n1 − 1) + s 2 (n2 − 1) 1 1 + n1 + n 2 − 2 n1 n2

=

( 2300 − 2450) − (0) (500) (19) + (540) 2 (24) 20 + 25 − 2 2

1 1 = -0.96 + 20 25

Because t = -0.96 > t.025,43 = -2.021, the decision is to fail to reject the null hypothesis. There is not enough evidence here to say that there is any difference in the average dollar amount of sales between this year and last.

Case Notes 25

2. Comparison of tractors at two plants using a confidence interval: n1 = 45 pˆ1 =

x1 = 18

n2 = 51

18 = .4000 45

pˆ 2 =

( pˆ 1 − pˆ 2 ) ± z

x2 = 12 12 = .2353 51

pˆ 1 qˆ1 pˆ 2 qˆ 2 + n1 n2

(.40 - .2352) ± 1.96

(.40)(.60) (.2353)(.7647) + = .1647 + .1845 45 51

-.0198 < p1 - p2 < .3492 The point estimate of the difference in quality of tractors at the two plants is 16.47%. However, due to the relatively small samples, the error of the interval is 18.45% which is greater than the point estimate. Combining the error of the interval with the point estimate results in the confidence interval shown above. Note that zero is in the interval indicating that there is a possibility that there is no difference in the quality ratings of tractors produced at the two plants. If this were a hypothesis testing problem, then the decision would be to fail to reject the null hypothesis based on the confidence interval’s inclusion of zero.

Case Notes 26

3. 2000 vs. 2001: t = -1.83 with a p-value of .07. This is not significant at α = .05. There is no significant difference in the mean ratings between 2000 and 2001. This is underscored by the confidence interval that includes zero. However, if α = .10 were used, there would be a significant difference. Examining the means reveals that the mean score for 2002 was higher. The sample sizes were 75 for 2000 and 93 for 2001. 4. Comparison of variances for week 1 and week 5: H0: σ12 = σ22 Ha: σ12 ≠ σ22 Let α = .05 F.025,5,6 = 5.99

α/2 = .025

s1 = 1.17

Week 5: n2 = 7

s2 = 1.68

F=

2

s2

2

=

df2 = n2 - 1 = 6

F.975,6,5 = 1/5.99 = 0.167

Week 1: n1 = 6

s1

df1 = n1 - 1 = 5

(1.17) 2 = 0.485 (1.68) 2

Since the observed value of F = 0.485 is > the left tail critical value of F = 0.167, the decision is to fail to reject the null hypothesis. The variances of product being produced these two weeks are not significantly different. Management would probably like this because this indicative of consistent production patterns. Wide swings in variance would be of concern because it would indicate that some weeks the variability is more out-of-control than others and a less consistent product is being produced.

Case Notes 27

Chapter 11 J. R. Clarkson Company 1. The two by three factorial design is analyzed using a two-way ANOVA. There are two independent variables, temperature and supplier. Temperature has three treatment levels: 70o, 110o, and 150o. Supplier has two classifications levels: supplier A and supplier B. The dependent variable is strength of the valve as measured in psi. Shown below is Excel output for this analysis. ANOVA Source of Variation Supplier Temperature Interaction Within Total

SS 20.056 800.111 84.778 217.333

df 1 2 2 12

1122.278

17

MS 20.056 400.056 42.389 18.111

F 1.11 22.09 2.34

P-value 0.313385 0.000095 0.138598

F crit 4.75 3.89 3.89

First, we examine the observed F for interaction which is 2.34 with a p-value of .1386. Since interaction is not significant at any commonly used alpha, we proceed to examine main effects. There is no significant difference between the two suppliers (F = 1.11, p-value = .31339). There is a significant difference in the strength of the valves by temperature at α = .0001. The mean psi for 70o is 159.83, for 110o is 158.33, and for 150o is 145 psi. It appears that at 150o the valves are not as strong. Using MINITAB, Tukey’s multiple comparison tests were done to determine if there were any significant differences in valve strength by temperature. The results are: Tukey's pairwise comparisons Family error rate = 0.0500 Individual error rate = 0.0203 Critical value = 3.67 Intervals for (column level mean) - (row level mean) 1 2

-5.444 8.444

3

7.890 21.777

2

6.390 20.277

These results show that there were significant differences between 70o and 150o and between 70o and 150o. The confirms what we observed above with the Excel output.

