Cases for Essentials of Statistics for Business & Economics 9th Edition by David R. Anderson, Dennis

02/Case 1 - Pelican Stores/BSTAT_02_Case1_PelicanStores.xlsx Sheet: &UnStack; Sheet: &DataIndices; Sheet: &DataCopy; Sheet: &GraphData; Sheet: &WorkArea; Sheet: &Miscel;_Area Sheet: Data

d of Payment

Items

Gender

Marital Status

Age

an Express

7

Female

Married

50

an Express

3

Female

Married

46

er

1

Male

Married

32

er

3

Female

Married

40

er

1

Female

Single

78

er

2

Female

Married

54

American Expr

Card

2

Female

Married

34

Discover

Card

1

Female

Married

44

MasterCard

Card

2

Female

Married

32

Proprietary Ca

Card

2

Female

Married

28

Visa

Card

5

Female

Married

56

Card

1

Male

Single

36

Card

1

Female

Married

46

Number of Item

Card

9

Female

Married

46

Bin

Card

2

Female

Married

32

1-3

3

Card

4

Female

Married

46

4-6

6

Card

4

Female

Married

32

7-9

9

Card

2

Female

Married

68

10 - 12

12

Card

6

Female

Single

48

13 - 15

15

Card

1

Female

Married

44

16 - 18

18

tary Card

1

Female

Married

36

tary Card

1

Female

Married

32

tary Card

5

Female

Married

28

tary Card

2

Female

Married

30

1. Percent frequency distribution for key variables

Method of pay

Type of custom

tary Card

2

Female

Married

46

Promotional

tary Card

1

Female

Married

36

Regular

tary Card

1

Female

Married

48

tary Card

1

Female

Married

40

tary Card

2

Male

Single

36

tary Card

3

Female

Single

22

American Expr

tary Card

6

Female

Married

56

Discover

s

2. Bar chart of customer purchases by method of payment

Sheet: Crosstab

Column Labels Promotional

Regular

Grand Total

39.5

39.5

Type of Customer

Net Sa

102.4

Promotional

5900.3

22.5

Regular

1859.7

102.4 22.5 100.4

100.4 54

54

44.5

44.5

78

78 22.5

56.52

22.5 56.52

44.5

44.5

29.5

29.5

31.6

31.6

160.4

160.4

64.5

64.5 49.5

49.5

71.4

71.4

94

94 54.5

54.5

38.5

38.5

44.8

44.8

31.6

31.6

70.82

70.82

266

266 74

74

3. Crosstabulation of Type of Customer vs Net Sales

7760.0

39.5

39.5

30.02

30.02 44.5

192.8

44.5 192.8

Sheet: ScatterDiagram Age

Net Sales

50

266

46

66

32

39.5

40

54.5

78

58

54

89

34

54

44

44.5

32

38.5

28

63.2

56

105.5

36

29.5

46

25

46

145.2

32

49.5

46

65.2

32

58

68

78

48

144

44

39.5

36

102.4

32

22.5

28

100.4

30

78

46

56.52

36

44.5

48

29.5

40

31.6

36

71.4

22

94

4. Scatter Diagram

56

44.8

02/Case 1 - Pelican Stores/BSTAT_02_Case1_PelicanStores_Report.docx Pelican Stores Pelican Stores, a division of National Clothing is a chain of women’s apparel stores operating throughout the country. The chain ran a promotion offering discounts to customers of other National Clothing stores. The discount coupons were not offered to regular customers. Sales made to people presenting the promotional coupons were considered sales that would not have otherwise been made.

Percent Frequency Distributions for Key Variables Table 1 presents a frequency distribution for method of payment. The proprietary card was used 70% of the time indicating purchases were made by regular customers. Figure 1 presents the method of payment as a histogram.

Table 2 presents a frequency distribution for number of items purchased. The majority of customers purchased 1 to 3 items (66%). Since 70% of customers were regular customers, they tended to purchase fewer items as there was no discount. Twenty-six percent (26%) of customers purchased between 4 and 6 items. The promotional discount may have encouraged these customers to purchase additional items they would not normally purchase. Only 8% of customers purchased 7 or more items. These 8 customers took full advantage of the promotional discount.

Table 3 presents a frequency distribution for type of customer. Seventy percent (70%) of customers were customers with the promotional discount. Since (as indicated in Table 1) 70% of customers used the proprietary card, some of the regular customers made purchases using a promotional discount.

Crosstabulation of Type of Customer versus Net Sales Examination of a cross tabulation of type of customer versus net sales (see Appendix A) reveals that promotional customers in general spent more on purchases than regular customers. In fact, 5900 of the 7760 (76%) of the total sales were made by customers with promotional discounts.

Scatter Diagram of Relationship Between Customer Age and Net Sales Figure 2 presents a scatter diagram of the relationship between customer age and net sales.

Customers range in age from 20 to approximately 80. There are more customers in the 25 to 45 age range. There does not appear to be a relationship between age and net sales.

This report presents results based on a preliminary investigation of customer purchases at Pelican Stores when a promotional discount was offered. Where appropriate suggestions have been made that could indicate the value of the promotional discount to sales. In summary, customers with promotional discounts spent more on purchases than regular customers. However, based on the number of proprietary card purchases there were regular customers also using the promotional discount.

Appendix A – Crosstabulation of Type of Customer versus Net Sales

02/Case 1 - Pelican Stores/BSTAT_02_Case1_PelicanStores_Rubric.docx Pelican Stores Case Study Chapter 2 (80 points total) RUBRIC

02/Case 1 - Pelican Stores/BSTAT_02_Case1_PelicanStores_TeachingTips.docx Pelican Stores – Teaching Tips Percent Frequency Distribution for Key Variables The Pelican Stores application is interested in the effect of promotional coupons on sales. Th fe key variables chosen for frequency distributions were method of payment, number of items, and type of customer.

Use the COUNTIF() function to compute frequencies for the method of payment. Using the Data Analysis Histogram tool is inappropriate here as it computes only cumulative percentages.

The number of items frequency distribution was created using 7 as the number of bins. This choice is fairly arbitrary. To determine the bin width, compute maximum value minus minimum value divided by the number of bins and the ROUNDUP() function: =ROUNDUP((MAX(range) – MIN(range)/# of bins),0) Use the COUNTIF() function to create the type of customer frequency distribution Bar Chart of Customer Payment by Method of Purchase Use the frequency distribution created earlier. Let method of purchase be the x axis data and customer payment the y axis data. Refer to p. 72 for general guidelines for effective displays. The Design tab can be used to format axis titles.

CrossTabulation of Type of Customer versus Net Sales The crosstabulation by customer reveals that promotional customers spent more per purchase than regular customers. Students may have trouble understanding which fields to drag to the pivot table fields area. Using type of customer for the column data makes the pivot table more manageable asthetically. Refer students to pages 98-101 in the Appendix for help.

Scatter Diagram Creation Scatter diagrams were created using the Insert tab on the ribbon from the charts category. For this exercise students should select the chart data prior to choosing the scatter chart from the chart group. Data are chosen in x, y coordinate order. For simplicity necessary data (age and net sales) were copied to a new worksheet with age as the x coordinate (in column A) and net sales as the y coordinate. Again, refer to p. 72 for general guidelines for effective displays. The Design tab can be used to format axis titles.

02/Case 2 - Movie Theatre Releases/BSTAT_02_Case2_MovieTheatreReleases_Report.docx Movie Theatre Releases The movie industry, a competitive business, uses a number of key variables to measure the success of a movie released to industry: opening weekend gross sales, total gross sales, number of theatres the movie was shown in, and the number of weeks the movie was in release are a few.

The too 100 movies of 2016 are summarized in this report Opening Gross Sales Table 1 presents a frequency distribution for opening gross sales. The majority of these 2016 movies (37%) gross up to 15 million in the opening weekend, followed by 33% grossing between 15 and 30 million. Eighty-three percent (83%) of movies gross up to 45 million on opening weekend.

Figure 1 presents opening gross sales as a histogram.

Total Gross Sales Table 2 presents a frequency distribution for total gross sales. The majority of these 2016 movies (59%) gross between 23 and 74 million dollars. Eighty-seven percent (87%) of movies indicate a total gross of 23 to 176 million dollars. Figure 2 presents total gross sales as a histogram.

Number of Theatres Table 3 presents a frequency distribution for the number of theatres the movie was released in. The majority of movies were released in between 3121 and 3279 theatres. Forty-five percent (45%) of movies were released in between 2803 and 3438 theatres. Twenty-five percent (25%) were released in between 3757 and 4233 theatres. Figure 3 shows a histogram of the number of theatres the movies were released in.

Number of Weeks in Release Table 4 presents a frequency distribution for the number of weeks in release for one hundred 2016 movies. The majority of movies were in release between 7 and 14 weeks. Ninety-two percent (92%) of movies were in release for up to 22 weeks. Figure 4 shows a histogram of the number of weeks in release.

Relationship Between Total Gross Sales and Number of Theatres Figure 5 presents a scatter diagram of total gross sales and number of theatres. Movies shown in fewer than 3000 theatres gross less than 100 million dollars. When movies are shown in more than 3000 theatres, total gross sales increase beyond the 100 million dollars threshold. Twelve percent (12%) movies grossing more than 200 million dollars are in a more than 3800 theatres.

Relationship Between Total Gross Sales and Number of Weeks in Release Figure 6 presents a scatter diagram of total gross sales and number of weeks in release. The relationship appears to be predominantly positive, albeit not in a linear form. When weeks in release is approximately between 7 and 18 total gross sales average 100 million dollars. Only 13 movies gross between 20 and 30 million. Rogue One: A Star Wars movie grossed the largest amount and was only in release for 20 weeks.

This report presents results based on examination of key variables used to measure success in the movie industry. These variables include: opening gross sales, total gross sales, number of theatres movies were released in, and the number of weeks in release.

02/Case 2 - Movie Theatre Releases/BSTAT_02_Case2_MovieTheatreReleases_Rubric.docx Movie Theatre Releases Case Study Chapter 2 (124 points total) RUBRIC

02/Case 2 - Movie Theatre Releases/BSTAT_02_Case2_MovieThea treReleases_TeachingTips.docx Movie Theatre Releases – Teaching Tips Tabular and Graphic Summaries Creating frequency distributions of quantitative data relies on determining the number of classes (between 5 and 20) to use to effectively summarize the data. Students that get stuck on this first subjective/arbitrary step often delay getting started or even give up. I used 12 for Opening Gross Sales, 10 for Total Gross Sales, 20 for number of theatres, and 5 for number of weeks. Perhaps give the students the number of classes for each table to avoid a slow start.

I recommend using either the Data Analysis tool or Pivot Table approach to avoid a discussion of control array formulas (and how to enter them).

The bin width is a computed value. Start point again is fairly arbitrary. Once width is established and first 2 ranges computed (For example, 0 – 15, 16 – 30), the bins column can be computed by establishing the pattern and then using the fill handle to complete the bins, working backwards then to complete the ranges for display (31 – 45…).

The majority of students in this level Statistics course may not have been exposed to the use of Pivot Tables. Students should be referred to the exercise in the Chapter 2 (pp. 98-101) as a reference.

The Pivot table for Opening Gross Sales will initially appear as shown below.

The Grouping option is the key to creating the Pivot table/chart. Right-click on any of the row label columns to select the Grouping option.

Both the Data Analysis Tool and Pivot Chart create the chart as a bar graph. To make the chart a histogram remove the gap (set to 0) in the Data Series Option.

Students should be reminded that frequency distributions/charts may differ from one another depending on values chosen for number of classes.

To create tables, suitable for reports, students should copy the pivot table and paste special just the values before formatting. Just getting through the Pivot Table steps is a challenge. Students may need to be reminded that just copying the Pivot Table or Histogram output into the report may not be acceptable. Perhaps offer an example: Table of Amount of Expenditures by Department and the Percentage of Expenditures by Department It may be helpful for students to generate a new worksheet and sort the data by department to help visualize the process while creating the Pivot Table. Remind students to select the table by clicking in A1 and using CTRL-SHIFT-Right Arrow then CTRL-SHIFT-Down Arrow to select the data (dragging the mouse to select this size data is inefficient). For this table, the Row category is department description (again visualize Department names in the rows in column A). The values field is again the Expenditures.

Once created the table summing expenditures can be copied to a new location (allowing for a table name), column titles added, and the expenditures data can be formatted in an exceptable style (comma, 2 decimals is appropriate). Remind students that the $ style clouds the data with extra symbols.

The original PivotTable can now be used to create the % of Expenditures table. To change the Expenditures values to percent, students will have to right-click on the expenditures data in the Pivot Table and choose Value Field Settings. They will most likely not be familiar with the Show Values As tab and should be instructed to select that tab and choose the desired calculation (% of Expenditures).

To combine all percent values less than 1%, PivotTable data should be copied (use the Paste Special to remove PivotTable features from the data). Formatted as a table, sorted, less than 1% data combined and and Other field created. Use the SUMIF() function to compute the Other data.

Students will have to be reminded that the criteria (<1%) must be put in quotes (“<1%”). Appropriate formatting (column headings, table name can then be added for reporting).

The Paste Special feature is often overlooked. For example, once the Other department description has been created, this result (21.20%) must be copied/paste special to remove the SUMIF function or the result will be lost.

Table of Amount of Expenditures by Fund and the Percentage of Fund by Department This table is created exactly as the expenditures by department tables. Use the previous Teaching Tips for help.

em

02/Case 2 - Movie Theatre Releases/BSTAT_02_Case2_Movies2016_Key.xlsx Sheet: Sheet1 Opening Gross Sales ($ millions)

Total Gross Sales ($ millions)

Number of Theaters

Weeks in Release

Opening Gross

155.08

532.18

4157

20

Max =

135.06

486.3

4305

25

Min =

179.14

408.08

4226

20

Number of Clas

104.35

368.38

4381

25

Class width =

103.26

364

4144

24

132.43

363.07

3856

18

Class

75.06

341.27

3959

22

0 - 15

166.01

330.36

4256

12

16 -30

133.68

325.1

4255

14

31 - 45

35.26

270.4

4029

20

46 - 60

56.63

248.76

3875

22

61 - 75

74.4

234.04

4144

19

76 - 90

85.06

232.64

3882

19

91 - 105

0.52

169.61

3416

46

106 - 120

59.22

162.43

4039

21

121 - 135

59.25

158.85

3928

13

126 - 150

65.77

155.44

4153

9

151 - 165

46.58

153.71

4066

21

166 - 180

0.88

151.1

3236

20

41.28

143.53

3987

25

Total Gross Sa

46.02

128.35

3963

17

Max =

35.54

127.44

3508

11

Min =

38.53

126.64

3591

11

Number of Clas

35.03

125.07

3955

20

Class width =

23.82

113.26

3215

13

38.16

107.51

3932

17

Class

41.04

103.14

4130

12

23 - 74

40.41

102.47

3356

11

75 - 126

24.07

100.55

3115

17

127 - 178

14.87

100.01

3478

17

179 - 230

34.26

97.69

3135

19

231 - 282

Sheet: Sheet4 Sheet: PivotTable Opening Gross Table 1 Frequency Distribution of Opening Gross Sales

nt of Opening Gross Sales ($ millions)

Opening Gross Sales 0-15 15-30 30-45 45-60 60-75 75-90 90-105 120-135 135-150 150-165 165-180 Grand Total

Sheet: PivotTable Total Gross Table 2 Frequency Distribution of Total Gross Sales

unt of Total Gross Sales ($ millions)

Opening Gross Sale 23-74 74-125 125-176 227-278 278-329 329-380 380-431 482-533

0

Grand Total

Sheet: PivotTable Number of Theatres

Table 3 Frequency Distribution of Number of Theatres Count of Number of Theaters

Number of Theatres

F

1

1213-1371

1

2

1372-1530

2

1

1531-1689

1

1

1690-1848

1

4

2167-2325

4

2

2326-2484

2

3

2485-2643

3

3

2644-2802

3

8

2803-2961

8

11

2962-3120

1

14

3121-3279

1

12

3280-3438

1

6

3439-3597

6

3

3598-3756

3

7

3757-3915

7

12

3916-4074

1

6

4075-4233

6

4

4234-4392

4

100

Grand Total

1

Sheet: PivotTable Weeks in Release Table 4 Frequency Distribution of Number of Weeks in Release

Count of Weeks in Release

Weeks in Release

8

7-14

4

15-22 23-30 39-46

00

Grand Total

Sheet: Scatter Diagrams Opening Gross Sales ($ millions)

Number of Theaters

Weeks in Release

Total Gross Sales ($ millions)

Number o

155.08

4157

20

532.18

4157

135.06

4305

25

486.3

4305

ustice

Find Them

e

179.14

4226

20

408.08

4226

104.35

4381

25

368.38

4381

103.26

4144

24

364

4144

132.43

3856

18

363.07

3856

75.06

3959

22

341.27

3959

166.01

4256

12

330.36

4256

133.68

4255

14

325.1

4255

35.26

4029

20

270.4

4029

56.63

3875

22

248.76

3875

74.4

4144

19

234.04

4144

85.06

3882

19

232.64

3882

0.52

3416

46

169.61

3416

59.22

4039

21

162.43

4039

59.25

3928

13

158.85

3928

65.77

4153

9

155.44

4153

46.58

4066

21

153.71

4066

0.88

3236

20

151.1

3236

41.28

3987

25

143.53

3987

46.02

3963

17

128.35

3963

35.54

3508

11

127.44

3508

38.53

3591

11

126.64

3591

35.03

3955

20

125.07

3955

23.82

3215

13

113.26

3215

38.16

3932

17

107.51

3932

41.04

4130

12

103.14

4130

40.41

3356

11

102.47

3356

24.07

3115

17

100.55

3115

14.87

3478

17

100.01

3478

34.26

3135

19

97.69

3135

02/Case 3 - Queen City/BSTAT_02_Case3_QueenCity-Key.xlsx Sheet: 2013ExpenseDB Expenditure