Case Notes 28

2. The data are analyzed using a one-way ANOVA. The independent variable is country with four classifications: Canada, Columbia, Taiwan, and the U.S. The dependent variable is the dollar cost to replace a seal. Shown below is the MINITAB output for this one-way analysis with multiple comparisons: One-way ANOVA: Cost versus Country Analysis of Variance for Cost Source DF SS MS Country 3 64331 21444 Error 24 15750 656 Total 27 80081 Level 1 2 3 4

N 7 7 7 7

Mean 244.29 323.57 195.00 222.14

Pooled StDev =

StDev 31.01 30.65 20.82 17.04

25.62

F 32.68

P 0.000

Individual 95% CIs For Mean Based on Pooled StDev -----+---------+---------+---------+(---*---) (---*---) (---*---) (---*---) -----+---------+---------+---------+200 250 300 350

Tukey's pairwise comparisons Family error rate = 0.0500 Individual error rate = 0.0110 Critical value = 3.90 Intervals for (column level mean) - (row level mean) 1

2

2

-117.0 -41.5

3

11.5 87.0

90.8 166.3

4

-15.6 59.9

63.7 139.2

3

-64.9 10.6

The results show that there is a significant difference in the cost of seal replacement between countries (F = 32.68, p-value = .000). An examination of the means shows that there is potential significant differences between countries (Canada - $244.29, Columbia - $323.57, Taiwan - $195.00, and U.S. - $222.14). The results of the Tukey multiple comparison analysis shows that there are significant pairwise differences between Canada and Columbia, between Canada and Taiwan, between Columbia and Taiwan, and between Columbia and the U.S. Clarkson might be very effective in marketing to Columbian companies because in Columbia the cost of replacing the seal is significantly higher than in other countries.

Case Notes 29 3. This is a randomized block design. The main independent variable of interest is type of valve. The blocking variable was day of the week. Below is Excel output for a two-way ANOVA without replication. ANOVA Source of Variation Day of the Week Type of Valve Error

SS 0.7767 6.0107 0.7793

df 4 5 20

Total

7.5667

29

MS 0.1942 1.2021 0.0390

F 4.98 30.85

P-value 0.00595206 0.000000010

F crit 2.87 2.71

A highly significant observed F value was obtained for type of valve (F = 30.85 with a p-value of .000000001). The various valve types and their associated mean lead times are: Safety – 1.64 Butterfly – 2.12 Clack – 1.32 Slide – 1.66 Poppet – 2.18 Needle – 0.88 Since there is a significant difference in valve type and since the means appear to be quite different, Tukey’s multiple comparisons were used to determine which pairs of means, if any, are significantly different. The MINITAB output for this analysis is given below: Tukey's pairwise comparisons Family error rate = 0.0500 Individual error rate = 0.00501 Critical value = 4.37 Intervals for (column level mean) - (row level mean) 1

2

3

4

2

-0.9776 0.0176

3

-0.1776 0.8176

0.3024 1.2976

4

-0.5176 0.4776

-0.0376 0.9576

-0.8376 0.1576

5

-1.0376 -0.0424

-0.5576 0.4376

-1.3576 -0.3624

-1.0176 -0.0224

6

0.2624 1.2576

0.7424 1.7376

-0.0576 0.9376

0.2824 1.2776

5

0.8024 1.7976

The results of these multiple comparison tests is that lead times for the following pairs of valves are significantly different:

Case Notes 30 Safety and Poppet, Safety and Needle, Butterfly and Clack, Butterfly and Needle, Clack and Poppet, Slide and Poppet, Slide and Needle, and Poppet and Needle.

The study was attempting to control for day of the week as a blocking variable. The blocking variable produced an F value that was significant at α = .01. Lead times differ by type of valve. The needle and clack valves have the shortest lead times and the butterfly and poppet valves have the longest lead times. A cursory examination of the mean lead times by day of the week indicates that Monday’s and Friday’s produce the longest lead times.

Case Notes 31

Chapter 12 Foot Locker in the Shoe Mix 1. Has the distribution of shoe sales by Price Category changed from the year 2000 to the year 2001? A chi-square goodness-of-fit test can be used to test this. Let the 2000 distribution be the expected frequencies and the 2001 values be the observed frequencies. Neither MINITAB nor Excel directly computes a chi-square goodnessof-fit test. Utilizing MINITAB’s calculator capability (could have used Excel’s), the following work is obtained. 2000

2001

115 38 37 30 22 21 11 17

126 40 35 27 20 20 11 18

chi-sq.

1.05217 0.10526 0.10811 0.30000 0.18182 0.04762 0.00000 0.05882 χ2 = 1.85380

The observed chi-square value is 1.85380 Using an α = .05, and df of k-1 = 7, the critical value is: χ7,.05 = 14.0671

Since the observed chi-square is less than the critical chi-square, the decision is to fail to reject the null hypothesis. There is not enough evidence to declare that the 2001 distribution of shoe sales is any different than the 2000 distribution of shoe sales. Implications to Foot Locker and Nike is that the target market has not changed. This might mean that marketing attempts to change the target market have not been effective. It might also mean that the production schedule for various types of shoes need not change. Demand for shoes at the various price levels remains constant. If the companies desire to sell more high end shoes, they need to make a renewed effort to effect changes in these distributions. This result basically tells them that they are where they were.

Case Notes 32

2. A chi-square test of independence is used to determine if gender is independent of gender. Shown below is MINITAB output for the analysis of this question. Chi-Square Test: Male, Female Expected counts are printed below observed counts Male 29 41.27

Female 43 30.73

Total 72

U.S. South

48 38.98

20 29.02

68

U.S. East

52 64.77

61 48.23

113

U.S. North

28 30.38

25 22.62

53

Europe

78 63.05

32 46.95

110

Australia

47 43.56

29 32.44

76

282

210

492

U.S. West

Total Chi-Sq =

3.647 + 4.898 + 2.090 + 2.806 + 2.517 + 3.380 + 0.186 + 0.250 + 3.545 + 4.761 + 0.272 + 0.365 = 28.716 DF = 5, P-Value = 0.000

An observed chi-square of 28.716 was obtained with an associated p-value of .000 on this question. This indicates that gender is not independent of location when it comes to the number of formal suggestions. A cursory examination of the observed values reveals that more suggestions were submitted by females than males in the U.S. West and the U.S. East. However, more suggestions were submitted by males than females in all other regions. In the U.S. South and Europe the ratio of formal suggestions of males to females was over two to one. What this result says is when it comes to making suggestions, the regional culture has an impact on who makes the suggestions. Foot Locker might make a concerted effort to increase participation by the gender with fewer suggestions in each region and perhaps make an attempt to change corporate culture so that regional tendencies do not effect the employee suggestion program. 3 According to sources, Foot Locker has a 19.4% share of the sneaker market. Foot Locker believes that it holds a higher share of the market in the U.S. West. To test this notion, Foot Locker hires a market research company that randomly samples 1,000 people in the U.S. West who have just purchased new sneakers. Of these, 230 have purchased their sneakers at Foot Locker. The hypotheses being tested are: H0: p = .194 Ha: p > .194

Case Notes 33

The sample information is:

n = 1,000

pˆ =

x = 230

230 = .230 1000

Let alpha be .01. Chapter 9 techniques can be used to analyze these data. However, the chi-square goodness-of-fit can also be used. Using the .194 as the expected proportion and .230 as the observed proportion, a 2x2 table can be set up and analyzed using the chi-square goodness-of-fit. The observed frequency is calculated by taking .194 of 1,000 (the market share that Foot Locker holds nationally). fe

fo

( fo − fe )2 fe

194

230

6.6804

Other Store 806

770

1.6079

Foot Locker

The resulting observed chi-square is 8.2884. With 1 degree of freedom, and an α = .01, the critical chi-square value is 6.6349. The decision is to reject the null hypothesis. This indicates that proportion of the market share in the west is significantly higher than in other locales. Perhaps Foot Locker ought to study the clientele, the stores, the sales people, and the sale methods to determine why they are doing better in the West and attempt to implement things that they learn in other locales.

Case Notes 34

Chapter 13 Delta Wire Uses Training as a Weapon 1. Shown below is MINITAB output for correlation and regression analysis for the education and sick day data. The correlation between hours of education and number of sick days is a relatively strong negative correlation. This indicates that the more hours of education received, the fewer the sick days. There may be several reasons for this. One explanation may be that as workers participate in an educational process about their work, they become more interested in what they are doing because they understand more about it and can see more potential. If the training is interesting, they may look forward to the time in the classroom. In addition, they may feel more a part of the team by having been included in training and feel more self worth for being selected for the training. Correlation of HrsEduc and SickDays = -0.773 The regression equation is SickDays = 7.75 - 0.0795 HrsEduc Predictor Constant HrsEduc s = 2.455

Coef 7.7455 -0.07946

Stdev 0.7980 0.01536

R-sq = 59.8%

t-ratio 9.71 -5.17

p 0.000 0.000

R-sq(adj) = 57.5%

Analysis of Variance SOURCE Regression Error Total

DF 1 18 19

SS 161.26 108.49 269.75

MS 161.26 6.03

F 26.75

p 0.000

The F test for the overall model is significant at .001 as is the t value testing the slope. The r2 of .598 indicates that almost 60% of the variation of the sick days is accounted for by the hours of education. The standard error of the estimate is 2.455 days which is a modest error. The regression equation has an intercept of 7.75 which indicates that the model predicts that a worker will have an average of 7.75 sick days if the worker has participated in no hours of education. Notice the negative slope. As the number of hours of education increase, the regression model is deduct days from the intercept resulting in a prediction of fewer sick days. Shown below is a regression plot of this line and the data:

Case Notes 35

2.