RPRISE TECHNOLOGY SOLUTIONS

5275

TAL PROJECTS

23909.68

TAL PROJECTS

315445.8

TAL PROJECTS

270042.5

TAL PROJECTS

29625

ROPOLITAN SEWER DISTRICT OF GREATER CINCINNATI

3081.6

ROPOLITAN SEWER DISTRICT OF GREATER CINCINNATI

282994.89

TAL PROJECTS

113240

TAL PROJECTS

5981.57

TAL PROJECTS

4482.18

TAL PROJECTS

7000

TAL PROJECTS

152440

TAL PROJECTS

201228.71

TAL PROJECTS

100000

TAL PROJECTS

3600

munity Development BLOCK GRANT

9372.76

E INVESTMENT TRUST

70520

SING OPPORTUNITIES FOR PERSONS WITH AIDS

100933.17

ORHOOD STABILIZATION PROGRAM

47919.4

TER PLUS CARE PROGRAM

35961

E INVESTMENT TRUST

2000

ORHOOD STABILIZATION PROGRAM

20199.86

A FEDERAL GRANTS

2938

TAL PROJECTS

113522.63

TAL PROJECTS

52482.22

TAL PROJECTS

36502.05

TAL PROJECTS

11782.32

munity Development BLOCK GRANT

14388.86

TAL PROJECTS

3713.4

TAL PROJECTS

7076.03

munity Development BLOCK GRANT

2344

Sheet: By Category PivotTable

Expenditures and % of Total Expenditures by Categor

Expenditures by Category

Percent of Expenditures by Category

Sum of Expenditure

Category

Expenditures

Category

Exp

0.2998117481453694

Capital

198365853.86999997

Capital

0.29

0.12853505055069492

Contractual Services

85043248.68000008

Contractual Services

0.12

0.13136248662753697

Debt Service

86913978.48

Debt Service

0.13

0.08121124528077109

Fixed Costs

53732177.31999997

Fixed Costs

0.08

0.0998335501641538

Fringe Benefits

66053340.28999993

Fringe Benefits

0.09

0.009663026576387717

Inventory

6393393.62

Inventory

0.00

0.03012948121998051

Materials and Supplies

19934710.04999999

Materials and Supplies

0.03

0.00027271091128447015

Payables

180435

Payables

0.00

0.21918070052382102

Payroll

145017555.44999987

Payroll

0.21

1

Grand Total

661634692.7599999

Grand Total

1

Sheet: By Department Data

Sheet: By Fund Data Expenditures by Fund Sum of Expenditure

Fund

ND

0.25483497219085594

050 - GENERAL FUND

KS

0.07903485533967962

101 - WATER WORKS

STEM FACILITIES

0.005302257073863178

102 - PARKING SYSTEM FACILITIES

Y CONVENTION CENTER

0.006126501851181434

103 - DUKE ENERGY CONVENTION CENTER

ORT

0.0010016543528505643

104 - LUNKEN AIRPORT

OLF ACTIVITIES

0.004256168578846696

105 - MUNICIPAL GOLF ACTIVITIES

R MANAGEMENT

0.003644963416201622

107 - STORMWATER MANAGEMENT

EMENT - CITY

0.023555168041427415

151 - BOND RETIREMENT - CITY

VENUE BONDS -TRUSTEE

0.008355009570220952

155 - USQUARE REVENUE BONDS -TRUSTEE

QUARE WEST REVENUE BONDS - TRUSTEE

0.0008444142305619081

156 - FOUNTAIN SQUARE WEST REVENUE BOND

ERVICE - TRUSTEE - MULTI SERIES

0.0001150042777874724

161 - MSD DEBT SERVICE - TRUSTEE - MULTI SER

EVENUE BONDS - TRUSTEE

0.0002696153662315704

166 - GRAETERS REVENUE BONDS - TRUSTEE

RVICES/ STORES

0.0018179000786409728

201 - PRINTING SERVICES/ STORES

CES

0.016617400674889003

202 - FLEET SERVICES

KS STORES - MATERIALS

0.0023068337962042747

204 - WATER WORKS STORES - MATERIALS

STORES

0.0001169864592153071

207 - RECREATION STORES

ANAGEMENT

0.0002483807330514504

209 - PROPERTY MANAGEMENT

INTENANCE STORES

0.0002596147570247008

210 - HIGHWAY MAINTENANCE STORES

AFETY AND RISK MANAGEMENT

0.0019599933682292537

212 - EMPLOYEE SAFETY AND RISK MANAGEMEN

OMPENSATION

0.0029592072051611864

213 - WORKER''S COMPENSATION

KS STORES - CHEMICALS

0.0026693317767734403

214 - WATER WORKS STORES - CHEMICALS

STRUCTION, MAINTENANCE AND REPAIR

0.007098043801798543

301 - STREET CONSTRUCTION, MAINTENANCE A

- INFRASTRUCTURE

0.010735188553022931

302 - INCOME TAX - INFRASTRUCTURE

velopment BLOCK GRANT

0.007177389308578127

304 - Community Development BLOCK GRANT

OTOR VEHICLE LICENSE TAX

0.001725733100900403

306 - MUNICIPAL MOTOR VEHICLE LICENSE TAX

CENTER EXPANSION TAX

0.002059290201842891

307 - CONVENTION CENTER EXPANSION TAX

NTS

0.00022671120731936991

314 - SPECIAL EVENTS

LOPMENT - PROPERTY OPERATIONS

0.0004292733786603683

317 - URBAN DEVELOPMENT - PROPERTY OPERA

NT

0.000185121021222551

318 - SAWYER POINT

ONS FOR RECREATION PURPOSES

1.9097003434466636e-05

319 - CONTRIBUTIONS FOR RECREATION PURPO

02/Case 3 - Queen City/BSTAT_02_Case3_QueenCity_Report.docx Queen City The city manager of Cincinnati, Ohio, also known as the Queen City recently created the Office of Performance and Data Analysis with the goal of improving the efficiency of city operations. This report summarizes previous year’s expenditures by category, department, and fund.

City Expenditures by Category Table 1 presents total expenditures by category and percent of total expenditures by category. The majority of expenses are a result of capital (29.98%) and payroll (21.92%) representing approximately 52% of total expenditures. Figures 1 and 2 present a bar chart of expenses by category and a bar chart of the percent of total expenditures by category.

City Expenditures by Department Appendix A presents total expenditures by department and percent of total expenditures by department. This data as presented in alphabetical order by department provides easy access to expenditures for each department. To further examine where the bulk of city expenditures are made, Table 3 presents total expenditures by departments as percent. Values with total expenditures of less than one percent are combined into another category placed at the end of the table. The largest percentage of expenditures (by less than 1%) is the Department of Police. The departments of Sewers and Transportation and Engineering, Engineering represent the next largest expenditures in the city.

City Expenditures by Fund Table 4 in Appendix B presents total expenditures by fund and percent of total expenditures by department. This data as presented in order by fund number provides easy access to expenditures for each department. To further examine where the bulk of city expenditures are made by fund, Table 5 presents total expenditures by fund as percent. Again, fund values with total expenditures of less than one percent are combined into another category placed at the end of the table. The General Fund presents the largest percent of expenditures by fund at 25.48%.

This report presents results requested by the City Manager based on an investigation by the Office of Performance and Data Analysis of previous year’s expenses in Cincinnati, Ohio. an Appendix A – Total Expenditures by Department Appendix B – Total Expenditures by Fund

02/Case 3 - Queen City/BSTAT_02_Case3_QueenCity_Rubric.docx Queen City Study Chapter 2 (120 points total) RUBRIC

02/Case 3 - Queen City/BSTAT_02_Case3_QueenCity_TeachingTips.docx Queen City – Teaching Tips General Students should be reminded about the use of absolute and relative cell references when copying formulas.

Category – Amount of Expenditures and Percent of Total Expenditures There are 5427 records. The size of the database may intimidate students in to sorting data by categories and performing multiple summations to create table data. While the PivotTable is the easiest approach, students may manually create a table with the categories and then use the SUMIF() function to compute expenditures data. If this approach is used, the COUNTIF() function can be used to confirm that the number of categories includes all 5427 records. This approach is used in the 2013ExpenseDB worksheet of the QueenCity_Key file.

Use the frequency distribution to create a bar chart by expenditure and % of Expenditures. Let category be the x axis data and expenditures or % of expenditures the y axis data. Refer to p. 72 for general guidelines for effective displays. The Design tab can be used to format axis titles. In the % of Expenditures table remove extra zeroes from the y axis titles (right-click on axis title – format axis scroll down to number 0 decimals).

The majority of students in this level Statistics course may not have been exposed to the use of Pivot Tables. Students should be referred to the exercise in the Chapter 2 (pp. 98-101) as a reference.

When the PivotTable field list appears, students should visualize what fields they want located on the rows and columns. Since a total of expenditures is required, the category labels are best represented as rows. For each category, the total expenditures are required. These are the values.

Once dragged to the values area of the report Sum of Expenditures appears.

Table of Amount of Expenditures by Department and the Percentage of Expenditures by Department It may be helpful for students to generate a new worksheet and sort the data by department to help visualize the process while creating the Pivot Table. Remind students to select the table by clicking in A1 and using CTRL-SHIFT-Right Arrow then CTRL-SHIFT-Down Arrow to select the data (dragging the mouse to select this size data is inefficient). For this table, the Row category is department description (again visualize Department names in the rows in column A). The values field is again the Expenditures.

Once created the table summing expenditures can be copied to a new location (allowing for a table name), column titles added, and the expenditures data can be formatted in an exceptable style (comma, 2 decimals is appropriate). Remind students that the $ style clouds the data with extra symbols.

The original PivotTable can now be used to create the % of Expenditures table. To change the Expenditures values to percent, students will have to right-click on the expenditures data in the Pivot Table and choose Value Field Settings. They will most likely not be familiar with the Show Values As tab and should be instructed to select that tab and choose the desired calculation (% of Expenditures).

To combine all percent values less than 1%, PivotTable data should be copied (use the Paste Special to remove PivotTable features from the data). Formatted as a table, sorted, less than 1%

data combined and and Other field created. Use the SUMIF() function to compute the Other data.

Students will have to be reminded that the criteria (<1%) must be put in quotes (“<1%”). Appropriate formatting (column headings, table name can then be added for reporting).

The Paste Special feature is often overlooked. For example, once the Other department description has been created, this result (21.20%) must be copied/paste special to remove the SUMIF function or the result will be lost.

Table of Amount of Expenditures by Fund and the Percentage of Fund by Department This table is created exactly as the expenditures by department tables. Use the previous Teaching Tips for help.

02/Case 4 - Cut-Rate Machining/BSTAT_02_Case4_Cut-RateMachining_Report.docx Cut-Rate Machining, Inc. Cut-Rate Machining, Inc. is attempting to decide on a vendor from whom to purchase a drilling machine. Each of four vendors was invited to ship one machine to the plant facility for a test. All machines were warmed up for two hours between 8:00am and 10:00am. One employee from first shift tested the first three machines in 2-hour time periods (10:00 – noon, Noon – 2:00, 2:00 – 4:00).

An employee from second shift completed. Sixteen (16) holes were drilled from each machine.

Measured Width – All Vendors Table 1 presents a scatter plot of the measured widths for all vendors. As indicated, the only 3cm widths were machined by the first two vendors. The trend line indicates a negative relationship between the number of holes drilled and the centimeter measurement.

Measured width - Hole-Maker, Inc. (HM) Table 2 presents a scatter plot of the measured widths for Hole-Maker, Inc. (HM). As indicated, only 3 of 16 holes are cut to specification. The trend line indicates that the majority of the holes are greater than 3 centimeters in diameter.

Measured width – Shafts & Slips, Inc. (SS) Table 3 presents a scatter plot of the measured widths for Shafts & Slips, Inc. (SS). Again, only 3 of 16 holes are cut to specification. The trend line indicates that the majority of the holes are smaller than 3 centimeters in diameter.

Measured width – Judge’s Jigs (JJ) Table 4 presents a scatter plot of the measured widths for Judge’s Jigs (JJ). None of the holes are cut to the 3-centimeter diameter designation. The trend line indicates a negative relationship between the hold width and centimeter diameter. There is no consistency in machining by this vendor.

Measured width – Drill-for-Bits, Inc. (DB) Table 5 presents a scatter plot of the measured widths for Drill-for-Bits, Inc. (DB). None of the holes are cut to the 3-centimeter diameter designation. The trend line indicates a positive relationship between the hold width and centimeter diameter. There is no consistency in machining by this vendor. Variability in centimeter width varies from less than 2 to greater than 4 centimeters.

Sources of Error for Vendor Assessment All 4 machines were warmed up at 8:00am for 2 hours. Machines were tested in the order of HM, JJ, SS, and DB. Only the HM machine was used immediately after warm up. The remaining machines were cooled down by 2, 4, and 6 hours before use. Each machine should have been warmed up for 2 hours immediately prior to use.

A first shift employee (Ms. Arnes) handled machining for the HM, JJ, and SS machines over 2-hour time periods. The variation in machining be a result of the number of hours worked. The only 3-centimeter diameters were cut between 10:00 and noon and noon to 2:00pm. Only 6 of the holes were cut to specification. By the end of her shift, Ms. Arnes machining diameters varied considerably. The closest was 2.71 diameters. The effect of machining over this 6-hour period by the same employee with declining centimeter accuracy should be considered.

A second shift employee (Ms. Silver) handled machining for the DB machine from 4:00 to 6:00pm.

This machine had not been warmed up since 10:00am that morning.

This report presents results based on machining of 3-centimeter diameter holes in 25-centimeter-thick stainless-steel sheets by four invited vendors. None of the 4 vendors machines produced acceptable results. Going forward the test process should consider consistency in warming up each machine 2 hours prior to use as well as consistency in employees chosen to complete the process. One suggestion would be to test the machines over 4 days between 10:00am and noon after each machine has been warmed up between 8:00 and 10:00am.

02/Case 4 - Cut-Rate Machining/BSTAT_02_Case4_Cut-RateMachining_Rubric.docx Cut-Rate Machining Study Chapter 2 (105 points total) RUBRIC

02/Case 4 - Cut-Rate Machining/BSTAT_02_Case4_Cut-RateMachi ning_TeachingTips.docx Cut-Rate Machining, Inc. – Teaching Tips Scatter Plots This exercise is an early introduction of the use of control charts. Control charts will be formally introduced in Chapter 19. Students may have trouble visualizing the count of holes drilled as the x-axis variable for creating the scatter plots. This is accomplished by simply selecting the centimeter column prior to inserting the scatter plot. Excel will automatically creating the count variable.

Suggest students add a trend line to show the variation associated with each machine. In addition, a mean 3cm line can be drawn using Insert – Illustrations – Shapes.

02/Case 4 - Cut-Rate Machining/BSTAT_02_Case4_CutRate-Key.xlsx Sheet: Data Shift

Time Period

Employee

Vendor

Measured Width (cm)

1

10:00 a.m. - noon

Ms. Airad

HM

3.5

1

10:00 a.m. - noon

Ms. Airad

HM

3.13

1

10:00 a.m. - noon

Ms. Airad

HM

3.39

1

10:00 a.m. - noon

Ms. Airad

HM

3.08

1

10:00 a.m. - noon

Ms. Airad

HM

3.22

1

10:00 a.m. - noon

Ms. Airad

HM

3.45

1

10:00 a.m. - noon

Ms. Airad

HM

3.32

1

10:00 a.m. - noon

Ms. Airad

HM

3.61

1

10:00 a.m. - noon

Ms. Airad

HM

3.1

1

10:00 a.m. - noon

Ms. Airad

HM

3.03

1

10:00 a.m. - noon

Ms. Airad

HM

3.67

1

10:00 a.m. - noon

Ms. Airad

HM

3.59

1

10:00 a.m. - noon

Ms. Airad

HM

3.33

1

10:00 a.m. - noon

Ms. Airad

HM

3.02

1

10:00 a.m. - noon

Ms. Airad

HM

3.55

1

10:00 a.m. - noon

Ms. Airad

HM

3

1

noon - 2:00 p.m.

Ms. Airad

SS

2.48

1

noon - 2:00 p.m.

Ms. Airad

SS

2.72

1

noon - 2:00 p.m.

Ms. Airad

SS

2.99

1

noon - 2:00 p.m.

Ms. Airad

SS

2.68

1

noon - 2:00 p.m.

Ms. Airad

SS

2.75

1

noon - 2:00 p.m.

Ms. Airad

SS

2.42

1

noon - 2:00 p.m.

Ms. Airad

SS

2.92

1

noon - 2:00 p.m.

Ms. Airad

SS

2.68

1

noon - 2:00 p.m.

Ms. Airad

SS

2.98

1

noon - 2:00 p.m.

Ms. Airad

SS

2.5

1

noon - 2:00 p.m.

Ms. Airad

SS

2.45

1

noon - 2:00 p.m.

Ms. Airad

SS

2.99

1

noon - 2:00 p.m.

Ms. Airad

SS

2.31

1

noon - 2:00 p.m.

Ms. Airad

SS

2.42

1

noon - 2:00 p.m.

Ms. Airad

SS

2.91

ems

03/Case 1 - Pelican Stores/BSTAT_03_Case1_PelicanStores_Key.xlsx Sheet: &UnStack; Sheet: &DataIndices; Sheet: &DataCopy; Sheet: &GraphData; Sheet: &WorkArea; Sheet: &Miscel;_Area Sheet: Data Net Sales

Method of Payment

Gender

Marital Status

Age

102.4

Proprietary Card

Female

Married

36

Descriptive Statistics on Net Sales

100.4

Proprietary Card

Female

Married

28

Net Sales

78

Proprietary Card

Female

Married

30

56.52

Proprietary Card

Female

Married

46

Mean

31.6

Proprietary Card

Female

Married

40

Standard Error

160.4

Visa

Female

Married

40

Median

64.5

Visa

Female

Married

46

Mode

71.4

Proprietary Card

Male

Single

36

Standard Deviation

94

Proprietary Card

Female

Single

22

Sample Variance

38.5

MasterCard

Female

Married

32

Kurtosis

44.8

Proprietary Card

Female

Married

56

Skewness

31.6

Proprietary Card

Female

Single

28

Range

70.82

Proprietary Card

Female

Married

38

Minimum

266

American Express

Female

Married

50

Maximum

39.5

Visa

Male

Married

48

Sum

30.02

Proprietary Card

Female

Married

60

Count

192.8

Proprietary Card

Female

Single

42

71.2

Proprietary Card

Female

Married

32

Descriptive Statistics on Net Sales Promotional Customer

18

Proprietary Card

Female

Married

70

Net Sales

63.2

MasterCard

Female

Married

28

63.2

Proprietary Card

Female

Married

44

Mean

105.5

MasterCard

Female

Married

56

Standard Error

117.5

Proprietary Card

Female

Married

50

Median

3

13.23

Proprietary Card

Female

Married

44

Mode

198.8

Proprietary Card

Female

Married

42

Standard Deviation

19.5

Visa

Female

Married

46

Sample Variance

62.4

Proprietary Card

Female

Married

54

Kurtosis

23.8

Proprietary Card

Female

Married

38

Skewness

39.6

Proprietary Card

Female

Married

60

Range

63.64

Proprietary Card

Female

Married

30

Minimum

14.82

Proprietary Card

Female

Married

32

Maximum

Sheet: Relationship-Age and Net Sales Age

Net Sales

Covariance

36

102.4

-7.334888888888898

28

100.4

30

78

Correlation

46

56.52

-0.010635891709415951

40

31.6

40

160.4

46

64.5

36

71.4

22

94

32

38.5

56

44.8

28

31.6

38

70.82

50

266

48

39.5

60

30.02

42

192.8

32

71.2

70

18

28

63.2

44

63.2

56

105.5

50

117.5

44

13.23

42

198.8

46

19.5

Negative linear association between age and net sales

Very weak negative linear association between age and net sales

54

62.4

38

23.8

60

39.6

30

63.64

32

14.82

03/Case 1 - Pelican Stores/BSTAT_03_Case1_PelicanStores_Report.docx Pelican Stores Pelican Stores, a division of National Clothing is a chain of women’s apparel stores operating throughout the country. The chain ran a promotion offering discounts to customers of other National Clothing stores. The discount coupons were not offered to regular customers. Sales made to people presenting the promotional coupons were considered sales that would not have otherwise been made. Descriptive statistics are reported here.

Descriptive Statistics – Net Sales Table 1 presents descriptive statistics on Net Sales. Mean, median, and mode are measures of central location of the data. In a perfect normal distribution these values would be the same. The mean or average sales are 77.60, median value is 59.71, and mode 31.6. Mode represents the most frequently occurring value. The skewness value of 1.71 (a value less than 2) indicates that the data distribution is considered approximately normal for analysis purposes.

Sales vary from a minimum of 13.23 to a maximum of 287.59. The range of sales is 274.36. For purposes of comparison, the standard deviation is a measure how the sales values vary about the mean. The standard deviation is 55.67.

The total for the 100 sales is 7760.05.

Descriptive Statistics – Net Sales for Promotional Customers Table 2 presents descriptive statistics on Net Sales to customers who received the promotional discount. The mean or average sales are 84.29 (a value higher than the average of all sales by approximately 7 dollars.), median value is 63.42, and mode 31.6. The skewness value of 1.52 (a value less than 2) indicates that this data distribution is considered approximately normal for analysis purposes as well.

Sales vary from a minimum of 13.23 to a maximum of 287.59. The range of sales is 274.36. These values are the same as values for the entire dataset. For purposes of comparison, the standard deviation, a measure of variability is 61.46.

The total for the 70 sales to promotional customers is 5900.30. Promotional customers sales represent 76% of total sales.

Descriptive Statistics – Net Sales for Regular Customers Table 3 presents descriptive statistics on Net Sales to regular customers, customers who frequently shop Pelican stores. The mean or average sales are 61.99 (a value lower than the average of all sales by approximately 15 dollars.), median value is 51, and mode 44.5. The skewness value of 1.35 indicates that this data distribution is considered approximately normal for analysis purposes as well.