The F value for the regression model indicates significant overall regression. The t-ratio, which here is the square root of F, also yields a p-value of .000 indicating that the population slope is different from zero. The value of r2 = 90.1% is extremely high denoting a strong predictability in the model. The regression equation has a positive slope which indicates that the higher the satisfaction scores, the more sales there are.

Case Notes 36

3. Shown below is Excel regression output for this problem. This output shows that hours of training is highly predictive of productivity (r2 = .976). The overall F value for this regression is extremely large and significant (F = 662.28, p-value of .000000…). In addition, the standard error (1005.644) is quite small in relation to the five-figure productivity numbers that have a range of 20,000. The t test of the slope is also highly significant (t = 179.65, p-value of .000000…) further underscoring the significance of hours of training as a predictor. Overall, this is a very strong regression model with much predictability. SUMMARY OUTPUT Regression Statistics Multiple R 0.988 R Square 0.976 Adjusted R Square 0.975 Standard Error 1005.644 Observations 18 ANOVA Regression Residual Total

Intercept Hours

df 1 16 17

SS MS 669781231.5 669781231.5 16181129.59 1011320.599 685962361.1

Coefficients Standard Error 70880.252 394.546 5.093 0.198

t Stat 179.65 25.73

F 662.28

P-value 7.138E-28 1.903E-14

Significance F 1.9027E-14

Case Notes 37 Chapter 14 Starbucks Introduces Debit Card 1. This model uses four independent variables in an effort to predict the amount of money people spend on their debit card. Overall, the model has modest to good predictability with an R2 of .755 and a standard error of $22.15. This standard error indicates that about 95% of the time, the model will be within +2($22.15) or +$44.30 of the actual figurewhich is not particularly good.. While the overall test of the model is significant (F = 15.38, p-value = .000007), an examination of the t tests and their associated p-values shows that only one of the predictors, income (t = 6.69, p-value = .000002) is significant. None of the other variables are even close. Had a simple regression model been developed using just income to predict the amount of the prepaid card, the R2 would be . 723, the t value for income would increase to 7.74, the standard error of the estimate would reduce to $21.96, and the overall F test would increase to 59.90. SUMMARY OUTPUT Regression Statistics Multiple R 0.869 R Square 0.755 Adjusted R Square 0.706 Standard Error 22.1483 Observations 25 ANOVA Regression Residual Total

Intercept Age Days Cups Income

df 4 20 24 Coefficients -83.826 0.237 1.190 1.422 2.407

SS 30175.0423 9810.9577 39986 Standard Error 22.494 0.576 1.474 2.631 0.360

MS 7543.7606 490.5479

t Stat -3.73 0.41 0.81 0.54 6.69

F 15.38

P-value 0.0013 0.6852 0.4291 0.5949 0.000002

Significance F 0.000007

Case Notes 38

2. This model attempts to predict the number of days per month that a customer frequents Starbucks. The predictor variables are age, income, and number of cups of coffee per day. The Excel results of this analysis are shown below. The model is modest to weak with an R2 of just .416. The standard error of 3.28 days indicates that the model would predict within + 2(3.28) or + 6.56 days about 95% of the time. A perusal of the data shows that the range of number of days is 16 days. The relatively large size of the standard error to this range is further evidence of the model’s weakness. A study of the t statistics reveals that the predictor variable, cups, is the only significant predictor (t = 3.40, p-value .0027). The number of cups of coffee that a person drinks per day seems to be a good predictor of the number of times per month the person frequents Starbucks. Heavy coffee drinkers come often (the coefficient indicates a positive relationship between cups and frequency). In attempting to increase store traffic, Starbucks could target their marketing efforts at the more heavy coffee drinkers or develop and market products that might lure lighter coffee drinkers to their outlets for different reasons. If a simple regression model is used to predict number of days by cups of coffee, the R2 is .345 and the standard error is 3.32.