Sales vary from a minimum of 22.50 to a maximum of 159.75. The range of sales is 137.25. The standard deviation is 35.07.

The total for the 30 sales to promotional customers is 1859.75, 24% of total sales.

Descriptive Statistics – Relationship Between Age and Net Sales The covariance value of -7.33 indicates a negative linear association between age and net sales.

The correlation value (-.01) suggests a very weak negative linear association between age and net

sales. This correlation suggests that it makes little sense to comment on the effect of age on net sales.

This report presents results based on a preliminary investigation of customer purchases at Pelican Stores when a promotional discount was offered. In summary, customers with promotional discounts spent more on purchases than regular customers, indicating that the promotional discount was of value in increasing net sales.

03/Case 1 - Pelican Stores/BSTAT_03_Case1_PelicanStores_Rubric.docx Pelican Stores Case Study Chapter 3 (60 points total) RUBRIC

03/Case 1 - Pelican Stores/BSTAT_03_Case1_PelicanStores_TeachingTips.docx Pelican Stores – Teaching Tips Descriptive Statistics – Net Sales, Net Sales by Promotional Customers, Net Sales by Regular Customers Indicate to students that a skewness measure less than 2 indicates an approximately normal distribution After indicating descriptive statistics measures, reports can compare regular and promotional customer values and perhaps compute percentage of total values.

Relationship between Age and Net Sale Make sure that students don’t attach any strength of relationship to the covariance measure.

Make sure that students don’t indicate any type of causality in the correlation measure.

03/Case 2 - Movies2016/BSTAT_03_Case2_Movie Releases_Report.docx Movie Releases in 2016 The movie industry, a competitive business, uses a number of key variables to measure the success of a movie released to industry: opening weekend gross sales, total gross sales, number of theatres the movie was shown in, and the number of weeks the movie was in release are a few.

The top 100 movies of 2016 are summarized in this report using descriptive statistics Descriptive Statistics – Opening Gross Sales Table 1 presents descriptive statistics on opening gross sales. Mean, median, and mode are measures of central location of the data. In a perfect normal distribution these values would be the same. The mean or average sales are 32.09 million, median value is 20.78 million, and mode 11.2 million. Mode represents the most frequently occurring value. The skewness value of 2.40 indicates that the data distribution is potentially skewed by high-performance outlier(s).

Sales vary from a minimum of .12 million to a maximum of 179.14 million. The range of sales is 179.02 million. For purposes of comparison, the standard deviation is a measure how the sales values vary about the mean. The standard deviation is 36.35.

The total opening gross sales is 3201.5 million.

Descriptive Statistics – Total Gross Sales Table 2 presents descriptive statistics on total gross sales. The mean or average sales are 103.82 million and median value is 62.99 million. There is no value that occurs more than once. The skewness value of 2.21 indicates that the data distribution is potentially skewed by high-performance outlier(s). An examination of the total gross sales data reveals that there were no gross sales values between 170 and 232 million dollars. Sales vary from a minimum of 23.59 million to a maximum of 532.18 million. The range of sales is 508.59 million. For purposes of comparison, the standard deviation, a measure of variability is 105.54 million.

The total gross sales are 10381.9 million. Opening gross sales represent 31% of total gross sales.

An examination of the data sorted by total gross sales indicates that Rogue One: A Star Wars Story, Finding Dory, and Captain America: Civil War are potential outliers. Using a z-score conversion to detect outliers requires a value less than -3 or greater than +3 for consideration. As shown in Table 2.1 Rogue One: A Star Wars Story is considered an outlier based on both opening and gross sales.

Descriptive Statistics – Number of Theaters Table 3 presents descriptive statistics based on number of theaters the movie is released in. The mean or average number of theaters is 3282, median value is 3285, and mode 4144. The skewness value of -.78 indicates that this data distribution is considered approximately normal for analysis purposes.

Number of theaters vary from a minimum of 1213 to a maximum of 4381. The range of sales is 3168. The standard deviation is 668.19.

The total for the number of theaters is 328,236 theaters.

Descriptive Statistics – Number of Weeks in Release

Table 4 presents descriptive statistics based on the number of weeks a movie is in release (shown in theaters). The mean or average number of theaters is 14.73, median value is 13, and mode 11.

The skewness value of 1.86 indicates that this data distribution is considered approximately normal for analysis purposes.

Number of weeks in release vary from a minimum of 7 to a maximum of 46. The range is 39. The standard deviation is 5.82 weeks.

The total for the number of weeks in release is 1473.

Descriptive Statistics – Relationship Between Total Gross Sales and All Other Variables An examination of the relationship between total gross sales and all other variables is now considered using covariance and correlation measures.

Covariance between opening and total gross sales is 3520.7, indicating a positive relationship between opening and total gross sales. The correlation value for this data is .91 indicating a very strong positive relationship between opening and total gross sales. It would be appropriate to use opening gross sales to predict total gross sales.

Covariance between number of theaters and total gross sales is 43265.3, indicating a positive relationship between number of theaters and total gross sales. The correlation value for this data is .61 indicating a moderately strong positive relationship between number of theaters and total gross sales.

Covariance between number of weeks in release and total gross sales is 290.84, indicating a positive relationship between number of weeks in release and total gross sales. The correlation value for this data is .47 indicating a moderate relationship between number of weeks in release and total gross sales.

This report presents results based on a preliminary investigation of movie releases in 2016. Results indicate that the Star Wars movie represents an outlier in the data. Further, opening gross sales is a strong predictor of total gross sales. It is suggested that further analysis remove the presence of outliers.

03/Case 2 - Movies2016/BSTAT_03_Case2_Movies2016-Key.xlsx Sheet: Sheet1 Gross Sales ($ millions)

Number of Theaters

Weeks in Release

18

4157

20

3

4305

25

08

4226

20

Mean

32.08500000000001

38

4381

25

Standard Error

3.63484324061415

4144

24

Median

20.775

07

3856

18

Mode

11.2

27

3959

22

Standard Deviation

36.3484324061415

36

4256

12

Sample Variance

1321.2085383838375

1

4255

14

Kurtosis

5.656292410332528

4

4029

20

Skewness

2.39785488639632

76

3875

22

Range

179.01999999999998

04

4144

19

Minimum

0.12

64

3882

19

Maximum

179.14

61

3416

46

Sum

3208.5000000000005

43

4039

21

Count

100

85

3928

13

44

4153

9

71

4066

21

1

3236

20

Mean

103.81900000000005

53

3987

25

Standard Error

10.553819029674093

35

3963

17

Median

62.985

44

3508

11

Mode

#N/A

64

3591

11

Standard Deviation

105.53819029674094

07

3955

20

Sample Variance

11138.309611111103

26

3215

13

Kurtosis

4.5715160020662715

51

3932

17

Skewness

2.2059687165956445

14

4130

12

Range

508.59

47

3356

11

Minimum

23.59

55

3115

17

Maximum

532.18

01

3478

17

Sum

10381.900000000005

9

3135

19

Count

100

Table 1 Opening Gross Sales ($ millions)

Table 2 Total Gross Sales ($ millions)

03/Case 2 Movies2016/BSTAT_03_Case2_Movies2016_Rubric.docx Movie Theater Releases Case Study Chapter 3 (90 points total) RUBRIC

03/Case 2 Movies2016/BSTAT_03_Case2_Movies2016_TeachingTips.docx Pelican Stores – Teaching Tips Descriptive Statistics – Net Sales, Net Sales by Promotional Customers, Net Sales by Regular Customers Indicate to students that a skewness measure less than 2 indicates an approximately normal distribution After indicating descriptive statistics measures, reports can compare regular and promotional customer values and perhaps compute percentage of total values.

Relationship between Age and Net Sale Make sure that students don’t attach any strength of relationship to the covariance measure.

Make sure that students don’t indicate any type of causality in the correlation measure.

itiion ($)

03/Case 3 - Business Schools of Asia-Pacific/BSTAT_03_Case3_Asian_Key.xlsx Sheet: Data Age

%Foreign

GMAT

English Test

Work Experience

Starting Salary ($)

28

47

Yes

No

Yes

71400

29

28

Yes

No

Yes

65200

22

0

No

No

No

7100

29

10

Yes

No

No

31000

28

60

Yes

Yes

No

87000

25

50

Yes

No

Yes

22800

Mean

23

1

Yes

No

No

7500

Standard Error

29

51

Yes

Yes

Yes

43300

Median

23

0

No

No

No

7400

Mode

30

80

Yes

Yes

Yes

46600

Standard Deviatio

32

20

Yes

No

Yes

49300

Sample Variance

32

26

No

No

Yes

49600

Kurtosis

26

37

Yes

No

Yes

34000

Skewness

34

27

No

No

Yes

60100

Range

25

6

Yes

No

Yes

17600

Minimum

30

30

Yes

Yes

Yes

52500

Maximum

29

90

No

Yes

Yes

25000

Sum

30

10

No

No

Yes

66000

Count

37

35

No

Yes

Yes

41400

32

30

No

Yes

Yes

48900

24

0

No

No

Yes

7000

29

43

Yes

No

Yes

55000

Mean

23

2.5

No

No

No

7500

Standard Error

32

15

No

Yes

Yes

16000

Median

28

3.5

Yes

No

Yes

13100

Mode

GMAT

English Test

Work Experience

Standard Deviatio

Yes

14

8

19

Sample Variance

No

11

17

6

Kurtosis

Table 1 Full-Time Enrollm

Table 2 Students per Facu

Skewness Table 8

Range

Mean Starting Salaries

Minimum

Sheet: Starting Salaries - Tuition Covariance

143409465.9999999

Correlation

0.7859085920378343

Local Tuition ($)

Foreign Tuition ($)

Starting Salary ($)

24420

29600

71400

19993

32582

65200

4300

4300

7100

11140

11140

31000

33060

33060

87000

7562

9000

22800

3935

16000

7500

6146

7170

43300

2880

16000

7400

20300

20300

46600

8500

8500

49300

16000

22800

49600

11513

11513

34000

Covariance

142871731.6666666

17172

19778

60100

Correlation

0.6667006516568184

17355

17355

17600

16200

22500

52500

18200

18200

25000

16426

23100

66000

13106

21625

41400

13880

17765

48900

1000

1000

7000

9475

19097

55000

11250

26300

7500

2260

2260

16000

3300

3600

13100

03/Case 3 - Business Schools of Asia-Pacific/BSTAT_03_Case3_Asian_Report.docx Business Schools of Asia-Pacific A survey shows that more and more Asians choose the master of business administration (MBA) degree route to corporate success. The number of applicants for MBA courses at Asia-Pacific schools continues to increase. This report uses data of the characteristics of the leading Asia-Pacific business schools and methods of descriptive statistics to summarize the data.

Full-Time Enrollment Table 1 presents descriptive statistics for full-time enrollment. Mean, median, and mode are measures of central location of the data. In a perfect normal distribution these values would be the same. Mean full-time enrollment is 165.16, median value is 126, and mode 30. Mode represents the most frequently occurring value. The skewness value of .76 (a value less than 2) indicates that the data distribution is considered approximately normal for analysis purposes.

Full-time enrollment varies from a minimum of 12 to a maximum of 463 (Indian Institute of Management). The range of enrollments is 451. For purposes of comparison, the standard deviation is a measure of how the enrollment values vary about the mean. The standard deviation is 140.8.

Total enrollment for the 25 schools is 4129.

Students per Faculty Table 2 presents descriptive statistics for the number of students per faculty. Mean students per faculty is 8.48 and median value is 7. The most frequently occurring students per faculty is 5. The skewness value of .76 again indicates that the data distribution is considered approximately normal for analysis purposes.

Number of students per faculty varies from a minimum of 2 to a maximum of 19 (University of Adelaide), indicating a range of 17. The standard deviation is a measure of how the number of students per faculty vary about the mean. The standard deviation is 5.06.

Local Tuition Table 3 presents descriptive statistics for local tuition in US dollars. Mean local tuition is $12,374.90. The median value is 11,513. The skewness value of .62 (a value less than 2) indicates that the data distribution is considered approximately normal for analysis purposes.

Local tuition varies from a minimum of 1000 dollars to a maximum of 33,060 (International University of Japan (Niigata)), representing a range of 32,060 dollars. For purposes of comparison, the standard deviation is a measure of how the local tuition varies about the mean. The standard deviation is 7,778.42.

Total local tuition for the 25 universities is 309,373 dollars. .

Foreign Tuition Table 4 presents descriptive statistics for foreign tuition, again in US dollars. Mean foreign tuition is $16,581.80. The median value is 17,765. Most frequently occurring value (the mode) is 16,000. The skewness value of -.01 (a value less than 2) indicates that the data distribution is considered approximately normal for analysis purposes.

Foreign tuition varies from a minimum of 1000 dollars to a maximum of 33,060 (International University of Japan (Niigata)), representing a range of 32,060 dollars. For purposes of comparison, the standard deviation is a measure of how the local tuition varies about the mean. The standard deviation is 9,134.85.

Total foreign tuition for the 25 universities is 414,545 dollars. Local and foreign tuition for 10 of the 25 schools is the same.

Age Table 5 presents descriptive statistics for students’ age. Mean age is 28.36. The median and mode value is 29. The skewness value of .12 indicates that the data distribution is considered approximately normal for analysis purposes.

Student age varies from a minimum of 22 to a maximum of 37 years, representing a range of 15 years. For purposes of comparison, the standard deviation is a measure of how the age varies about the mean. The standard deviation is 3.78. % Foreign Table 6 presents descriptive statistics for the percentage of foreign students enrolled at a university, that is, students from another country attending a particular university. Mean foreign percentage is 28.08. The median is 29 percent and mode 0 percent. Three (3) of the 25 schools indicate no foreign student enrollment. The skewness value of .87 indicates that the data distribution is considered approximately normal for analysis purposes.

Percentage of foreign enrollment varies from a minimum of 0 to a maximum of 90 percent (Asian Institute of Management (Bangkok)). For purposes of comparison, the standard deviation is a measure of how this percentage varies about the mean. The standard deviation is 25.01.

Starting Salary Table 7 presents descriptive statistics for the starting salary of Asians obtaining MBA degrees.

Mean starting salary is $37,292. The median is 41,400 dollars and mode 7500. The skewness value of .22 indicates that the data distribution is considered approximately normal for analysis purposes.

Starting salary varies from a minimum of 7000 to a maximum of 87,000 dollars (International University of Japan (Niigata)), an 80,000 range. For purposes of comparison, the standard deviation is a measure of how this percentage varies about the mean. The standard deviation is 23,459.30.

Difference between Local and Foreign Tuition Costs Ten (10) of the 25 schools indicate no difference in local and foreign tuition costs. Remaining tuition differences are of foreign tuition costs greater than local tuition costs. The largest difference ($15,050) is at Lahore University of Management Sciences. Only 4 schools report differences of more than $10,000.

Difference between Mean Starting Salaries and Work Experience Table 8 presents a comparison of the difference between mean starting salaries when work experience is required. When work experience is required as part of participation in the MBA program, salaries average 16,721.93 more than MBA programs with no work experience requirement.

Difference between Mean Starting Salaries and English Test Requirement Table 9 presents a comparison of the difference between mean starting salaries when an English test requirement is in place. When an English test is required as part of participation in the MBA

program, salaries average 11.463.97 more than MBA programs with no English test requirement.

Relationship between Starting Salaries and Local Tuition Figure 1 presents a scatter diagram of the relationship between local tuition and starting salary indicating a positive relationship. As local tuition increases, starting salary increases. The correlation is .79 indicating a moderately strong positive relationship.

Relationship between Starting Salaries and Foreign Tuition Figure 1 presents a scatter diagram of the relationship between foreign tuition and starting salary.

The correlation is .67 indicating a moderately strong positive relationship. For comparison purposes, the relationship between starting salaries and local tuition appears to be stronger.

This report presents results based on examination of key variables used to describe MBA enrollment in Asian universities. Starting salary information provided is used to examine differences in salary when work experience and English test requirements are enforced. Programs with work experience and English test requirements indicate larger starting salaries. There is a positive relationship between starting salary and tuition.

03/Case 3 - Business Schools of Asia-Pacific/BSTAT_03_Case3_Asian_Rubric.docx Business Schools of Asia-Pacific Case Study Chapter 3 (130 points total) RUBRIC

03/Case 3 - Business Schools of Asia-Pacific/BSTAT_03_Case3_Asian_TeachingTips.docx Business Schools of Asia-Pacific– Teaching Tips Descriptive Statistics – Full-time enrollment, students per faculty, local tuition, foreign tuition, age, % foreign, starting salary Indicate to students that a skewness measure less than 2 indicates an approximately normal distribution After indicating descriptive statistics measures, reports can compare regular and promotional customer values and perhaps compute percentage of total values.

Differences between Starting Salary/Work Experience and Starting Salary/English Test Requirement Students will most likely not be familiar with the AVERAGEIF() function.

AVERAGEIF(range, criteria, [average range]). Indicate that the first range searches for a criteria (Work Experience = Yes or No; English Test Requirement = Yes or No). The average range (when entered) indicates the range of values for which to compute an average.

Relationship between Tuition and Starting Salary If covariance measures are reported, make sure that students don’t attach any strength of relationship to the covariance measure. Since the value provided is in “squared” terms of the variable of interest, a business report may not be the place to report this value.

Make sure that students don’t indicate any type of causality in the correlation measure.

03/Case 5 - African Elephant Populations/BSTAT-03_Case5_AfricanElephants-Key.xlsx Sheet: Data Table 2 Mean Annual Change 1989-2007

n ual nge

Table 3 Mean A 2007-20

Country

(x1)(x2)…

xbarg

Mean Annual Change

Angola

0.20403225806451614

0.9154822647876368

-0.08451773521236317

Angola

813066190076514

Botswana

3.440921568627451

1.0710636125645567

0.07106361256455673

Botswan

7264037634174176

Cameroon

0.7258018867924528

0.9823532172613675

-0.017646782738632516

Camero

1296387889485116

Cen African Rep

0.1754736842105263

0.9078451493674056

-0.09215485063259443

Cen Afr

458639672643791

Chad

2.075806451612903

1.0414093968327907

0.04140939683279066

Chad

55005932903824

Congo

0.31574285714285716

0.9379618971129879

-0.06203810288701206

Congo

3855571511522002

Dem Rep of Congo

0.27898823529411765

0.9315350838214812

-0.0684649161785188

Dem Re

951761968497483

Gabon

0.9294342105263158

0.9959427389172543

-0.004057261082745733

Gabon

1573176496443827

Kenya

1.6650526315789473

1.028730352167171

0.028730352167170903

Kenya

0241985209819882

Mozambique

1.4025806451612903

1.018972956180238

0.018972956180238087

Mozamb

305302084476896

Somalia

0.011666666666666667

0.7809228938716606

-0.21907710612833942

Somalia

2843640915469334

Tanzania

2.0875375

1.0417354907161778

0.04173549071617777

Tanzani

2164665454859758

Zambia

0.7129512195121951

0.9813787648674297

-0.018621235132570302

Zambia

6656129915926705

Zimbabwe

2.304813953488372

1.0474816833548488

0.047481683354848814

Zimbabw

Decline 10% or more 1 Experienced growth 6

Sheet: Box Plot

Country

03/Case 5 - African Elephant Populations/BSTAT_03_Case5_Afric anElephantPopulations_Report.docx African Elephant Populations Elephant populations in African nations have been devastated by poaching since the mid-1980s. As important to African ecosystems, strong measures have been taken to effectively protect elephant populations. Data from elephant populations for several African nations was collected in 1979, 1989, 2007, and 2012. This report uses reported data and the geometric mean to analyze changes over time in the populations of elephants in several African nations.

Mean Annual Change in Elephant Population – 1979-1989 Table 1 presents the mean annual change of elephant populations over a 10-year period from 1979 to 1989. Elephant populations in eight countries declined at a rate of ten percent or more. These countries are: Central African Republic, Chad, Democratic Republic of the Congo, Kenya, Mozambique, Somalia, Tanzania, and Zambia. During the same period Botswana, Congo, and Gabon experienced growth in their elephant populations.