Case Notes 39

SUMMARY OUTPUT Regression Statistics Multiple R 0.645 R Square 0.416 Adjusted R Square 0.332 Standard Error 3.2791 Observations 25 ANOVA Regression Residual Total

Intercept Cups Income Age

df 3 21 24

SS 160.7593 225.8007 386.56

Coefficients Standard Error 5.9684 3.0651 1.0644 0.3127 0.0716 0.0509 -0.0785 0.0835

MS 53.5864 10.7524

F 4.98

t Stat 1.95 3.40 1.41 -0.94

P-value 0.0650 0.0027 0.1742 0.3578

Significance F 0.0091

Case Notes 40

3. Shown below is the output from an Excel multiple regression analysis to predict sales revenue by number of stores, number of drinks, and average weekly earnings. The predictability is extremely high with an R2 of .9998. In predicting sales revenues that range from 400 to 2600, the standard error of the estimate is only 16.69. The overall F of 4539.21 is significant at alpha = .00001. While number of stores is not a significant predictor (t = -0.95, p-value = .41145), both number of drinks (t = -7.47, p-value = .00497) and average weekly earnings (t = 13.70, p-value .00084) are significant at α = .01. Notice that for the predictor, number of drinks, both the t value and the coefficient are negative. This indicates that, at least in this model with other variables in the model, there is a negative relationship between number of drinks and sales revenue. However, a cursory examination of the raw data shows that as sales revenues increase so do the number of drinks. This points out one of the dangers in over interpreting the regression coefficients (discussed in Chapter 15 in section on multicollinearity). When a simple regression model is run using number of drinks as the sole predictor of sales revenue, the r2 is .929, and more importantly the regression coefficient is positive as is the t statistic. This might serve as an informal/intuitive introduction to the notion of collinearity. The correlation between the two significant predictors in the multiple regression model, number of drinks and weekly earnings, is .984. SUMMARY OUTPUT Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations

0.9999 0.9998 0.9996 16.689 7

ANOVA Regression Residual Total

Intercept Stores Drinks Earnings

df 3 3 6

SS MS 3792735.879 1264245.293 835.55 278.52 3793571.429

Coefficients Standard Error -13500.237 946.164 -0.0264 0.0278 -75.20448212 10.07005905 38.989 2.846675638

t Stat -14.27 -0.95 -7.47 13.70

F 4539.21

P-value 0.00075 0.41145 0.00497 0.00084

Significance F 0.000006

Case Notes 41

Chapter 15 Virginia Semiconductor 1. Shown below is a multiple regression analysis which contains a model to predict the size of a company’s purchase by four other predictors. Below that is a stepwise regression analysis of the same data. The full multiple regression model has an R2 of 77.3%. However, the adjusted R2 is only 69% indicating that there are some nonsignificant predictors in the model. The stepwise regression analysis confirms this by showing that an R2 of 77.0% can be obtained with only two predictors, size of company and whether or not the company has a central purchasing agent. The overall F value is significant at alpha = .01. The standard error of the estimate is 94.43. An examination of the p-values of the t ratios to test the slopes confirms the stepwise regression output. In the full multiple regression model, only size of the company has a significant t value with a p-value of .01. The variable, central purchasing agent, has a p-value of .081 which is significant if alpha α = .10. In the stepwise regression analysis, size of company was the strongest single predictor yielding an R2 of 69.44%. The inclusion of central purchasing agent as step two brings the R2 up to 77%. These are the only two significant predictors included in the stepwise analysis. Both analyses result in positive regression coefficients for these two significant predictors. The larger the company, the larger the size of purchase tends to be which makes sense. Also, having a central purchaser tends to result in a larger size of purchase. The distance that a customer company is away from Virginia Semiconductor does not seem to impact size of purchase nor does the percent of a customer company’s imports. Regression Analysis The regression equation is: CustPurc = - 1.8 + 1.37 SizeCo - 0.32 Co%Imp + 0.111 MilesAwy + 110 CentPur? Predictor Constant SizeCo Co%Imp MilesAwy CentPur?

Coef -1.78 1.3735 -0.321 0.1110 110.43

s = 94.43

R-sq = 77.3%

Stdev t-ratio 69.22 -0.03 0.4412 3.11 2.065 -0.16 0.3789 0.29 57.45 1.92

p 0.980 0.010 0.879 0.775 0.081

R-sq(adj) = 69.0%

Case Notes 42

Analysis of Variance SOURCE Regression Error Total

DF 4 11 15

SS 333650 98097 431748

MS 83413 8918

F 9.35

p 0.002

Stepwise Regression F-to-Enter:

4.00

F-to-Remove:

4.00

Response is CustPur? on 4 predictors, with N = 16 Step 1 Constant 24.316 SizeCo T-Ratio

1.82 5.64

CentPur? T-Ratio S R-Sq

2.