Mean Annual Change in Elephant Population – 1989-2007 Table 2 presents the mean annual change of elephant populations over a 10-year period from 1979 to 2007. Only one country (Somalia) continues to experience average annual declines of 10% or more in its elephant population. Most other nations experienced relatively small mean annual changes in their elephant population.

Mean Annual Change in Elephant Population – 2007-2012 Table 3 presents the mean annual change of elephant populations over a 5-year period from 2007 to 2012. Two countries (Chad and the Democratic Republic of the Congo) experienced average annual declines in their elephant population of 10% or more from 2007-2012. Most other nations had relatively small mean annual changes in their elephant populations, with the exception of Congo which experienced a 17.4% growth in their elephant population.

Comparison of Mean Population Changes Table 4 presents a comparison of the three analyses. Figure 1 provides a box plot of elephant populations to support this analysis. From these box plots we can see that the population of elephants declined dramatically from 1979 to 1989. Between 1989 and 2007 the populations began to come back. Between 2007 and 2012 stabilization of the populations has for the most part been maintained. Central African Republic and Democratic Republic of Congo continue a trend of declining population although improved from the initial 1979-1989 period. Chad, while experiencing population growth between 1989-2007 had since then returned to its earlier population decline (more than 14%). Tanzania, as well, experienced growth in 1989-2007 but has since returned to a 6.8% decline in population.

This report presents results over time of changes in elephant populations in several African countries. Several nations appear to have reversed the declines experienced in 1979-1989 of elephant populations. Growth rates, however, are still generally low and in some countries still negative. The rates of change in 2007-2012 indicate that it will take many decades for the populations to recover to their 1979 levels.

03/Case 5 - African Elephant Populations/BSTAT_03_Case5_Afric anElephantPopulations_Rubric.docx African Elephant Populations - Case Study Chapter 3 (95 points total) RUBRIC

03/Case 5 - African Elephant Populations/BSTAT_03_Case5_Afric anElephantPopulations_TeachingTips.docx African Elephant Populations– Teaching Tips Geometric Mean It may take some coaxing to urge students to use the geometric mean for this problem. Reference page 110, last paragraph in the section, for encouragement.

Applying the geometric mean in Excel is facilitated by the POWER function. It is my experience that most business students will not have been exposed to this function. To compute xbarg demonstrate the use of the POWER function with =POWER(cell address, 1/10) for calculating the 10th root for the 1979-1989 range.

You might suggest that students use freeze panes in B3 to be able to reference countries by name while completing calculations.

Suggest that students create tables for their report from the Excel analysis AFTER they have completed calculations and copy those tables into the report (see provided Excel key) In Excel, the box plot is known as a box and whisker and found by seeing All Charts in the Insert Charts tab.

04/Case 1 - Hamilton County Judges/BSTAT_04_Case1_HamiltonCountyJudges_Report.docx Hamilton County Judges Hamilton County judges try thousands of cases per year. For purposes of evaluating the performance of the judges, The Cincinnati Enquirer conducted a study of cases handled by Hamilton County judges over a three-year period. This report uses probability analysis to evaluate the performance of the judges in each of three courts: Common Pleas Court, Domestic Relations Court, and Municipal Court.

Probability of an Appeal Appeal probabilities were computed for judges of each of three courts.

Common Pleas Court Table 1 presents the probability of an appeal for Common Pleas court. The probability of an appeal is .0401. In the Common Pleas court, the judge with the lowest probability of an appeal is Ralph Winkler.

Domestic Relations Court Table 2 presents the probability of an appeal for Domestic Relations court. The probability of an appeal is .0035. In the Domestic Relations court, the judge with the lowest probability of an appeal is Ronald Panioto.

Municipal Court Table 3 presents the probability of an appeal for Municipal court. The probability of an appeal is .0046. In the Municipal court, the judge with the lowest probability of an appeal is Carla Grady.

The lowest probability of an appeal for all three courts is in Domestic Relations court.

Probability of a Reversal Reversal probabilities were computed for each of three courts.

Common Pleas Court Table 4 presents the probability of a reversal for Common Pleas court. The probability of an appeal is .0045. In the Common Pleas court, the judge with the lowest probability of a reversal is Ralph Winkler.

Domestic Relations Court Table 5 presents the probability of a reversal for Domestic Relations court. The probability of an appeal is .0005. In the Domestic Relations court, the judge with the lowest probability of a reversal is Ronald Panioto.

Municipal Court Table 6 presents the probability of a reversal for Municipal court. The probability of an appeal is .0009. In Municipal court, the judges with the lowest probability of a reversal are equally Karla Grady and Deidra Hair.

Probability of a Reversal Given Appeal

The probability of a reversal given an appeal were computed for each of three courts.

Common Pleas Court Table 7 presents the probability of a reversal given appeal for Common Pleas court. The probability of an appeal is .1129. In Common Pleas court, the judge with the lowest probability of a reversal given an appeal is Thomas Nurre.

Domestic Relations Court Table 8 presents the probability of a reversal given appeal for Domestic Relations court. The probability of an appeal is .1604. In Domestic Relations court, the judge with the lowest probability of a reversal given an appeal is Ronald Panioto.

Municipal Court Table 9 presents the probability of a reversal given appeal for Municipal court. The probability of a reversal given an appeal is .2080. In Municipal court, the judges with the lowest probability of a reversal given an appeal are equally Karla Grady and Deidra Hair.

Judge Rankings For purposes of evaluating the performance of judges, rankings were created for each judge for each of the three courts. The lowest ranking of 1 was assigned to the judge with the least probability by court for an appeal, a reversal, and a reversal given an appeal. The sum of the rankings was then computed with the inference that the lowest combined rating is associated with the most effective performance. The highest combined rating represents the judge with the least effective performance.

Common Pleas Court Table 10 presents the judge rankings for Common Pleas court. Ralph Winkler has the overall best performance with a sum ranking of 5 (1, 1, 3). J. Howard Sundermann, Jr. has the overall worst performance with a sum ranking of 45 (16, 16, 13).

Domestic Relations Court Table 11 presents the judge rankings for Domestic Relations court. Ronald Panioto has the overall best performance with a sum ranking of 3 (1, 1, 1). Deborah Gaines has the overall worst performance with a sum ranking of 11 (4, 4, 3).

Municipal Court Table 12 presents the judge rankings for Municipal court. Karla Grady has the overall best performance with a sum ranking of 3 (1, 1, 1) followed closely by Deidra Hair with a sum ranking of 5 (3, 1, 1). Leslie Isaiah Gaines has the overall worst performance with a sum ranking of 57 (19, 20, 18).

This report presents results based on an examination of appeals, reversals, and reversals given appeals used to describe the performance of judges in Hamilton County. Provided tables can be examined for a comparison of individual judges’ performance.

04/Case 1 - Hamilton County Judges/BSTAT_04_Case1_HamiltonCountyJudges_Rubric.docx Hamilton County Judges Case Study Chapter 4 (110 points total) RUBRIC

04/Case 1 - Hamilton County Judges/BSTAT_04_Case1_Hamilton CountyJudges_TeachingTips.docx Hamilton County Judges – Teaching Tips Indicate to students the value of cut and paste when similar processes are being described. This also requires attention to detail when reusing previous text. To avoid moving back and forth between text and Excel file, suggest that students use the snipping tool to copy managerial report requirements into the text file.

Suggest that students build the tables required for the report in a separate worksheet of the Excel file and use the snipping tool to copy tables into text.

Attention to detail Time to build tables for reports Use of Rank.Eq formula Use of VLOOKUP to find judge Probability of Appeal For each judge computed as the number of appeals divided by the number of cases disposed.

Students will most likely not be familiar with the RANK.EQ() function. The RANK.EQ() function requires the use of the optional [order] function. Enter 1 for ascending order.

Probability of Reversal For each judge computed as the number of reversals divided by ,the number of cases disposed.

The RANK.EQ() function requires the use of the optional [order] function. Enter 1 for ascending order.

Probability of Reversal Given Appeal For each judge, computation is given by P(A■ B) is the probability of a reversal or (# of reversals / # of disposed) The RANK.EQ() function requires the use of the optional [order] function. Enter 1 for ascending order.

Ranking of Judges The VLOOKUP() function could be used to establish the judges name associated with an associated rank. At this point in the processing, most students will not be receptive to “another” function that they are unfamiliar with.

04/Case 1 - Hamilton County Judges/BSTAT_04_Case1_Judge_Key.xlsx Sheet: Data

ditional Probability of Reversal n Appeal

Rank

Sum of Ranks

759124087591241

5

25

403361344537816

4

12

181818181818185

15

30

666666666666667

9

16

511811023622047

2

15

780219780219777

16

38

181818181818182

14

34

267175572519084

12

38

200000000000002

8

22

678832116788322

10

31

9586776859504134

1

11

302325581395349

6

25

413793103448274

11

37

666666666666669

13

45

Common Pleas C

236220472440945

7

24

Domestic Relation

818181818181818

3

5

Municipal Court

293984108967084

1a. Probability of cases being appealed

1b. Probability of cases being reversed

Common Pleas C

Domestic Relation

ditional Probability of Reversal n Appeal

Rank

Sum of Ranks

285714285714285

2

6

05263157894737

4

10

75

3

11

375

1

3

Municipal Court 2. Probability of a case being appealed for each judge

Column E of each 3. Probability of a case being reversed for each judge

Column G of each

037735849056603

4. Probability of a reversal given an appeal for each judge

ditional Probability of Reversal n Appeal

Rank

Sum of Ranks

30232558139535

4

31

Column I of each

64705882352941

10

30

Sheet: Tables Created for Report

Table 7 Probabili Common

Table 4 Probability of a Reversal - Common Pleas Court Judge

Disposed

Reversed

Probability of Reversal

Judge

46625

Fred Cartolano

3037

12

0.003951267698386566

Fred Car

9881

Thomas Crush

3372

10

0.0029655990510083037

Thomas

11444

Patrick Dinkelacker

1258

8

0.006359300476947536

Patrick D

65915

Timothy Hogan

1954

7

0.0035823950870010235

Timothy H

7833

Robert Kraft

3138

7

0.0022307202039515616

Robert K

5265

William Mathews

2264

18

0.007950530035335688

William M

3958

William Morrissey

3032

22

0.007255936675461741

William M

9483

Norbert Nadel

2959

20

0.006759040216289287

Norbert N

5952

Arthur Ney Jr.

3219

14

0.004349176762969867

Arthur Ne

3334

Richard Niehaus

3353

16

0.0047718461079630185

Richard N

3333

Thomas Nurre

3000

6

0.002

Thomas

869

John O'Connor

2969

12

0.0040417649040080834

John O'C

869

Robert Ruehlman

3205

18

0.005616224648985959

Robert R

9005

J. Howard Sundermann Jr.

955

10

0.010471204188481676

J. Howar

9287

Ann Marie Tracey

3141

13

0.004138809296402419

Ann Mari

7776

Ralph Winkler

3089

6

0.0019423761735189381

Ralph Wi

7076

Totals

43945

199

0.004528387757424053

Totals

Table 8 Probabili Domestic

Table 5 Probability of an Reversal - Domestic Relations Court Judge

Disposed

Reversed

Probability of Reversal

Judge

78014

Penelope Cunningham

2729

1

0.00036643459142543056

Penelope

37194

Patrick Dinkelacker

6001

4

0.0006665555740709882

Patrick D

99966

Deborah Gaines

8799

9

0.0010228435049437436

Deborah

16962

Ronald Panioto

12970

3

0.0002313030069390902

Ronald P

651695

Totals

30499

17

0.0005573953244368668

Totals

Table 6 Probability of a Reversal - Municipal Court

Table 9 Probabili Municap

eal

Judge

Disposed

Reversed

Probability of Reversal

Judge

06993

Mike Allen

6149

4

0.0006505122784192552

Mike Alle

26933

Nadine Allen

7812

6

0.0007680491551459293

Nadine A

461

04/Case 2 - Rob's Market/BSTAT_04_Case2_MarketBasket_Key.xlsx Sheet: Data 1. Probability that a random customer does not purchase any of the products 278

=COUNTIFS($A$

164

375

P(none of the products) =

1000

348

78

574

2. Probability that a random customer purchases white bread

1000

0.461 3. Probability that a random customer purchases wheat bread 0.164

228

101

10. Probability that a random customer purchases white bread, grape jelly, creamy peanut butter

671

124

1000

P(white,grape,creamy) =

Sheet: JellyProbabilities

4. Probability that a random customer purchases grape jelly given that he/she purchases white bread White Grape

306

Not Grape

155

Total

461

White Grape

0.306

Not Grape

0.155

Total

0.460999999999

5. Probability that a random customer purchases strawberry jelly given that he/she purchases white bread White Strawberry

38

Not Strawberry

423

Total

461

Joint Probability Table for Bread - STRAWBERRY White Strawberry

0.038

Not Strawberry

0.423

Total

0.461

berry

White Grape

306

Strawberry

38

None

117

Total

461

Sheet: PeanutButterProbabilities

6. Probability that a random customer purchas

7. Probability that a random customer purchas

white

wheat

none

Total

8. Probaility that a random customer purchase

186

17

25

228

9. Probability that a random customer purchas

54

16

31

101

221

131

319

671

P(creamy | white)

461

164

375

1000

P(natural | white)

white

wheat

none

Total

P(creamy | wheat)

0.186

0.017

0.025

0.228

P(natural | wheat)

0.054

0.016

0.031

0.101

0.221

0.131

0.319

0.671

0.46099999999999997

0.164

0.375

1

obability Table for Peanut Butter

04/Case 2 - Rob's Market/BSTAT_04_Case2_RobsMarket_Report.docx Rob’s Market Rob’s Market (RM), a regional food store chain in the southwest United States, initiated a study of the purchase behavior of customers who use the RM loyalty card. Point-of-sale data was collected of customer purchases as they check out of the market. A market basket analysis was conducted to investigate customer buying behavior with regard to bread, jelly, and peanut butter. A data set of purchases by 1000 customers over a one-week period was provided. Details provided here present the results of this analysis.

Probability that a Customer does not purchase any of the three products Two hundred seventy-eight (278) of the 1000 customers purchased none of the products for a probability of .278 Probability of Bread Type Purchases Table 1 presents a count of bread purchases by type (white, wheat, none). White bread was purchased by 461 customers, a probability of .461. Approximately 46% of customers purchased white bread. Wheat bread was purchased by 164 customers, a probability of .164 customers.

Approximately 16% of customers purchased wheat bread. The majority of customers prefers white bread.

Probability of Jelly Type Purchases Table 2 presents the joint probability table for jelly and bread type. The probability that a customer purchases grape jelly given that they purchased white bread is .664 (64%), computed from the table as .306/.461. The probability that a customer purchases strawberry jelly given white bread is .082 (8.2%), computed from the table as .038/.461. White bread customers prefer grape to strawberry jelly.

Probability of Peanut Butter Type Purchases Table 3 presents the joint probability table for peanut butter and bread type. The probability that a customer purchases creamy peanut butter given white bread is .4035 (40%), computed from the table as .186/.461). The probability that a customer purchases natural peanut butter given white bread is .117 (12%), computed from the table as .054/.461. Customers who purchase white bread are more likely to purchase creamy peanut butter. The probability that a customer purchases creamy peanut butter given a wheat bread purchase is .104 (~10%), computed from the table as .017/.164. The probability that a customer purchases natural peanut butter given a wheat bread purchase is .098 (~10%). There appears to be very little difference in peanut butter type purchase for customers purchasing wheat bread.

Probability of White Bread, Grape Jelly, and Creamy Peanut Butter Purchase The probability of customers purchasing white bread, grape jelly, and creamy peanut butter, a scenario that describes the traditional peanut butter and jelly sandwich, is .124 (12.4%). It appears that eating habits for Rob’s Market customers have steered away from the traditional peanut butter and jelly sandwich This report presents results based on a market basket analysis to investigate customer buying behavior with regard to bread, jelly, and peanut butter. Provided tables can be used to investigate buying behavior of additional buying behaviors not discussed in this report.

04/Case 2 - Rob's Market/BSTAT_04_Case2_RobsMarket_Rubric.docx Rob’s Market Case Study Chapter 4 (75 points total) RUBRIC

04/Case 2 - Rob's Market/BSTAT_04_Case2_RobsMarket_TeachingTips.docx Rob’s Market – Teaching Tips Creating joint probability tables can be created easily using the COUNTIFS() function. This requires initial understanding of COUNTIF. Indicate to students that the COUNTIFS() function allows you to consider multiple criteria across different ranges, i.e., P(no bread, no jelly, no peanut butter). The criteria are the word “none”. Suggest maybe NOT using a cell reference to the word none to avoid a conversation about absolute/relative cell references.

The joint probability table for bread and jelly table was created in the key by creating separate data sets for white, wheat, and none bread (see JellyProbabilities worksheet). Students struggling with COUNTIFS() may prefer this format. Remember the goal here is not advanced Excel skills but the understanding of conditional probabilities. The formula given below is most easily understand with reference to cells in a joint probability table. It may be worth spending time helping students build a joint probability table.

The peanut butter probabilities table is built (in the PeanutButterProbabilites worksheet) using =COUNTIFS($A$4:$A$1003,F$3,$B$4:$B$1003,$E4). Depending on how much students are struggling with probabilities, this could be a place to review/introduce the use of a mixed cell reference.

05/Case 1 - Go Bananas!/BSTAT_05_Case1_GoBananas_Key.xlsx Sheet: Sheet1

hat a weekly sample will result in a shutdown of the production process is working properly 25

=n

0.08

=p

boxes fail to meet the standard fail to

ht Standard (x)

P(X = x) 0

0.12436428680229492

1

0.2703571452223804

2

0.28211180371030975

3

0.1880745358068735

4

0.08994869103806968

5

0.032850826292164514

6

0.009521978635410044

7

0.002247423528854431

8

0.00043971329912384505

9

7.222344043578133e-05

10

1.0048478669255623e-05

11

1.1915192098532756e-06

12

1.2087876033461953e-07

13

1.051119657979882e-08

14

7.834433102260618e-10

15

4.9958703840502494e-11

16

2.7150504067208203e-12

17

1.2501111257279263e-13

18

4.773959005888173e-15

19

0

20

0

21

0

22

0

23

0

24

0

=BINOMDIST(B6,$E$3,$E$4,TRUE)

P(p>=5) =

0.04514

Sheet: Sheet2 Table 2 Design Process Improvement Probabilites p

P(x >=5)

0.08

0.04514353742007149

0.07

0.027447356813656842

0.06

0.01504927019903013

0.05

0.007164947902586594

05/Case 1 - Go Bananas!/BSTAT_05_Case1_GoBananas_Report.docx Go Bananas Great Grasslands Grains, Inc. (GGG) manufactures and sells a wide variety of breakfast cereals.

GGG has produced a new cereal, Go Bananas!, with banana-flavored marshmallows and has determined that consumers are most happy when the cereal contains at least 1.6 ounces and no more than 2.4 ounces of banana-flavored marshmallows. GGG intends to sample 25 boxes of Go Bananas! on a weekly basis and shut down production if at least five boxes fail to meet the standard weight desired. The production process is designed so that only 8% of all boxes fail to meet the established standard. This report examines probabilities associated with these specifications and investigates the potential for improving the design process to 1%.

Probability of Meeting Standards Table 1 presents the probability that an indicated number of boxes will fail to meet the weight standard. There is a .12 (12%) chance that no boxes will fail to meet the standard; .28 (28%) chance that 2 boxes will fail to meet the standard. Given that the process will shut down if 5 or more boxes fail to meet the standard, these results indicate that the process will be shut down 4.5% (.0451) of the time (this value is given by the sum of the probabilities from 5 to 25).