2 -7.771 1.45 4.26 109 2.07

97.1 69.44

87.4 77.00

The regression analysis attempts to develop a model which can be used to predict sales figures with the predictors of average hours worked per week and number of customers. Shown below are scatter plots of each of these predictors with sales. Note the number of customers has a slight linear shape but is more like the upper left quadrant in Tukey’s 4quadrant approach. Hours per week produces a graph that is quite scattered. In an effort to examine each of these predictors together and with squared and interaction terms included, a “full” quadratic model was develop. The results follow. Note that none of the predictors are significant as measured by the t test for slope. Yet the R2 value is over 90%. The stepwise regression analysis that follows reveals that there is significant interaction between the two predictors. The interaction predictor variable accounts for over 68% of the variance of sales by itself. In combination with the interaction variable, the variable, squared number of customers is also significant bringing the total R2 to over 88% with just these two variables. This analysis demonstrates how predictors by themselves may not be significant by that the interaction of 2 or more predictors might be.

Case Notes 43

Case Notes 44

Regression Analysis The regression equation is Sales = 41 - 2.22 Hrs/Wk + 0.67 No.Custs + 0.0316 HrsSq - 0.0196 CustSq + 0.0093 IntHrCus Predictor Constant Hrs/Wk No.Custs HrsSq CustSq IntHrCus

Coef 41.5 -2.221 0.668 0.03163 -0.01965 0.00933

Stdev 117.9 5.099 3.373 0.05644 0.02981 0.06959

s = 1.097

R-sq = 90.1%

t-ratio 0.35 -0.44 0.20 0.56 -0.66 0.13

p 0.739 0.681 0.851 0.599 0.539 0.899

R-sq(adj) = 80.3%

Analysis of Variance SOURCE DF SS Regression 5 55.026 Error 5 6.016 Total 10 61.042

MS 11.005 1.203

F 9.15

p 0.015

Stepwise Regression F-to-Enter:

4.00

F-to-Remove:

4.00

Response is Sales on 5 predictors, with N = 11 Step Constant

1 7.089

2 3.139

IntHrCus T-Ratio

0.0115 4.41

0.0247 6.27

CustSq T-Ratio S R-Sq

-0.0132 -3.71 1.46 68.41

0.942 88.38

Case Notes 45

3. Shown below is a MINITAB scatter plot of sales and number of employees. Notice how the graph rises and then levels out. This fits fairly closely with the upper left quadrant of Tukey’s 4-quadrant approach. From this, we attempted to use the log of number of employees as a second predictor. The result is shown below in a stepwise regression analysis. Stepwise Regression F-to-Enter:

4.00

F-to-Remove:

4.00

Response is Sales2 on 2 predictors, with N = 10 Step Constant

1 -94.68

2 -883.10

logEmpl T-Ratio

25.40 3.58

220.40 4.79

No.Empl. T-Ratio S R-Sq

-1.24 -4.26 3.89 61.57

2.19 89.29

Notice that log of number employees enters the analysis first and accounts for almost 62% of the variation of sales. At the second step, number employees enters the process adding another 28%. It was likely worthwhile to recode the data using logs. According to this, sales increases are associated with the log of number employees and not as much with the straight linear increase. The company should exercise caution in merely hiring more people as sales increase.

Case Notes 46

Chapter 16 DeBourgh Manufacturing Company 1. The decomposition analysis shows several things. A study of the original data shows that there is a general upward trend. This is underscored by the graph showing the trend line fit through the data. There appear to be important seasonal effects in these data. Note that for January, February, and March, the seasonal indices are less than 100. From April through August, seasonal indices are more than 100 peaking in the summer months of June, July, and August. This would indicate that these are the strongest months of locker sales. This might make sense because schools might be purchasing lockers and installing them in the summertime in preparation for the opening of school. The months of October, November, and December have low seasonal indexes. The upward trend marks good news for DeBourgh demonstrating a consistent growth. The seasonal indices can be used to help DeBourgh plan for both production and shipping. By examining lead time and production times, DeBourgh can more closely schedule raw materials from suppliers and do human resource planning. 2. Using MINITAB, several forecasting techniques were explored. Three different moving averages were used. These were unweighted moving averages of 4 years, 3 years, and 2 years producing MADs of 5.1825, 5.2461, and 4.0717. The 2-year moving average produced the least MAD and seemed to make the best fit. The MINITAB graphical output for the 2-year moving average model is shown below. Next, single exponential smoothing models were examined with various values of α. For α = .3, the MAD was 7.3010. For α = .6, MAD was 7.0447. For α = .9, MAD was 3.6254. Thus, the higher the value of alpha, the better the forecast with a big improvement as alpha moved from .6 to .9. Since alpha weights the actual previous value, the exponential smoothing works better here when the model actually shadows the previous value. Since there seems to be a downward overall trend in the data, tracking the previous value may be a good strategy. An optimal search for alpha resulted in an alpha of .987. The value of MAD for this model was 2.9569. The MINITAB graphical output for this optimal value of alpha is shown below. Trend analysis was preformed on these data by fitting a line through the data. MINITAB trend analysis resulted in MAD of 4.7155. A quadratic fit resulted in MAD = 2.8426.

Case Notes 47

Overall, the best fitting model to these data was the quadratic trend analysis model. The model produced by this analysis is 91.6691 - 6.21916 t + 0.266113 t2. Using t = 15 to represent the year 2002, the predicted per-unit labor cost is: 91.6691 - 6.21916(15) + 0.266113 (15)2 = 58.26. Shown below is the graphical analysis for the quadratic trend model produced using MINITAB. DeBourgh can become more productive and produce better quality lockers by reducing costs (particularly those do to poor work, scrap, rework, poor raw materials, etc.).

Moving Average

Actual

85

Predicted Actual Predicted

Cost

75

Moving Average

65

55 0

5

10

Time

15

Length:

2

MAPE:

6.4362

MAD:

4.0717

MSD:

51.6762

Case Notes 48

Single Exponential Smoothing

Actual

85

Predicted

Labor Cost

Actual Predicted

75

Smoothing Constant

65

55 0

5

10

Alpha:

0.987

MAPE:

4.6176

MAD:

2.9569

MSD:

37.2189

15

Time

Trend Analysis for Labor Cost Quadratic Trend Model Yt = 91.6691 - 6.21916*t + 0.266113*t**2 Actual

85

Fits

Labor Cost

Actual Fits

75

65

MAPE: MAD: MSD:

55 0

5

10

Time

15

3.9710 2.8426 16.7767

Case Notes 49

Chapter 17 Schwinn 1. Since two independent samples are being compared and the shape of the population distribution is unknown, the Mann-Whitney U test is used to analyze the data rather than the t test for two independent samples. The null hypothesis is that there is no difference between the age of Schwinn customers in Colorado Springs (near the mountain) and the age of Schwinn customers in St. Louis. The MINITAB computer output for the Mann-Whitney test is shown below. The analysis reveals that the median age for a customer in Colorado Springs was 31 years of age and the median age for a customer in St. Louis is 14 years. It appears that the target market in Colorado Springs is young adult versus St. Louis where it is children and early teens. Since one of the sample sizes is greater than 10, a large sample Mann-Whitney U test is the appropriate test. As noted in the text, the MINITAB Mann-Whitney test does not yield a z statistic but rather gives the value of W and a p-value. Since the p-value is .0030, the null hypothesis is rejected at α = .01. There is a significant difference between the age of Schwinn customers in Colorado Springs and those in St. Louis. The median ages support the theory that a much older group of customers purchase Schwinn bikes in Colorado Springs perhaps to be used in mountain biking. The St. Louis population appears to be a youth market. Mann-Whitney Test and CI: C.S., St. Louis C.S. N = 11 Median = 31.00 St. Loui N = 9 Median = 14.00 Point estimate for ETA1-ETA2 is 17.00 95.2 Percent CI for ETA1-ETA2 is (9.00,23.00) W = 155.0 Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.0030 The test is significant at 0.0030 (adjusted for ties)

2. Since three independent groups are being compared, it would appear that this is a completely randomized design and that a one-way ANOVA could be used to analyze the data. However, it is uncertain whether the data are normally distributed or not, so a Kruskal-Wallis test is used to analyze the data. The independent variable is supplier with three classifications: supplier 1, supplier 2, and supplier 3. The dependent variable is the weight of the handle bar. The null hypothesis is that there is no difference in the weight of handle bars supplied by the three suppliers. The alternative hypothesis is that there is a difference in the weights of handle bars by supplier. MINITAB was used to analyze these data. The results, given below, show that there is no significant difference in the weights of handle bars according to supplier. The H value (3.09) is MINITAB’s equivalent to the Kruskal-Wallis K value. The associated p-value (.213) denotes that there is no significant difference even at α = .10. The differences in the medians, shown in the MINITAB output, are merely due to chance. To Schwinn this might mean that, at least on handle bar weight, these suppliers are interchangeable.