Improving the Shutdown Design Standard Management investigated the potential for improving the design process so that production processes are only shut down 1% of the time. Since 5 boxes exceeds this standard, management examined the probability at 6 boxes (.0123). This value exceeds the 1% standard. When 7 or more boxes fail to meet the standard weight, the probability is .0028. To shut down the process no more than 1% of the time, the production process should be shut down when 7 or more boxes fail to meet the standard weight of banana-flavored marshmallows.

Improving the Design Process The VP for Production of GGG has indicated that with additional resources the production process could be redesigned to reduce the percentage of boxes that fail to meet the standard weight to the 1% standard. Table 2 presents the probability that the process will be shut down when 5 or more boxes fail to meet the standard at improved production processes (.07, .06, .05). Evaluation of potential design improvement indicates that when the process is working properly to between .05 and .06, the probability that at least 5 of the sampled boxes will fail to meet the standard is .01 or less.

In this report probability calculations revealed the potential of the current process meeting the standard (5 or more boxes fail to meet the banana-flavored marshmallow content) to be 4.5%. In order to shut down the process no more than 1% of the time, seven or more boxes would have to fail to meet the standard. If with additional resources the production process were redesigned so that between 5% and 6% of all boxes failed to meet the standard, the production process would shut down no more than 1% of the time.

05/Case 1 - Go Bananas!/BSTAT_05_Case1_GoBananas_Rubric.docx Go Bananas! Case Study Chapter 5 (70 points total) RUBRIC

05/Case 1 - Go Bananas!/BSTAT_05_Case1_GoBananas_TeachingTips.docx Go Bananas! – Teaching Tips This case does not provide a data file. However, data provided in the case description is used in many calculations. I would suggest that students compute results in an Excel file (see BSTAT_05_Case1_GoBananas_Key.xlsx) Many students may still have and prefer to work with calculators. Suggest trying the use of the BINOM.DIST function. If they decide to create a table like the figure, the function process is not trivial.

You may want to demonstrate the use of the SUM function where the range contains an absolute and a relative reference and explain how this causes the values to accumulate in the column.

Describe #2 as: if 5 or more boxes yields 4.5%, what number of boxes would result in a 1% error. 6 boxes is still greater than 1%. At 7 boxes the value is less than 1%. #3 requires computing binomial distributions with numbers < .08 until P(x>=5) is less than 1%. It may be helpful for students to set up columns in the worksheet like the following figure and then proceed with processing.

05/Case 2 - Grievance Committee at Tuglar Corporation/BSTAT_05_Case3_GrievanceCommittee.xlsx Sheet: Sheet1

n Trials are not independent Probability of success changes from trial to trial number of elements in population population size number of employees on grievance committee

r is elected to serve =HYPGEOM.DIST(0,10,r,n,FALSE) Table 1 Hypergeometric Probabilities Total # of Volunteers

# of Volunteers in Accounting Departm

Year 1

29

1

Year 2

31

1

Year 3

23

1

Year 4

26

2

Year 5

28

1

Total # of Volunteers

# of Volunteers in Accounting Departm

Year 1

29

2

Year 2

31

2

Year 3

23

2

Year 4

26

3

r has been selected from accounting department =PRODUCT(E12:E16) 0.05954442133771825

ng Department representation Fewer volunteers from accounting department

Table 2 Increased Accounting Volunteers

re volunteer per year

Year 5

28

2

05/Case 2 - Grievance Committee at Tuglar Corporation/BSTAT_0 5_Case3_GrievanceCommittee_Report.docx Grievance Committee at Tuglar Corporation Employees in the Accounting Department are distressed that no member of their department has served on the Tuglar Corporation grievance committee in the past five years. Each year management issues a call for volunteers to serve on the committee and randomly selects 10 respondents to serve. This report investigates whether the lack of members of the Accounting Department on the grievance committee for the past five years is suspicious.

In any year the accounting departments concern is the number of successes x (members of the accounting department who are included on the grievance committee), from the r (members of the accounting department who volunteered for the committee). This number is randomly selected from a population (N) of employees who volunteered to serve on the grievance committee. Ten (n) employees are selected randomly. The number of members of the accounting department who are selected to serve on the grievance committee is hypergeometrically distributed.

Table 1 presents the hypergeometric probabilities for the past five years when there were no accounting department members selected. These probabilities in the range .37 to .68 are reasonably high. The probability that no members of the accounting department have been selected to serve is the product of the probabilities or .0595. This number is not unreasonably low.

The probability that at least one member of the accounting department would be selected to serve on the grievance committee becomes .94 (1-.0595).

The lack of accounting department representation results from a lack of volunteers from the accounting department. Table 2 presents hypergeometric probabilities if only one more member of the department had volunteered in each f the past five years, .22 to .45. While these numbers appear to still be relatively high, they result in a product probability of .0051. Thus, the probability that at least one member of the accounting department would be selected to serve on the grievance committee becomes .9949 (1 - .0051).

Accounting department representation on the grievance committee is a result of the number of volunteers to serve. More volunteers from the accounting department would increase the probability of inclusion on the committee.

05/Case 2 - Grievance Committee at Tuglar Corporation/BSTAT_0 5_Case3_GrievanceCommittee_Rubric.docx Grievance Committee at Tuglar Corporation Case Study Chapter 5 (50 points total) RUBRIC

05/Case 2 - Grievance Committee at Tuglar Corporation/BSTAT_0 5_Case3_GrievanceCommittee_TeachingTips.docx Grievance Committee at Tuglar Corporation – Teaching Tips This case does not provide a data file. However, data provided in the case description is used in many calculations. I would suggest that students compute results in an Excel file (see BSTAT_05_Case3_GrievanceCommittee_Key.xlsx) The hypergeometric probability function is daunting. Provide and relate each variable to the arguments in the function.

The need for the product of the 5-year probabilities is not discussed in the text. Explain the need to compute this product to obtain a final probability.

05/Case 3 - McNeil's Auto Mall/BSTAT_05_Case2_McNeilsAutoMall.xlsx Sheet: Sheet1 Table 1 Poisson Cumulative Probabilities

P(x > 7) =

0.1485694020874615

x

Cumulative Probability

0.15293909038415154

0

0.0011137751478448032

0.13494625622131015

1

0.008687446153189466

0.09922518839802222

2

0.03443792757136132

0.05836775788118954

3

0.09280568545255086

0.025750481418171853

4

0.19203087385057308

0.007573671005344663

5

0.3269771300718832

0.0011137751478448032

6

0.47991622045603477

7

0.6284856225434963

8

0.7547696143178385

9

0.8501841858806749

10

0.9150660945434037

11

0.9551749108076361

12

0.9779032400240344

13

0.9897919045372273

14

0.9955663987293496

15

0.9981841694297784

0.37151437745650373

0.37151437745650373

05/Case 3 - McNeil's Auto Mall/BSTAT_05_Case2_McNeilsAutoMall_Report.docx McNeil’s Auto Mall Saturday mornings (8am – noon) are the busiest time of the week for McNeil’s Auto Mall. Ms. McNeil wants to determine how many salespeople she should have on her lot on Saturday mornings in order to achiever her goal of having two more customers on her lot than can be served at any time. Realizing that occasionally the number of customers on her lot will exceed the number of salespersons by more than two, she is willing to accept such an occurrence no more than 10% of the time. This report examines probabilities associated with these specifications and determines the minimum number of salespeople that should be on the lot.

The arrival of customers on Saturday morning is random at a constant rate with a mean of 6.8 customers per hour. The probability of an arrival, then, is the same for any two intervals of equal length and arrival or non-arrival of customers is independent from arrival or non-arrival in other intervals. The number of customers who arrive on the lot on Saturday morning is Poisson distributed.

Ms. McNeil currently uses five salespeople on her lot on Saturday morning. The probability that the number of customers will exceed 7 (the more than two objective) is .3715. This probability does not satisfy her stated objective of no more than 10% (.3715 > .10).

In order to determine the minimum number of salespeople Ms. McNeil should have on her lot on Saturday mornings to achieve her objective that the number of customers on her lot exceed the number the number of salespersons by more than two no more than 10% of the time, a Poisson table of cumulative probabilities was created. Table 1 presents the results of these calculations.

Ten customers in an hour is the smallest value of x for which the cumulative probability exceeds .90. At x = 10 the number of arrivals is less than 10%. Therefore, in order to ensure that the number of customers exceeds the number of salespersons by more than two no more than 10% of the time, Ms. McNeil should employee eight salespeople on her lot on Saturday mornings.

05/Case 3 - McNeil's Auto Mall/BSTAT_05_Case2_McNeilsAutoMall_Rubric.docx McNeil’s Auto Mall Case Study Chapter 5 (points total) RUBRIC

05/Case 3 - McNeil's Auto Mall/BSTAT_05_Case2_McNeilsAutoMall_TeachingTips.docx McNeil’s Auto Mall – Teaching Tips This case does not provide a data file. However, data provided in the case description is used in many calculations. I would suggest that students compute results in an Excel file (see BSTAT_05_Case2_McNeilsAutoMall_Key.xlsx) Many students may still have and prefer to work with calculators. Suggest trying the use of the POISSON.DIST function. Cumulative probabilities must after x = 0 be the sum of the previous probability plus the current.

06/Case 1 - Specialty Toys/BSTAT_06_Caase1_SpecialtyToys_Rubric.docx Specialty Toys Case Study Chapter 4 (80 points total) RUBRIC

06/Case 1 - Specialty Toys/BSTAT_06_Case1_SpecialToys_Key.xlsx Sheet: Sheet1

z=

1.9599639845400536

σ=

5102.134569246541

The normal distribution is described as µ = 20000 σ = 5102

z value

Probability of a stockout

15000

-0.9800078400627205

0.8364588756667601

18000

-0.3920031360250882

0.652472051667122

24000

0.7840062720501764

0.21651821502699364

28000

1.5680125441003527

0.05843910244110373

=1-NORM.S.DIST(E26

06/Case 1 - Specialty Toys/BSTAT_06_Case1_SpecialtyToys_Report.docx Specialty Toys Specialty Toys, Inc. sells a variety of new and innovative children’s toys. Specialty plans to introduce a new product called Weather Teddy this coming season in time for the holiday market and has chosen an October market entry date. Facing the decision of how many Weather Teddy units to order, member of the management team suggest the following order quantities: 15,000, 18,000, 24,000, or 28,000. This report provides an analysis of the stock-out probabilities for various order quantities, an estimate of the profit potential, and an order quantity recommendation.

Specialty expects to sell Weather Teddy for $24 based on a unit cost of $16. Surplus inventory will be sold for $5 per unit. Specialty’s senior sales forecaster predicted an expected demand of 20,000 units with a .95 probability that demand would be between 10,000 units and 30,000 units. Figure 1 presents the normal distribution as predicted by the sales forecaster.

Probability of a Stock Out for Suggested Order Quantities Table 1 presents the probability of a stockout for suggested quantities. The minimum probability of a stockout is 5.8% at an order quantity of 24,000. At the minimum order of 15,000 there is a 63% change of a probability stockout. An order quantity decision must also take into account profit/loss possibilities at the suggested quantities.

Profit Projections for Order Quantities Project projections for suggested order quantities are presented in Table 2. If unit sales are 10,000, the optimal order quantity is 15,000. At all other quantities (18,000, 24,000, and 28,000) for a profit of 25,000. At all other order quantities when there are only 10,000-unit sales, the company experiences a loss.

When unit sales are 20,000 the optimal order quantity among quantities considered is 18,000. Profit of 144,000 is greatest at this order quantity. At 24,000 and 28,000 profits are 116,000 and 72,000, respectively.

When unit sales are 30,000, the largest profit is at an order quantity of 28,000. At a 28,000-order quantity there is also the potential for the largest loss in profits if unit sales are as low as 10,000 (118,000).

Profit Potential at 70% Market Demand 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, thus only a 30% chance of any stock-outs. Figure 2 presents a graph of the normal distribution describing this scenario. As indicated in the figure the quantity associated with a 70% chance of meeting demand is 22,653. Project profit under the three sales scenarios (10,000, 20,000, and 30,000 sales) are presented in Table 3. At an order quantity of 22,653, the highest profit is realized if 30,000 units are sold (181,224). More realistically, a 20,000-unit sales results in a profit of 130,817 with excess units (2,653) sold at $5 after the season.

Order Quantity Recommendation Based on this analysis, the suggested order quantity is between 18,000 and 20,000. At 18,000 production, the worst-case scenario of 10,000 sales results in a loss of 8000, but a larger profit margin of 144,000 if all 20,000 are sold. Production units of 24,000 yield only a profit of 116,000 if all units are sold. Quantities produced at 70% of market demand (22,653) yield only 130,817 dollars in profit if 20,000 units are sold.

In this report an analysis of the stock-out probabilities for various order quantities, an estimate of the profit potential, and an order quantity recommendation. Analysis results indicate an optimal order quantity of between 18,000 and 20,000 Weather Teddy’s.

06/Case 1 - Specialty Toys/BSTAT_06_Case1_SpecialtyToys_TeachingTips.docx Specialty Toys – Teaching Tips This case does not provide a data file. However, data provided in the case description is used in many calculations. I would suggest that students compute results and build tables/figures for the report in an Excel file (see BSTAT_06_Case1_SpecialtyToys_Key.xlsx) Use online pictures to find a simple image of a normal distribution along with Excel shapes and text boxes to create the normal distribution figures required. The Snipping Tool app can then be used to capture the entire figure as a single unit.

If not already introduced this would be a good place to provide additional examples of the use of NORM.S.INV() to compute a z value at a given alpha and NORM.S.DIST() to compute the area under a curve.

NORM.S.DIST(1.57,TRUE) finds the cumulative area under the curve from left to right. Students often need help understanding that if they are interested in describing the area under the curve to the right of the indicated z-value, the formula takes the form 1-NORM.S.DIST(1.57,TRUE).

06/Case 3 - Gebhardt Electronics/BSTAT_06_Case2_GebhardtElectronics_Report.docx Gebhardt Electronics Gebhardt Electronics produces a wide variety of transformers that it sells directly to manufacturers of electronics equipment. For one component in development of these transformers, Gebhardt wants to use primarily flawless lengths of wire in making this component. Gebhardt is considering a new supplier for copper wire. This report considers the supplier’s claims in order to determine if these claims meet Gebhardt’s standards requirements.

Probability of meeting Gebhardt’s standard with the new supplier.

The variable of consideration is the distance in inches between consecutive flaws on a provided length of copper wire. The new supplier provides spools of wire that average 50 inches between flaws. This data represents an exponential distribution which models the time between events (in this case flaws in the wire). The probability the distance in inches between consecutive flaws on a length of wire is less than or equal to 36 (the 3-foot length) is .5133. Thus, the probability the distance in inches between consecutive flaws on a length of copper wire exceeds 36 is .4867 (1 .5133). Gebhardt is only willing to accept a probability of .05 (1 in 20 inches). Gebhardt’s criteria will not be met by the new supplier.

Mean number of inches that meet Gebhardt’s 1/20 (.05) requirement In order to determine the mean number of inches between consecutive flaws that will meet Gebhardt’s criteria, values of the mean starting with 50 are computed until the probability of .05 is met. Gebhardt’s criteria are not met until the mean number of inches between consecutive flaws is 702. (1-P(x ≤ 36)).

Mean number of inches to accomplish 1/100 (.01) requirement Suppose Gebhardt would like to accomplish a 1 in 100 chance of finding no flaws in consecutive lengths of wire. Continuing past 702 inches to compute probabilities until the .01 requirement is met results in a minimum length of 3582 inches (298.5 feet of wire).

Discussion The new supplier does not meet Gebhardt’s quality standards.

06/Case 3 - Gebhardt Electronics/BSTAT_06_Case2_GebhardtElectronics_Rubric.docx Gebhardt Electronics Study Chapter 4 (60 points total) RUBRIC

06/Case 3 - Gebhardt Electronics/BSTAT_06_Case2_GebhardtEle ctronics_TeachingTips.docx Gebhardt Electronics – Teaching Tips This case does not provide a data file. However, data provided in the case description is used in many calculations. Excel makes the many calculations trivial. (see BSTAT_06_Case2_GebhardtElectronics_Key.xlsx) Use EXPON.DIST() and the file handle. Students may need to be reminded that they are looking for a value <= .95 for the 1/50 requirement. Establish the counting pattern in column x. Use EXPON.DIST with relative cell reference for f(x). Remind students to use the fill handle to drag the 3 cells formula down until they arrive at the desired value To satisfy 1/100 requirement, worksheet will grow to over 3500 rows.

06/Case 3 - Gebhardt Electronics/BSTAT_06_Case2_Gebhardt_Key.xlsx Sheet: Sheet1

chance that length is flawless

inches between flaws

x = distance between flaws - 50 inches

0.5132477440400283

=EXPON.DIST(36,1/50,TRUE)

Searching for probability <= .05

x

f(x)

1- f(x)

50

0.5132477440400283

0.48675225595997174

51

0.5063272116108697

0.4936727883891303

52

0.49958007963942996

0.50041992036057

53

0.49300051086709495

0.506999489132905

54

0.486582880967408

0.513417119032592

55

0.4803217737150016

0.5196782262849984

56

0.4742119755742201

0.5257880244257799

57

0.46824846986942936

0.5317515301305706

58

0.4624264306688763

0.5375735693311237

59

0.4567412164890582

0.5432587835109418

60

0.4511883639059735

0.5488116360940265

61

0.4457635811426331

0.554236418857367

62

0.44046274168816185

0.5595372583118381

63

0.43528187799224083

0.5647181220077592

64

0.430217175269077

0.569782824730923

65

0.42526496543723546

0.5747350345627645

66

0.42042172121519045

0.5795782787848096

67

0.41568405038716016

0.5843159496128398

68

0.41104869024944657

0.5889513097505534

69

0.40651250224397206

0.5934874977560279

4. using a 1/100 chance continu

07/Case 1 - Marion Daires/BSTAT_07_Case_MarionDairies.xlsx Sheet: Sheet2

σxbar =

Standard error of the mean

σ / SQRT(n)

n=

50

xbar =

80

z=

σ=

25

Area under the curve =

n=

150

xbar =

80

z=

σ=

25

Area under the curve =

3.5355339059327373

-1.41

0.079269

2.041241452319315 -2.45

0.007142

07/Case 1 - Marion Daires/BSTAT_07_Case_MarionDairies_Report.docx Marion Dairies Marion Dairies entered the yogurt market last year with a blueberry flavored yogurt called Blugurt.

Consumer ratings of the product have a mean of 80 and a standard deviation of 25 on a 100-point scale. Past experience indicates that a consumer who rates a product with a score of greater than 75 on this scale will consider purchasing the product.

Marion’s marketing department is pressing to extend the product line with the introduction of strawberry-flavored yogurt call Strawgurt. The goal is to increase Marion’s market share rather than cannibalize existing sales of Blugurt. A simple random sample of 50 customers will be asked to rate the two yogurts on the 100-point scale to assess whether Strawgurt will appeal to potential customers who do not like Blugurt.

This report summarizes probabilities associated with this sample and considers an increase in sample size.

Probability that the mean score of Blugurt will be 75 or less We know that the expected value of the sample mean is the population mean, thus the mean of all possible values is 80. With a sample size of 50, the standard error of the mean is 3.5355 (25/√50).

The sampling distribution of the mean can be approximated by a normal distribution. With the sample mean equal to 75, the z score is -1.41 ((75-80)/3.5355). As indicated in Figure 1, the area under the curve to the left of z = -1.41 is .0793. Thus, the probability the mean score of Blugurt given by a random sample of 50 Marion Dairies customers will be 75 or less is .0793.

Probability that the mean score of Blugurt will be 75 or less is sample size is 150 We know that the expected value of the sample mean is the population mean, thus the mean of all possible values is 80. With a sample size of 150, the standard error of the mean is 2.0412 (25/√150). The sampling distribution of the mean can be approximated by a normal distribution.

With the sample mean equal to 75, the z score is -2.45 ((75-80)/2.0412). The area under the curve to the left of z = -2.45 is .0071. Thus, the probability the mean score of Blugurt given by a random sample of 150 Marion Dairies customers will be 75 or less is .0071.