Case Notes 50

Kruskal-Wallis Test: Weight versus Supplier Kruskal-Wallis Test on weight C5 Supplier 1 Supplier 2 Supplier 3 Overall H = 3.09

N 8 6 5 19 DF = 2

Median 201.9 198.2 195.2

Ave Rank 12.4 9.5 6.8 10.0

Z 1.57 -0.26 -1.48

P = 0.213

3. The MINITAB output for the Runs Test contains a p-value of .9083 that is not significant (underscored by MINITAB’s statement – Cannot reject at alpha – 0.05). This indicates that the paint flaws are occurring in a random manner. In this sample, there are 19 observations above K and 56 below (out of 75 bicycles). Since the data were coded with a 1 to indicate at least one flaw and a 0 to indicate no flaws, there are 19 bicycles with at least one flaw out of the 75 bicycles. The proportion of flawed bicycles is

19 75

= .2533 (shown as K in the output). While Schwinn management

may by happy to know that there appears to be no systematic pattern of flaws, they are likely to be unhappy about having at least one flaw in 25.33% of the bikes. Perhaps, an intensive, manufacturer-wide quality effort could be made to reduce the percentages of flaws. After, studying chapter 18 (Statistical Quality Control), the student might be able to recommend the use of a Pareto Chart to prioritize the types of flaws that are occurring and a Fishbone diagram to assist managers and production engineers in searching for causes of the flaws which might include such things as raw materials, machinery, technique, and workers.

Case Notes 51

Chapter 18 Robotron 1.

Produce → → → → → Shipping → Billing

Item ↑ ↓ ↑ ↑ Stop Beginning ↑ → → → → → → → →of Assembly ↑ ↑ Line ↑ ↑ ↑(no) ↑ ↑ Being Tag ↑ ↑ Manufactured?(yes) → Item → ↑ ↑ ↑ ↑ ↑ Plant ↑ ↑ Clerk ↑ ↑ (no) ↑ ↑ Mail Order Standard ↑(no) ↑ Room → Processing → Item? (yes) → Warehouse → Available? (yes) → → → →→

2. Part 173: The xbar chart looks pretty good with the exception of sample number 13 which has a mean that is above the upper control limit. Most of the rest of the means seem to fluctuate randomly around the centerline mostly in the inner 1/3 of the limit area. The R chart indicates that the ranges are in control. None of the ranges are outside the control limits and most lie near the centerline with no obvious pattern occurring. Part 248: The x bar chart indicates a process that is in control. None of the points are outside of the limits. In fact, only two of the spikes are in the outer 2/3 of the limits. There seems to be little or no pattern with points occurring randomly. However, the range chart indicates some potential problems. The ranges for samples 9 and 17 are above the upper control

Case Notes 52 limits. These samples and their origin should be investigated. However, the rest of the range values seem to be in order and well under control. 3. p chart: The centerline of the p chart is .03640. On average, 3.64% of the items in each sample are in nonconformance. Yet, there are 4 of the 100 samples that produced sample proportions above the upper control limit which is established at .1159 or 11.59%. Management has to be concerned about samples that have almost 12% nonconforming items when the overall average is only 3.64%. There are many samples that are on the lower control limit. However, since the lower values of p for nonconformance indicate few items in nonconformance, these values are good. In fact, 19 of the 100 samples had no items nonconforming. Overall, the patterns seem to be random. Managers should investigate to determine why, seemingly out-of-the-blue, samples contain unacceptable proportions of nonconforming items.

Case Notes 53

Chapter 19 Fletcher-Terry: On the Cutting Edge 1. There are several decisions that management had to make during this time including 1) whether or not to invest in technology, 2) expand its line of offerings through imports, 3) conduct a significant planning process, 4) attempt to increase market share, 5) develop new products, 6)create greater employee involvement, 7) invest in employee education, 8) invest in plant improvements, and 8) implement a participatory management system. Several states of nature occurred which could have effected Fletcher-Terry’s outcomes. Some of these include: 1) its largest customers decided to introduce their own privatelabel cutters made overseas, 2) the technology that Fletcher-Terry invested in would not work, 3) dollar weakened, 4) slow-down in demand for cutters, 5) employees do not respond to company efforts.

2.

Decision Import Alternative Not Import

$ Up (.25) $350,00 0 -$22,700

$ Same (.35) $275,00 0 -$22,700

$ Down (.40) -$555,000 -$22,000

The expected monetary value (EMV) for each of these alternatives is: Import: (.25)($350,000) + (.35)($350,000) + (.40)(-$555,000) = -$63,000 Not Import: (.25)(-$22,700) + (.35)(-$22,700) + (.40)(-$22,700) = -$22,700 The EMV’er would choose the highest of these alternatives which is to not import and take a $22,700 loss. The risk avoider would also choose to not import. However, a risk taker might decide to import gambling that the dollar does not go down.

Case Notes 54

The decision table and the expected values are displayed below in a decision tree.