Discussion Whenever the sample size is increased, the standard error of the sample mean decreases. Since the sampling distribution of the sample mean at n = 150 has a smaller standard error, the potential values of the sample mean that the sample can yield have less variation and tend to be closer to the population mean than the potential values of the sample mean when n = 50. A smaller standard error decreases the likelihood that a sample will have a mean Blugurt score of 75 or lower.

07/Case 1 - Marion Daires/BSTAT_07_Case_MarionDairies_Rubric.docx Marion Dairies Case Study Chapter 7 (65 points total) RUBRIC

07/Case 1 - Marion Daires/BSTAT_07_Case_MarionDairies_TeachingTips.docx Marion Dairies – Teaching Tips This case does not provide a data file. However, data provided in the case description is used in many calculations. I would suggest that students compute results and build tables/figures for the report in an Excel file (see BSTAT_07_Case_MarionDairies_Key.xlsx) Use online pictures to find a simple image of a normal distribution along with Excel shapes and text boxes to create the normal distribution figures required. The Snipping Tool app can then be used to capture the entire figure as a single unit.

If not already introduced this would be a good place to provide additional examples of the use of NORM.S.INV() to compute a z value at a given alpha and NORM.S.DIST() to compute the area under a curve.

NORM.S.DIST(1.57,TRUE) finds the cumulative area under the curve from left to right. Students often need help understanding that if they are interested in describing the area under the curve to the right of the indicated z-value, the formula takes the form 1-NORM.S.DIST(1.57,TRUE).

08/Case 1 - Young Professionals Magazine/BSTAT_08_Case1_Yo ungProfessionalMagazine_Report.docx Young Professional Magazine Young Professional magazine was developed for a target audience of recent college graduates in their first 10 years in a business/professional career. The publisher is interested in expanding the magazine’s advertising base. This report summarizes findings from a survey commissioned to develop a profile of its subscribers.

Descriptive Statistics – Quantitative Data Table 1 presents descriptive statistics for audience age. Average age of readers is 30. The youngest reader is 19 and the oldest reader 42. The 410 responses are may be considered normally distributed.

Table 1 presents descriptive statistics for value of investments. The average of investments made is $28,538. Minimum investment is zero. Maximum investment is 133,400. The distribution of investment data may be considered normally distributed.

Table 3 presents descriptive statistics for the number of transactions. The average number of transactions is 6. Minimum number of transactions is zero. Maximum number of investments by a respondent is 21. The distribution of number of transactions data may be considered normally distributed.

Table 4 presents descriptive statistics for household income. The average household income is $74,460. Minimum household income is 16,200. Maximum household income by a respondent is 322,500. The skewness value (2.01) indicates that the distribution may not be considered normally distributed. The distribution is skewed slightly right.

Descriptive Statistics – Qualitative Data Fifty-six percent (56%) of respondents are males. Fifty-six percent (56%) of respondents have made no real estate purchases. Sixty-two percent (62%) of respondents have broadband access.

Fifty-three percent (53%) of respondents have children.

Confidence Interval The 95% confidence interval calculation for age indicates that with 95% confidence the mean age is between 29.7 and 30.5.

The 95% confidence interval calculation for household income indicates that with 95% confidence the mean household income is between 71,079 and 77,840 dollars.

The 95% confidence interval for the proportion of subscribers who have broadband access indicates that with 95% confidence the proportion of subscribers who have broadband access is between .58 and .67.

The 95% confidence interval for the proportion of subscribers who have children indicates that with 95% confidence the proportion of subscribers who have children is between .44 and .53.

Discussion The average age of respondents (30) indicates that the magazine is reaching its’ goal of attracting young professionals within 10 years of college graduation. One hundred fifty-four (154) respondents are older than 31; 38%, indicating perhaps an opportunity to target older professionals.

Fifty-six percent (56%) of respondents have not made real estate purchases. An opportunity exists to target 1st time real estate purchasers. In addition, 53% of respondents have children; an opportunity for single-family home purchases.

Fifty-six percent (56%) of respondents are male and 62% have broadband access.

This magazine would be a good outlet for online brokers to advertise and provide a link for broadband access users. This advertising should attract respondents who have not made a real estate purchase (56%) as well as single-family home advertising for families with children.

Over half of respondents have children; with an average age of 30, possible young children. Men (56%) and women professionals would likely be interested in educational software and computer games for young children.

Articles of interest to readers may include articles for first time investors and investors considering large investments (than 45,000). Real estate articles for first time buyers and buyers of single-family homes may be of interest. Additionally, articles about childcare options for young professionals and the value of available educational learning tools for children may be of interest as well.

This report presents results based on a survey of recent college graduates in a business/professional career for purposes of the magazines’ intent to expand its reader base.

08/Case 1 - Young Professionals Magazine/BSTAT_08_Case1_Yo ungProfessionalMagazine_Rubric.docx Young Professional Magazine Chapter 8 (90 points total) RUBRIC

08/Case 1 - Young Professionals Magazine/BSTAT_08_Case1_Yo ungProfessionalMagazine_TeachingTips.docx Young Professional Magazine – Teaching Tips Descriptive Statistics Make sure students use the Data Analysis Toolkit and run the Descriptive Statistics tool. There is no default automatically computed. If the student fails to choose one of the available check boxes, the following dialog box appears.

Choosing the confidence level for the mean provides the margin of error used to compute a confidence interval when sigma is unknown.

95% Confidence Intervals Remind students to save computation time with the Descriptive Statistics – Confidence Interval option when sigma is unknown. Students should avoid use of the terms “the household income is between A and B 95% of the time”. Instead use “we are 95% confident that the income is between A and B” To obtain a positive value for z used to compute the margin of error, use 1-.05 (for alpha = .05) =NORM.S.INV(1-0.05/2)

sactions

08/Case 1 - Young Professionals Magazine/BSTAT_CH8_Case1_Professional_Key.xlsx Sheet: Data Broadband Access?

Household Income ($)

Have Children?

Yes

75200

Yes

Yes

70300

Yes

Yes

48200

No

No

95300

No

No

73300

Yes

Yes

123400

Yes

Yes

73900

Yes

Yes

54300

Yes

Yes

93100

No

Yes

60100

No

No

48600

No

No

43500

No

Yes

73600

Yes

Yes

68200

Yes

Mean

30.1121951219

No

61900

Yes

Standard Error

0.19873232672

No

57600

No

Median

30

No

82300

No

Mode

32

No

64600

Yes

Standard Deviation

4.02402291877

Yes

61100

Yes

Sample Variance

16.1927604508

Yes

31200

Yes

Kurtosis

0.02271196040

Yes

92600

No

Skewness

-0.0255549824

Yes

68300

No

Range

23

Yes

35100

No

Minimum

19

Yes

85700

No

Maximum

42

Yes

140300

Yes

Sum

12346

Yes

108200

Yes

Count

410

No

61100

No

No

33900

Yes

Yes

54400

Yes

Yes

61200

No

Mean

28538.2926829

No

58000

Yes

Standard Error

780.841273518

.

Descriptive Statistics - Quantitative Data

Table 1 Age

Table 2 Value of Investments ($)

Sheet: Confidence Interval Estimates Household Income ($)

30.11219512195122

Mean

74459.5121951

0.19873232672355037

Standard Error

1719.54885914

30

Median

66050

32

Mode

60300

4.024022918775681

Standard Deviation

34818.2106717

16.19276045083195

Sample Variance

1212307794.38

0.022711960403613674

Kurtosis

7.49563383920

-0.025554982485954107

Skewness

2.00790048535

23

Range

306300

19

Minimum

16200

42

Maximum

322500

12346

Sum

30528400

410

Count

410

0.3906642456534155

Confidence Level(95.0%)

3380.25659436

22

+-

0.3906642456534155

805

to

30.502859367604636

95

+-

3380.2565943685363

41

to

77839.76878949048

n

95.0%)

nterval Age

ce Level

t the mean age is between 29.7 and 30.5

nterval Household Income

ce Level

t the household income is between 71,079.25 and 77,839.77

nterval for proportion of subscribers who have broadband 410

/broadband =

256

Sheet: Investments Value of Investments ($)

Min =

0

Bins

Bins

Frequency

12200

Max =

133400

15000

15000

58

12400

# of Bins =

10

30000

30000

208

26800

Bin width=

15000

45000

45000

89

19600

60000

60000

34

15100

75000

75000

17

39700

90000

90000

2

21900

105000

105000

1

41900

120000

120000

0

16100

135000

135000

1

18400

150000

150000

0

More

0

33800 15500 17300 47900 28200 19400 31000 21300 21300 21300 34100 32600 0 20800 23100 39800 17900 33300 21800 54000 34800

08/Case 2 - Gulf Real Estate Properties/BSTAT_08_Case2_GulfRe alEstateProperties_Report.docx Gulf Real Estate Properties Gulf Real Estate Properties, Inc., a real estate firm located in southwest Florida monitors condominium sales of Gulf View and No Gulf View properties. Gulf View properties are located directly on the Gulf of Mexico. This report summarizes recently collected sales data for 40 Gulf View condominiums and 18 No Gulf View condominiums.

Descriptive Statistics Tables 1 and 2 present descriptive statistics for Gulf View and No Gulf View. As anticipated, Gulf View condominiums are the most expensive. A Gulf View condominium lists for a mean price of $474,000 and a median price of $437,000. A No Gulf View condominium lists for a mean price of $212,800 and a median price of $212,500. Gulf View condominiums are more than twice as expensive as No Gulf View condominiums.

The standard deviation of the list price for Gulf View condominiums is $197,300 and the standard deviation of the list price for No Gulf View condominiums is $48,900. The Gulf View condominiums have a greater variation in price.

The most expensive list price is for a Gulf View condominium at $975,000 and the least expensive list price is a No Gulf View condominium at $148,000. The highest four Gulf View list prices are $885,000, $895,000, and two at $975,000. Ten percent (10%) of the Gulf View condominiums have an unusually high list price.

The mean number of days to sell a condominium is slightly better for Gulf View condominiums than No Gulf View condominiums (106 days vs 135 days). On average, it takes a little over three months to sell a Gulf View condominium and a little over four months to sell a No Gulf View condominium.

The quickest sale was for a Gulf View condominium that old in less than a month (28 days). The slowest sale was for a No Gulf View condominium that took 338 days to sell.

Gulf View condominiums with a mean list price of 474,000 and a mean sale price of 454,200 sell on average 4.2% below list price. No Gulf View condominiums with a mean list price of 212,800 and a mean sale price of 203,300 appears to sell on average 4.5% below list price. List and sale prices for the two types of condominiums present only a small percentage difference.

In summary, Gulf View condos are substantially more expensive and tend to sell slightly faster. The discount off the list price is 4.0 to 4.5% for both.

Confidence Intervals Table 3 presents confidence intervals for sale price and number of days to sell for Gulf View condominiums. With 95% confidence, we can expect Gulf View condominiums to sell for between $392,700 and $515,800. The mean days to sell a Gulf View condominium is between 89 and 123 days.

Table 4 presents confidence intervals for sale price and number of days to sell for No Gulf View condominiums. With 95% confidence, we can expect No Gulf View condominiums to sell for between $181,400 and $225,000. The mean days to sell a No Gulf View condominium is between 97 and 173 days.

Sample Sizes The branch manager requested estimates of the mean selling price of Gulf View condominiums with a margin of error of $40,000. At a 95% confidence lever, this estimate would require a sample

size of 89 Gulf view condominiums. Additionally, the branch manager requested estimates of the mean selling price of No Gulf View condominiums with a margin of error of $15,000. At a 95% confidence level, the recommended sample size is 33.

Estimating a Selling Price Gulf Real Estate Properties just signed contracts for two new listings: a Gulf View condominium with a list price of $589,000 and a No Gulf View condominium with a list price of $285,000. We have been asked to estimate the final selling price and number of days required to sell each of these units. Table 5 presents these results. The estimated selling price for the Gulf View condominium is $564,415 and is expected to sell in about 106 days. The estimated selling price for the No Gulf View condominium is $272,121 and is expected to sell in about 106 days.

This report presents statistical results designed to help a real estate agent understand the condominium market on the Southwest Florida Gulf coast.

08/Case 2 - Gulf Real Estate Properties/BSTAT_08_Case2_GulfRe alEstateProperties_Rubric.docx Gulf Real Estate Properties Chapter 8 (115 points total) RUBRIC

08/Case 2 - Gulf Real Estate Properties/BSTAT_08_Case2_GulfRe alEstateProperties_TeachingTips.docx Gulf Real Estate Properties – Teaching Tips Descriptive Statistics Make sure students use the Data Analysis Toolkit and run the Descriptive Statistics tool. There is no default automatically computed. If the student fails to choose one of the available check boxes, the following dialog box appears.

Choosing the confidence level for the mean provides the margin of error used to compute a confidence interval when sigma is unknown.

95% Confidence Intervals Remind students to save computation time with the Descriptive Statistics – Confidence Interval option when sigma is unknown. Students should avoid use of the terms “the mean sales price is between A and B 95% of the time”. Instead use “we are 95% confident that the mean selling price is between A and B” Sample Size To obtain a positive value for z used to compute the margin of error, use 1-.05 (for alpha = .05) =NORM.S.INV(1-0.05/2) Planning value for sigma is the sample standard deviation (see page 395 for additional examples of planning values)

08/Case 2 - Gulf Real Estate Properties/BSTAT_CH8_Case2_GulfProp_Key.xlsx Sheet: Data Sale Price

Days to Sell

Mean

454.2225

Mean

106

Standard Error

30.439729539782824

Standard Error

8.25607779231885

Median

417.5

Median

96

Mode

305

Mode

85

Standard Deviation

192.51775341044547

Standard Deviation

52.21602072652435

Sample Variance

37063.08537820509

Sample Variance

2726.5128205128203

Kurtosis

1.183621479174095

Kurtosis

2.0220258704051735

Skewness

1.1593879144156714

Skewness

1.0776421228790343

Range

810

Range

254

Minimum

165

Minimum

28

Maximum

975

Maximum

282

Sum

18168.9

Sum

4240

Count

40

Count

40

Confidence Level(95.0%)

61.57016454849345

Confidence Level(95.0%)

16.69949358564047

Sale Price

Days to Sell

Mean

203.1888888888889

Mean

135

Standard Error

10.345378460111261

Standard Error

17.984016142155433

Median

203.5

Median

126

Mode

179

Mode

#N/A

Standard Deviation

43.89172357851549

Standard Deviation

76.29971860251864

Sample Variance

1926.4833986928124

Sample Variance

5821.64705882353

Kurtosis

-0.4640682712774109

Kurtosis

2.21530122383285

Skewness

0.30418060957270504

Skewness

1.3605469573578461

Range

157

Range

290

Minimum

135.5

Minimum

48

Maximum

292.5

Maximum

338

6. Sample sizes - Gulf V

Sheet: Tables for Report Table 2 Descriptive Statistics - No Gulf View Condominiums Sale Price

Days to Sell

454.2225

106

Mean

212.8055555

30.439729539782824

8.25607779231885

Standard Error

11.53651342

417.5

96

Median

212.5

305

85

Mode

279

910267243

192.51775341044547

52.21602072652435

Standard Deviation

48.94528123

58333334

37063.08537820509

2726.5128205128203

Sample Variance

2395.640555

068212183

1.183621479174095

2.0220258704051735

Kurtosis

-0.08523964

39309995

1.1593879144156714

1.0776421228790343

Skewness

0.543955061

810

254

Range

174

165

28

Minimum

148

975

282

Maximum

322

18168.9

4240

Sum

3830.500000

40

40

Count

18

61.57016454849345

16.69949358564047

Confidence Level(95.0%)

24.33991573

580267522

35907718

List Price

08/Case 3 - Metropolitan Research/BSTAT_08_Case3_MetropolitanResearch_Report.docx Metropolitan Research, Inc. Metropolitan Research, Inc., a consumer research organization, conducted a research study to evaluate consumer satisfaction with the performance of automobiles produced by a major Detroit manufacturer. To learn more about transmission failures, Metropolitan examined a sample of 50 vehicles repaired for transmission failure. This report summarizes findings for these vehicles at the time of transmission failure.

Descriptive Statistics – Miles Driven Table 1 presents descriptive statistics for miles driven until transmission failure. The mean miles driven is 73,340 and median is 72,705. The first quartile of 59,881 indicates that 25% of the repairs occurred with less than 60,000 miles on the vehicle. In Figure 1, the histogram indicates that eight (16 %) of repairs occurred with less than 45,000 miles. These data support a conclusion of early transmission failures for this automobile manufacturer.

Confidence Interval The 95% confidence interval for the population mean of miles driven before transmission failure is 66,264 to 80,416 indicating that with 95% confidence the interval contains the population mean.

Sample Size Consideration The research firm wants the population of mean number of miles driven until transmission failure to be estimated with a margin of error of 5000 miles. This goal can be accomplished with a larger sample size. The necessary sample size in this case is 96 cars with transmission failure.

Discussion To more fully evaluate the transmission problem, additional information is necessary not limited to but including: Transmission failure data for other automobile manufacturers The proportion of all automobiles that experience transmission failures Industry standards for transmission failures With this additional information comparative statements could be made about how the manufacturer in question compares to other manufacturers as well as to industry standards.

This report presents results based on a of sample of vehicles with transmission repairs. Using the data available only, the manufacturer in question appears to have problems with early transmission failures.

08/Case 3 - Metropolitan Research/BSTAT_08_Case3_MetropolitanResearch_Rubric.docx Metropolitan Research Chapter 8 ( 75 points total) RUBRIC

08/Case 3 - Metropolitan Research/BSTAT_08_Case3_Metropolita nResearch_TeachingTips.docx Metropolitan Research – Teaching Tips Descriptive Statistics Make sure students use the Data Analysis Toolkit and run the Descriptive Statistics tool. There is no default automatically computed. If the student fails to choose one of the available check boxes, the following dialog box appears.

Choosing the confidence level for the mean in the Descriptive Statistics dialogue box provides the margin of error used to compute a confidence interval when sigma is unknown.

Because histograms and frequency distributions were discussed in an earlier Chapter (chapter 2), suggest reminding students that these tools can be used to further describe transmission failure incidence as part of discussion 95% Confidence Intervals Remind students to save computation time with the Descriptive Statistics – Confidence Interval option when sigma is unknown. Students should avoid use of the terms “the mean miles driven is between A and B 95% of the time”. Instead use “we are 95% confident that the mean miles driven is between A and B” Sample Size To obtain a positive value for z used to compute the margin of error, use 1-.05 (for alpha = .05) =NORM.S.INV(1-0.05/2) Planning value for sigma is the sample standard deviation (see page 395 for additional examples of planning values)

08/Case 3 - Metropolitan Research/BSTAT_CH8_Case3_Auto_Key.xlsx Sheet: Data 1st Quartile

59880.75

73340.3 3521.210059201298 72705 #N/A

ion

24898.715108435223

e

619946014.0510204 0.1671262115990304 0.2601179743207555 113048 25066 138114 3667015

Sample Size

50

el(95.0%)

7076.13653970131

z= Planning value

of Error

9869

E= +-

7076.13653970131

to

80416.43653970132

n= Sample size is

hat the mean number of miles to transmission failure is and 80,416.

Sheet: Histogram Bins

Frequency

Cumulative %

25000

0

0

35000

4

0.08

45000

4

0.16

55000

2

0.2

65000

8

0.36

75000

10

0.56

85000

5

0.66

95000

10

0.86

105000

2

0.9

115000

0

0.9

125000

4

0.98

135000

0

0.98

More

1

1

09/Case 1 - Quality Associates, Inc_/BSTAT_09_Case1_QualityAssociates_Report.docx Quality Associates, Inc. Quality Associates, Inc. advises its clients about sampling and statistical procedures that can be used to control their manufacturing processes. For a recent client’s sample of 800 observations, a population standard deviation was assumed to be .21 when the process was operating satisfactorily. This report summarizes findings of analysis of four random samples (n = 30) of the process to see if the process was operating satisfactorily. The hypothesis test suggested by Quality Associates is: H0: µ = 12 Ha: µ ≠ 12 Hypothesis Tests Statistical analysis was completed on each of four samples at a significance level of .01 (α = .01).

For sample 1, with a mean of 11.96, the p-value was computed as .28. Since the p-value is not less than the significance level, the null hypothesis cannot be rejected and the process may be assumed to be operating satisfactorily.

For sample 2, with a mean of 12.03, the p-value was computed as .45. Since the p-value is not less than the significance level, the null hypothesis cannot be rejected and the process may be assumed to be operating satisfactorily.

For sample 3, with a mean of 11.89, the p-value was computed as .003. Since the p-value is less than the significance level, the null hypothesis is rejected and corrective action should be taken.

For sample 4, with a mean of 12.08, the p-value was computed as .28. Since the p-value is not less than the significance level, the null hypothesis cannot be rejected and the process may be assumed to be operating satisfactorily.

Only sample 3 leads to the rejection of the hypothesis H0: µ = 12. Corrective action is warranted for sample 3. For all other samples, the process appears to be operating satisfactorily.

Sample Standard Deviations Sample standard deviations are in the .206 to .220 range (.220, .220, .207, .206, respectively). The population standard deviation assumption of .21 is good.

Control Limits Control limits can be established for this process by computing a confidence interval at a significance level of .01 using a mean of 12 as established. The confidence interval appears in Figure 1. Thus, the upper control limit for the process is 12.10 and the lower control limit is 11.90.

As long as the sample mean is between these two limits, the process is in control and no corrective action is required. Sample 3 with a mean of 11.89 revealed that corrective action is necessary.

Discussion – Implications of a Larger Significance Level Increasing the level of significance will cause the null hypothesis to be rejected more often. This would lead to quicker corrective action when the process is out of control, but also indicates a higher probability of stopping the process and attempting corrective action when the process is operating satisfactorily,

This report presents results of hypothesis tests conducted by Quality Associates, Inc. to establish process control limits for a process with a mean value of 12 at a .01 level of significance.

09/Case 1 - Quality Associates, Inc_/BSTAT_09_Case1_QualityAssociates_Rubric.docx Quality Associates Case Chapter 9 (70 points total) RUBRIC

09/Case 1 - Quality Associates, Inc_/BSTAT_09_Case1_QualityAssociates_TeachingTips.docx Quality Associates, Inc. – Teaching Tips Hypothesis Testing The authors offer a template for computing p-values: It may help to reinforce the location of these p-value calculations with normal distribution figures. It may be helpful as well to indicate to students that the NORM.S.DIST() function works from left to right on the normal distribution figure. Thus, an upper tail test requires the form 1 – NORM.S.DIST() Lower tail test (NORM.S.DIST(x, True) Upper tail test (1 – NORM.S.DIST(x, True)) Two-tail test (2 * the smallest value calculated value) Control Limits It may help to understand the control limits by illustrating the Sample 3 data.

999998

09/Case 1 - Quality Associates, Inc_/BSTAT_CH9_Case1_Quality_Key.xlsx Sheet: Data n 30 xbar 12.081333333333337 Sample 4 12.02 12.02 12.05 12.18 12.11 12.07 12.05 11.64 12.39

1. Hypothesis Test

11.65

Sample 1

Sample 2

Sample 3

12.11

α = .01

α = .01

α = .01

11.9

σ = .21

σ = .21

σ = .21

12.22

z=

-1.0780570973116832

z=

0.7476847610388779

z=

-2.8

11.88

p-value =

0.14050413807871312

p-value =

0.7726748374570538

p-value =

0.00

12.03

0.8594958619212869

0.2273251625429462

0.99

12.35

0.28100827615742624

0.4546503250858924

0.00

12.09

DO NOT

DO NOT

REJ

11.77

REJECT

REJECT

12.2 11.79 12.3 12.27 12.29 12.47 12.03 12.17 11.94

3. Compute limits

09/Case 2 - Bayview University/BSTAT_09_Case2_Bayview_Report.docx Ethical Behavior of Students at Bayview University The Dean of the College of Business at Bayview University has been concerned with cheating among business students for many years. The Chronicle of Higher Education reported that 56% of business students admitted to cheating at some time during their academic career as compared to 47% of nonbusiness students. The dean commissioned a study to assess the current ethical behavior of business students at Bayview. An anonymous exit survey was administered to a sample of 90 graduating business students. Three types of cheating were considered: copying from the Internet, copying answers from another student’s exam, and collaborating with others on individual projects. This report summarizes an assessment of cheating by business student at Bayview University.

Descriptive Statistics Table 1 presents descriptive statistics for all business students. These results show that 16 of the 90 business students (17.78%) presented work copied from the Internet as their own; 18 of the 90 (20%) copied answers off another student’s exam; and 29 of the 90 (32.22%) collaborated with other students on projects that were supposed to be completed individually. Overall, 37 of the 90 (41.11%) were involved in some type of cheating.

Table 2 presents descriptive statistics for male students. For the 48 male students, 14.58% presented work copied off the Internet as their own; 18.75% copied answers off another student’s exam; and 37.50% collaborated with other students on projects that were supposed to be completed individually. Overall, 21 of the 48 male students (43.75%) were involved in some type of cheating.

Table 3 presents descriptive statistics for female students. For the 42 female students, 21.43% presented work copied off the Internet as their own; 21.43% copied answers off another student’s exam; and 26.19% collaborated with other students on projects that were supposed to be completed individually. Overall, 16 of the 42 female students (38.10%) were involved in some type of cheating.

The proportion of students involved in some type of cheating is higher for male students (43.75%) than for female students (38.10%). The most common form of cheating involved collaborating on projects that were supposed to be completed individually. Approximately 32% of all business students were involved in this type of cheating. The proportion of students that presented work copied from the Internet as their own is higher for female students (21.43%) than for male students (14.58%).

95% Confidence Intervals Table 4 presents a confidence interval at a 95% level of confidence for the proportion of all students who participate in some form of cheating. This interval indicates that the population proportion of all students who cheat is between .31 and .51.

Table 5 presents a confidence interval at a 95% level of confidence for the proportion of male students who participate in some form of cheating. This interval indicates that the population proportion of male business students who cheat is between .297 and .578.

Table 6 presents a confidence interval at a 95% level of confidence for the proportion of female students who participate in some form of cheating. This interval indicates that the population proportion of female business students who cheat is between .23 and .53.

Hypothesis Tests

The Chronicle of Higher Education reported that the proportion of business students who admitted to some form of cheating was .56. The dean at Bayview University is interested in testing to see if the population proportion of business students at Bayview who admitted to some form of cheating is less than business students at other universities. Thus, the appropriate hypothesis test is H0: p ≥ .56 Ha: p < .56 indicating a lower tail test. The test statistic is z ≤ -2.85. The p-value associated with this test statistic is .0022. Since the p-value is less than the .05 level of significance indicated, we conclude that the ethical behavior, in terms of cheating, at Bayview University is better than for business students at other universities.

In the same article, the Chronicle of Higher Education reported that the proportion of non-business students who admitted to some form of cheating was .47. The dean at Bayview University is interested in testing to see if the population proportion of business students at Bayview who admitted to some form of cheating is less than non-business students at other universities. Thus, the appropriate hypothesis test is H0: p ≥ .47 Ha: p < .47 indicating a lower tail test. The test statistic is z ≤ -1.119. The p-value associated with this test statistic is .1315. Since the p-value is not less than the .05 level of significance indicated, we conclude that there is not sufficient evidence to conclude that the ethical behavior, in terms of cheating, at Bayview University is better than for non-business students at other universities.

Discussion The proportion of all business students at Bayview that are involved in some type of cheating is much less than the proportion for business students at other universities. However, the fact that 41% of all Bayview business students are involved in some type of cheating indicates a serious problem for the college. The dean needs to take steps to deal with the problem of cheating in the college. Possible steps the dean might consider include: Forming a committee of students and faculty to investigate how students and faculty can work together to reduce the incidence of cheating Requiring course syllabi to include a statement regarding expectations of the instructor regarding cheating incidents and the penalty for dealing with these incidents when discovered Sponsoring seminars where business executives can talk about the importance of ethical behavior Continuing to administer the exit survey to obtain data that can be used to determine if efforts to reduce cheating are working.

09/Case 2 - Bayview University/BSTAT_09_Case2_Bayview_Rubric.docx Bayview University Case Chapter 9 (120 points total) RUBRIC

09/Case 2 - Bayview University/BSTAT_09_Case2_Bayview_TeachingTips.docx Ethical Behavior of Students at Bayview University Descriptive Statistics To avoid students entering many more formulas then necessary, review use of absolute and relative references.

Introduce COUNTIFS() for Male and Female as likely many students may never have used it.

For example, If the person is a Male, did they copy from the Internet To ascertain any incidence of cheating requires a new column of data. Using the COUNTIF(), I counted the incidents of cheating in columns B through D. A result of 0 indicates no evidence of cheating. Students may be more comfortable using Yes and No responses in this column. Using numbers seemed to me to make the COUNTIF formula easier, ie, =COUNTIF(E2:E91,”0) for the No count and =COUNTIF(E2:E91,”<>0” Students may need to be reminded that even the conditional statement <> 0 must be contained in quotes.

Confidence Intervals Caution students about cutting and pasting to create data for a new interval and forgetting to change either cell addresses or values Hypothesis Testing The authors offer a template for computing p-values: It may help to reinforce the location of these p-value calculations with normal distribution figures. It may be helpful as well to indicate to students that the NORM.S.DIST() function works from left to right on the normal distribution figure. Thus, an upper tail test requires the form 1 – NORM.S.DIST() Lower tail test (NORM.S.DIST(x, True) Upper tail test (1 – NORM.S.DIST(x, True)) Two-tail test (2 * the smallest value calculated value)

09/Case 2 - Bayview University/BSTAT_CH9_Case2_Bayview_Key.xlsx Sheet: Data n=

90

Percent

Collaborated on Individual Project

Count

Percen

72

0.8

No

61

0.6777

Yes

18

0.2

Yes

29

0.3222

n=

48

Count

Percen

Percent

Copied on Exam

Count

0.8222222222222222

No

0.17777777777777778

Percent

Copied on Exam

Count

Percent

Collaborated on Individual Project

0.8541666666666666

No

39

0.8125

No

30

0.625

0.14583333333333334

Yes

9

0.1875

Yes

18

0.375

n=

42

Count

Percen

Percent

Copied on Exam

Count

Percent

Collaborated on Individual Project

0.7857142857142857

No

33

0.7857142857142857

No

31

0.7380

0.21428571428571427

Yes

9

0.21428571428571427

Yes

11

0.2619

Standard Error =

0.052323778320929555

0.4111111111111111

z=

1.9599639845400538

0.56

Margin of Error =

0.10255272104407957

11

+-

0.10255272104407957

15

to

0.5136638321551906

Sheet: Tables for Report

Percent

Collaborated on Individual Project

Count

P

72

0.8

No

61

0

18

0.2

Yes

29

0

Count

Percent

Collaborated on Individual Project

Count

P

No

39

0.8125

No

30

0

0.14583333333333334

Yes

9

0.1875

Yes

18

0

Percent

Copied on Exam

Count

Percent

Collaborated on Individual Project

Count

P

0.7857142857142857

No

33

0.7857142857142857

No

31

0

0.21428571428571427

Yes

9

0.21428571428571427

Yes

11

0

4111111111111111

+-

0.10255272104407957

3085583900670315

to

0.5136638321551906

ount

ount

ount

Percent

Copied on Exam

Count

0.8222222222222222

No

0.17777777777777778

Yes

Percent

Copied on Exam

0.8541666666666666

4375

+-

0.14042609662344283

29707390337655715

to

0.5779260966234429

38095238095238093

+-

0.1468658491835961

10/Case 1 - Par Inc/Case 1 - Par Inc_/BSTAT_10_Case1_ParInc _Report.docx Par, Inc. Par, Inc., is a major manufacturer of golf equipment. Management has been investigating a new golf ball coating designed to resist cuts and provide a more durable ball. This report investigates the effect of the new coating on driving distances by comparing driving distances for the two balls. Forty (40) balls of both the new and current models were subjected to distance tests performed with a mechanical hitting machine.

Hypothesis Testing The hypothesis test conducted in this study is based on the information that the new golf ball is design to resist cuts and yet still offer good driving distances. The hypotheses were structured so that rejection of the null hypothesis corresponds to the conclusion that the newer golf ball has lower mean driving distances; a conclusion that indicates that the cut resistance advantage may be offset by the loss of distance. The hypothesis test is: H0: µcurrent – µnew ≤ 0 H0: µcurrent – µnew > 0 Table 1 presents the results of a two-sample t-test assuming unequal variances. Using the .05 level of significance, a p-value of .094 indicates that the null hypothesis cannot be rejected. The data does not provide statistical evidence that the new golf ball has a lower mean driving distance than the current golf ball.

Descriptive Statistics Table 2 presents descriptive statistics for a sample of 40 current technology golf balls. The sample mean for the current golf is 270.275 and the sample variance is 76.615.

Table 3 presents descriptive statistics for a sample of 40 new technology golf balls. The sample mean for the current golf is 267.5 and the sample variance is 97.95.

Comparing the descriptive statistics between the current and the new golf balls reveals that the means are fairly close. In fact, the max driving distance is the same. The sample variances (76.6 and 97.95, respectively) indicate that the hypothesis test assuming unequal variances is appropriate.

95% Confidence Intervals The 95% confidence interval for the current technology golf ball is 267.5 to 273.1 indicating that with 95% confidence, the population mean driving distance is between 267.5 and 273.1 yards. The 95% confidence interval for the new technology golf ball is 264.3 to 270.1 indicating that with 95% confidence, the population mean driving distance is between 264.3 and 273.1 yards. These results are consistent with the hypothesis test conclusion that indicate insufficient evidence to conclude that the new technology golf ball has a lower driving distance.

Discussion In failing to reject the null hypothesis (H0: µcurrent – µnew ≤ 0) the research findings are inconclusive. The researcher should not be ready to conclude that the mean driving distance for the new ball is equal to or better than the current golf ball. Given the mechanical hitting machine continued study should be relatively easy. More data (a larger sample size) is indicated to give Par a better idea of the difference between the two golf balls.

10/Case 1 - Par Inc/Case 1 - Par Inc_/BSTAT_10_Case1_ParInc_Rubric.docx Quality Associates Case Chapter 9 (70 points total) RUBRIC

10/Case 1 - Par Inc/Case 1 - Par Inc_/BSTAT_10_Case1_ParInc_TeachingTips.docx Par, Inc. – Teaching Tips Hypothesis Testing Entertain discussion about what hypothesis test should be conducted. Formulation used in this example is based on information that the new golf ball is being designed to “resist cuts nd yet still offer good driving distances”. Hypotheses are not designed to prove the new golf ball drives further than the current ball. Perhaps the cut resistance of the ball is offset by a loss of distance.

Suggest using subscripts other than the numbers 1 and 2 to help with remembering which population is which. µcurrent indicates the population mean of the current golf ball.

At this point, suggest that students avoid doing mathematical calculations when there are tools available to do the job. Students may have trouble picking the appropriate test from the Data Analysis toolkit. Go over the difference between z and t (sigma known, sigma unknown) and rules for when variances can be assumed equal/unequal.

It may help to reinforce the type of test and location of these p-value calculations with normal distribution figures.

Lower tail test (NORM.S.DIST(x, True) Upper tail test (1 – NORM.S.DIST(x, True)) Two-tail test (2 * the smallest value calculated value)

10/Case 1 - Par Inc/Case 1 - Par Inc_/BSTAT_Chapter10_Case_Golf_Key.xlsx Sheet: Data

ces

Current

New

270.275

267.5

76.61474358974361

97.94871794871794

40

40

0 Table 2 Descriptive Statistics - Current Golf Balls

77 1.3283615935245674 0.09399121064039592 1.6648845372582084

Mean

270.275

0.18798242128079184

Standard Error

1.383968420789864

1.9912543953883848

Median

270

Mode

272

Standard Deviation

8.752984838884597

Sample Variance

76.61474358974361

nce - current

+-

2.799340358349271

Kurtosis

-0.7625872223598078

to

273.0743403583493

Skewness

0.30616937109063574

Range

34

Minimum

255

Maximum

289

nce - new

+-

3.1651836026015

Sum

10811

to

270.6651836026015

Count

40

Confidence Level(95.0%)

2.799340358349271

11/Case 1 - Air Force Training Program/BSTAT_Chapter11_Case1 _AirForceTraining_Report.docx Air Force Training Program An Air Force introductory course in electronics uses a personalized system of instruction where each student views a videotaped lecture and is given a programmed instruction text. A proposed alternative involves the use of computer-assisted instruction. This report compares the proposed and current methods of instruction whereby 122 students were randomly assigned to one of the two methods.

Descriptive Statistics Tables 1 and 2 present descriptive statistics for the current and proposed training programs. The sample mean 75.066 and 75.426 hours show that both methods have approximately the same mean completion time. The standard deviations of 3.945 hours for the current method and 2.506 hours for the proposed method show the proposed method has less variation.

Hypothesis Tests Table 3 presents results of a test of the difference between population means. The tested null hypothesis is H0: µ1 – µ2 = 0 indicating that there is no difference in the population means. The t value of -.60 is associated with a corresponding p-value of .548. Thus, at a level of significance of .05, the null hypothesis cannot be rejected. Sample evidence does not indicate a difference between methods in terms of completion times.

The standard deviations and variances of the two methods are as follows: Current Proposed Standard Deviations 3.945 2.506 Variance 15.562 6.282 The tested null hypothesis is H0: σ12 = σ22. The F value of 1.84 is associated with a corresponding p-value of .0006. Thus, at a level of significance of .05, the null hypothesis is rejected and we can conclude that the two methods differ in terms of variance. The data indicate that the proposed method has the smaller variance indicating that students trained under this method are more consistent in terms of completion time.

Discussion Under the proposed method, students are more likely to complete the training in approximately the same amount of time. There should be less chance of faster students waiting for slower students to complete the training.

A final decision should consider the amount of learning under the two methods. Is the quality of the training with the proposed method the same or better than the quality of training with the current method? Both groups could be administered an exam at the end of the training program. Analysis of the examination scores would determine if the programs were similar in terms of the amount of learning provided. This analysis should be included in the final decision process.

This report presents initial investigation results comparing current and proposed Air Force training program methods.

11/Case 1 - Air Force Training Program/BSTAT_Chapter11_Case1 _AirForceTraining_Rubric.docx Air Force Training Program Chapter 11 (60 points total) RUBRIC

11/Case 1 - Air Force Training Program/BSTAT_Chapter11_Case1 _AirForceTraining_TeachingTips.docx Air Force Training Program – Teaching Tips Suggest that students avoid doing mathematical calculations when there are tools available to do the job. Students may have trouble picking the appropriate test from the Data Analysis toolkit. Go over the difference between z and t (sigma known, sigma unknown) and rules for when variances can be assumed equal/unequal.

It may help to reinforce the type of test and location of these p-value calculations with normal distribution figures.

Lower tail test (NORM.S.DIST(x, True) Upper tail test (1 – NORM.S.DIST(x, True)) Two-tail test (2 * the smallest value calculated value)

11/Case 1 - Air Force Training Program/BSTAT_Chapter11_Case1_Training_Key.xlsx Sheet: Data Table 2 Proposed Training

75.06557377049181

Mean

75.42622950819673

0.5050936462752816

Standard Error

0.32090975221618767

76

Median

76

76

Mode

76

3.9449074871240284

Standard Deviation

2.5063852882417645

15.562295081967216

Sample Variance

6.281967213114752

-0.06932527643823505

Kurtosis

0.5869404138741925

-0.22053355811077843

Skewness

-0.28748624289474806

19

Range

13

65

Minimum

69

84

Maximum

82

4579

Sum

4601

61

Count

61

1.0103377205576873

Confidence Level(95.0%)

0.6419150784211763

H0: µcurrent = µproposed

ual Variances

Ha: µcurrent ≠ µproposed Current

Proposed

α = .05

75.06557377049181

75.42622950819673

t = -.603

15.562295081967216

6.281967213114752

p-value = .548

61

61

Since the p-value is not less than .05,

0

hypothesis and conclude that the mea

102 -0.6026831811868915

H0: σ2current = σ2proposed

0.27402854167658686

Ha: σ2current ≠ σ2proposed

1.659929975970338

α = .05

0.5480570833531737

F=

1.9834952585628811

p-value =

Since the p-value is less than alpha re

and conclude that the population varia

11/Case 2 - Meticulous Drill and Reamer/BSTAT_Chapter11_Case 2_MeticulousDrillandReamer_Report.docx Meticulous Drill and Reamer Meticulous Drill and Reamer (MD&R;) specializes in drilling and boring precise holes in hard metals. The company recently contracted to drill holes with 3-centimeter diameters in large carbon-steel alloy disks and requires a special drill to complete this job. Davis Drills’ T2005 and Worth Industrial Tools’ AZ100 have agreed to allow MD&R; to use each drill to drill holes with a diameter of 3 cm. This report considers the accuracy (closeness of the diameter to 3 centimeters) and the precision (the variance of the diameter) of the holes drilled in order to make a purchase decision.

Descriptive Statistics Tables 1 and 2 present descriptive statistics for holes drilled by T2005 and AZ100. The sample mean 3.0223 and 3.0229 cm show that both methods have approximately the same mean diameter, indicating accuracy. The standard deviations of .113 cm for the T2005 and .190 cm for the AZ100 show the T2005 has less variation indicating that the T2005 may be the best choice in terms of precision.

Hypothesis Test The tested null hypothesis is H0: σ12 = σ22. The F value of 2.85 is associated with a corresponding p-value of .0027. Thus, at a level of significance of .05, the null hypothesis is rejected and we can conclude that the two drills differ in terms of variance. The data indicate that the T2005 has the smaller variance indicating that the T2005 drill is the most precise.

Discussion Given that the accuracy of the 2 drills (in terms of sample means) is approximately the same and that the T2005 is most precise, the T2005 is the drill of choice. This report presents investigation results that Meticulous Drill and Reamer can use to make a purchase decision.

11/Case 2 - Meticulous Drill and Reamer/BSTAT_Chapter11_Case 2_MeticulousDrillandReamer_Rubric.docx Meticulous Drill and Reamer Chapter 11 (50 points total) RUBRIC

11/Case 2 - Meticulous Drill and Reamer/BSTAT_Chapter11_Case 2_MeticulousDrillandReamer_TeachingTips.docx Meticulous Drill and Reamer – Teaching Tips Remind students that calculation of the F statistic is an upper-tail test and requires the created ratio to place the larger variance in the numerator.

11/Case 2 - Meticulous Drill and Reamer/BSTAT_Chapter11_Case2_MeticulousDrills_Key.xlsx Sheet: Data Table 2 Holes Drilled by AZ100

3.0222580645161297

Mean

3.022903225806451

0.020228318686049455

Standard Error

0.03417284826866097

3.02

Median

2.97

3.04

Mode

3.31

0.11262651190015475

Standard Deviation

0.19026636676664813

0.0126847311827957

Sample Variance

0.03620129032258066

3.607021249743643

Kurtosis

-1.0396111329490751

-0.8243431923796838

Skewness

0.20683967953028795

0.6200000000000001

Range

0.6800000000000002

2.69

Minimum

2.71

3.31

Maximum

3.39

93.69000000000003

Sum

93.71

31

Count

31

0.04131173808980245

Confidence Level(95.0%)

0.06979026677244773

1. Which drill produces holes with

T2005 mean diameter closer to 3

2. Which model has a smaller va 2.853926488539266 0.0026586344890992975

null hypothesis not the same

T2005 produces holes with a sm

12/Case 1 - BiPartisan Agenda for Change/BSTAT_12_Case1_BiP artisanAgendaforChange_TeachingTips.docx BiPartisan Agenda for Change – Teaching Tips Descriptive Statistics Introduce COUNTIFS() as likely many students may never have used it.

Tests of Independence To avoid students entering many more formulas then necessary, review use of absolute and relative references.

To compute the EXPECTED table of values required for the Chi Square uses mixed cell references.

It may be worth describing in detail how $J3 indicates that when the formula is copied the column reference remains the same and H$6 indicates that the row reference remains the same. The figure below may be helpful to show that $J3 stays in the same column when copied and H$6 stays in the same row.

Remind students when using the CHISQ.TEST function that only the observed and expected data is used; no totals.

12/Case 1 - BiPartisan Agenda for Change/BSTAT_12_Case1_Bip artisanAgendaforChange_Report.docx A Bipartisan Agenda for Change In a study conducted by Zogby International for the Democrat and Chronicle, New Yorkers were polled to determine whether the New York state government works in a number of areas.

Respondents were asked questions involving pay cuts for state legislators, restrictions on lobbyists, term limits for legislators, and whether state citizens should be able to put matters directly on the ballot for a vote. Results regarding these proposed reforms had broad support, crossing all demographic and political lines.

A follow-up survey of 100 individuals from the western region of New York was conducted. In this survey respondents reported their party affiliation as well as their opinion about legislative pay cuts, restrictions on lobbyists, and legislators term limits. This report summarizes findings from the follow-up survey.

Descriptive Statistics Table 1 presents descriptive statistics for the research question “Should legislative pay be cut for every day the state budget is late?” The percentage of respondents responding yes are as follows: Democrat – 61.1% Independent – 52.6% Republican – 86.7% Preliminary conclusions indicate that since the percentage of Republicans that answered yes is much greater than the percentage of Democrats or Independents that answered yes, the classifications do not appear to be independent.

Table 2 presents descriptive statistics for the research question “Should there be restrictions on lobbyists?” The percentage of respondents responding yes are as follows: Democrat – 58.3% Independent – 78.9% Republican – 75.6% Preliminary conclusions indicate that since the percentage of Democrats that answered yes is much less than the percentage of Independents or Republicans that answered yes, the classifications do not appear to be independent.

Table 3 presents descriptive statistics for the research question “Should there be term limits requiring legislators serve a fixed number of years?” The percentage of respondents responding yes are as follows: Democrat – 47.2% Independent – 52.6% Republican – 71.1% Preliminary conclusions indicate that since the percentage of Republicans that answered yes is much greater than the percentage of Democrats or Independents that answered yes, the classifications do not appear to be independent.

Tests of Independence In order to test statistically for the independence of responses in each category a chi-square test of independence is conducted for each survey question. For each question, we are interested in whether the answer to the question is independent of political affiliation. A generic form of the hypotheses is: H0: A Yes response to the question is independent of party affiliation Ha: A Yes response to the question is not independent of party affiliation.

Should legislative pay be cut for every day the state budget is late?

The hypothesis test was conducted at a .05 level of significance. The p-value associated with this test for independence is .006. Since the p-value is less than the level of significance, the null hypothesis is rejected indicating that a yes response is not independent of party affiliation.

Should there be restrictions on lobbyists?

The hypothesis test was conducted at a .05 level of significance. The p-value associated with this test for independence is .156. Since the p-value is not less than the level of significance, the null hypothesis can NOT be rejected indicating that a yes response may be independent of party affiliation. Thus, the preliminary conclusion based upon descriptive statistics was incorrect in this case.

Should there be term limits requiring legislators serve a fixed number of years?

The hypothesis test was conducted at a .05 level of significance. The p-value associated with this test for independence is .078. Since the p-value is not less than the level of significance, the null hypothesis can NOT be rejected indicating that a yes response may be independent of party affiliation. Thus, the preliminary conclusion based upon descriptive statistics was incorrect in this case.

Discussion There appears to be broad support across all party lines with regard to restrictions on lobbyists and requiring legislators serve a fixed number of years. With regard to the issue of a legislative pay cut, however, the response (yes and no) and party affiliation are not independent. On this issue, Republican respondents have a much different perspective than the Democrats or the Independents.

This report summarizes findings of 100 respondents from the western region of New York.

Responses to three yes/no questions were captured as well as the party affiliation of the respondent in order to examine differences in response based on party affiliation.

12/Case 1 - BiPartisan Agenda for Change/BSTAT_12_Case1_Bip artisanAgendaforChange_Rubric.docx A Bipartisan Agenda for Change Chapter 12 (105 points total) RUBRIC

OBSERVED

12/Case 1 - BiPartisan Agenda for Change/BSTAT_12_Case1_NYReform_Key.xlsx Sheet: Data Table 1 Should legislative pay be cut?

EXPECTED

Should legislati

Yes

No

Totals

% Response

Democrat

22

14

36

0.6111111111111112

Democrat

Independent

10

9

19

0.5263157894736842

Independent

Republican

39

6

45

0.8666666666666667

Republican

Totals

71

29

100

Totals p-value =

Table 2 Should there be restrictions on lobbyists?

Should there be Yes

No

Totals

% Response

Democrat

21

15

36

0.5833333333333334

Democrat

Independent

15

4

19

0.7894736842105263

Independent

Republican

34

11

45

0.7555555555555555

Republican

Totals

70

30

100

Totals p-value =

Table 3 Should there be term limits on legislators?

Should there be Yes

No

Totals

Democrat

17

19

36

0.4722222222222222

Democrat

Independent

10

9

19

0.5263157894736842

Independent

Republican

32

13

45

0.7111111111111111

Republican

Totals

59

41

100

Totals p-value =

12/Case 2 - Fuentes Chips/BSTAT_12_Case2_FuentesChips_Key.xlsx Sheet: Data

EXPECTED Candied Bacon Availability

Total

c = pDeepSouth = pMountain = pPacific

0.0013546956656406365 23.57142857142857

Yes

No

Total

27

13

40

New England

29

11

40

Mid-Atlantic

28

12

40

Midwest

33

7

40

Great Plains

32

8

40

South Atlantic

33

7

40

Deep South

25

15

40

Mountain

17

23

40

Pacific

224

96

320

T

12/Case 2 - Fuentes Chips/BSTAT_12_Case2_FuentesSaltySnacks_Report.docx Fuentes Salty Snacks Fuentes Salty Snacks, Inc. added a new flavor to its line of potato chips, candied bacon. Senior management is interested in determining if penetration of Fuentes Salty Snacks Candied Bacon is consistent across eight defined U.S. Sales regions: New England, Mid-Atlantic, Midwest, Great Plains, South Atlantic, Deep South, Mountain, and Pacific. This report summarizes results of analysis of selected random samples of 40 grocery stores in each of its eight U.S. sales regions.

Descriptive Statistics Table 1 presents sample proportions for each region. Table 2 presents observed and expected frequencies for each region. Preliminary conclusions indicate that penetration appears t bee highest in the Great Plains and Deep South. Penetration in the Pacific region is less than that of other regions followed by Mountain and New England which are only slightly less than expected.

Goodness of Fit Hypothesis Test for Region In order to test statistically for consistent penetration across all regions a chi-square goodness of fit test is conducted. The hypothesis is that the proportion of grocery stores that currently carries Fuentes’ Candied Bacon Potato Chips is equal across its eight U.S. sales regions. The hypothesis is stated as: The hypothesis test was conducted at a .05 level of significance. The p-value associated with this test for independence is .001. Since the p-value is less than the level of significance, the null hypothesis is rejected indicating that not all population proportions are equal.

Marascuilo Pairwise Comparison Procedure Results of hypothesis testing indicate that penetration of grocery stores across Fuentes’ eight U.S. sales regions is not equal. The Marascuilo pairwise comparison procedure was used at a significance level of .05 to test for differences between regions. Results are available in Appendix A of this report. Significant differences were discovered between the Great Plains and Pacific region and the Deep South and Pacific region.

Discussion Penetration in the Pacific region is less than other regions, with statistically significant differences between the Great Plains and the Pacific and Deep South and Pacific regions. Penetration is higher than expected in the Great Plains, Deep South, and South Atlantic regions. Penetration is slightly higher than expected in the Mid-Atlantic region and slightly lower than expected in the New England region. These results however are not statistically significant.

This report summarizes findings of analysis of market penetration of Fuentes’ Candied Bacon Potato Chips across eight defined U.S. Sales regions.Appendix A

12/Case 2 - Fuentes Chips/BSTAT_12_Case2_FuentesSaltySnacks_Rubric.docx Fuentes Salty Snacks Chapter 12 ( 85 points total) RUBRIC

12/Case 2 - Fuentes Chips/BSTAT_12_Case2_FuentesSaltySnack s_TeachingTips.docx Fuentes Salty Snacks – Teaching Tips Remind students that the CHISQ.TEST function is for use with versions later than Excel 2007.

CHITEST is available for versions of Excel prior to 2007.

The chapter appendix reports computation of expected frequencies as indicated below. Using this format each of these formulas must be entered separately.

It may be worth the time to show students how to take advantage of mixed and absolute cell references. The formula required is row total * column total / grand total. Using $E7*B$9/$E$9 allows the students to copy the formula across and down the table, requiring only one “correct” formula entry. Use the following to describe the process: $E7 indicates that when this formula is copied across or down the row may change but not the column reference B$9 indicates that when this formula is copied the column designation may change but not the row $E$9 is an absolute reference to the grand total which is always in the same location.

Marascuilo Pairwise Comparisons These 28 pairwise comparisons appear daunting. Remind students to take advantage of cut and paste and absolute and mixed cell references to avoid entering each formula from scratch. For example, having entered the New England comparison with an absolute reference allows copying the formula to begin second set of comparisons and changing 9 to 10 in $X$9 and changing X16 to X11. Then this set of comparisons can be copied to complete Mid-Atlantic to all others, etc.

Requires considerable Excel intuition, but is a major time saver.

12/Case 3 - Fresno Board Games (BBG)/BSTAT_12_Case3_BBG_Key.xlsx Sheet: Data

Die 3

Die 4

Die 5

96

93

73

75

98

83

85

74

94

63

78

70

87

80

87

94

77

93

500

500

500

333333333333

83.33333333333333

83.33333333333333

83.33333333333333

042156569664

12.436505404118439

9.750213672872361

10.053191864610321

Use of mixed cell addresses to enter one form

O

E

106

83.33333333333333

86

83.33333333333333

73

83.33333333333333

69

83.33333333333333

75

83.33333333333333

91

83.33333333333333

O

E

85

83.33333333333333

12/Case 3 - Fresno Board Games (BBG)/BSTAT_12_Case3_FresnoBoardGames_Report.docx Fresno Board Games Fresno Board Games manufactures and sells several different board games. The most popular game, Cabestrillo Cinco is played with 5 six-sided dice. Fresno is considering a move to Big Boss Gaming, Inc. (BBG), a new supplier that has offered to sell dice to Fresno at a substantially lower price. Management wants to ensure that each of the six-sided dice is fair. In this report, the results of five randomly selected dice, each rolled 500 times are examined to ascertain fairness (thus quality) of the dice.

Descriptive Statistics Table 1 presents sample descriptive statistics for each of the five die after 500 rolls each. The expectation may be that each number should appear exactly the same number of times. The sample mean for each die indicates the proportion average of 83.33 suggesting that the dice may be fair. Standard deviations vary from 6.38 to 13.87 suggesting a variability in the dice.

Hypothesis Tests In order to test statistically for the fairness of dice a chi-square goodness of fit test is conducted.

The hypothesis for each die is that the proportion of each number occurring is 1/6. The hypothesis is stated as: The hypothesis test was conducted at a .01 level of significance. The p-value associated with this test for the first die tested is .04. Since the p-value is not than the level of significance, there is insufficient evidence to suggest that the die is not fair. Results of hypothesis testing for all 5 die are available in Appendix A of this report.

The p-value associated with this test for the second die tested is .79. Since the p-value is not than the level of significance, there is insufficient evidence to suggest that the die is not fair.

The p-value associated with this test for the third die tested is .30. Since the p-value is not than the level of significance, there is insufficient evidence to suggest that the die is not fair.

The p-value associated with this test for the fourth die tested is .34. Since the p-value is not than the level of significance, there is insufficient evidence to suggest that the die is not fair.

The p-value associated with this test for the fifth die tested is .30. Since the p-value is not than the level of significance, there is insufficient evidence to suggest that the die is not fair.

Discussion Results of the hypotheses tests to not indicate that the dice are not fair, suggesting that Fresno may want to purchase dice from BBG. Appendix A

12/Case 3 - Fresno Board Games (BBG)/BSTAT_12_Case3_FresnoBoardGames_Rubric.docx Fresno Board Games – Teaching Tips Notice the use of the mixed cell reference which allows copying formula for die 2 through 5.

COUNTIF(A$3:A$503,"1") It may be worth the time to show students how to take advantage of mixed and absolute cell references The formula to compute the chi-square test statistic is Suggest to students that they complete computations in parts as indicated below to be able to more quickly find errors in computations.

12/Case 3 - Fresno Board Games (BBG)/BSTAT_12_Case3_Fresn oBoardGames_TeachingTips.docx Fresno Board Games Chapter 12 ( 140 points total) RUBRIC

13/Case 1 - Wentworth Medical Center/BSTAT_13_Case1_Medical1_Key.xlsx Sheet: Data

rida

New York

North Carolina

an

5.55

Mean

8

Mean

7.0

ndard Error

0.47834693421583074

Standard Error

0.4920419323166418

Standard Error

0.6

dian

6

Median

8

Median

7.5

de

7

Mode

8

Mode

8

ndard Deviation

2.1392325234704352

Standard Deviation

2.200478416880723

Standard Deviation

2.8

mple Variance

4.576315789473687

Sample Variance

4.842105263157895

Sample Variance

8.0

tosis

-1.0621928182910465

Kurtosis

0.6264316690759513

Kurtosis

-0.9

ewness

-0.27355708978785215

Skewness

0.6256873885428021

Skewness

-0.0

nge

7

Range

9

Range

9

imum

2

Minimum

4

Minimum

3

ximum

9

Maximum

13

Maximum

12

m

111

Sum

160

Sum

141

unt

20

Count

20

Count

20

nfidence Level(95.0%)

1.0011916396662026

Confidence Level(95.0%)

1.0298556001162762

Confidence Level(95.0%)

1.3

ups

Count

Sum

Average

Variance

rida

20

111

5.55

4.576315789473687

w York

20

160

8

4.842105263157895

th Carolina

20

141

7.05

8.050000000000002

urce of Variation

SS

df

MS

F

P-v

ween Groups

61.033333333333246

2

30.516666666666623

5.240885808978599

0.0

hin Groups

331.90000000000003

57

5.822807017543861

ova: Single Factor

MMARY

OVA

13/Case 1 - Wentworth Medical Center/BSTAT_13_Case1_Medical2_Key.xlsx Sheet: Data

ida

New York

North Carolina

an

14.5

Mean

15.25

Mean

1

ndard Error

0.7089651464968982

Standard Error

0.9230242056471359

Standard Error

0

dian

14.5

Median

14.5

Median

1

de

17

Mode

14

Mode

1

ndard Deviation

3.170588522490323

Standard Deviation

4.1278897374094825

Standard Deviation

2

mple Variance

10.052631578947368

Sample Variance

17.039473684210527

Sample Variance

8

osis

-0.3407994814383639

Kurtosis

-0.030136700406970363

Kurtosis

-0

wness

0.2807214968790258

Skewness

0.5253524940223095

Skewness

-0

ge

12

Range

15

Range

1

mum

9

Minimum

9

Minimum

8

ximum

21

Maximum

24

Maximum

1

m

290

Sum

305

Sum

2

nt

20

Count

20

Count

2

fidence Level(95.0%)

1.4838811053551184

Confidence Level(95.0%)

1.9319118652205771

Confidence Level(95.0%)

1

ups

Count

Sum

Average

Variance

ida

20

290

14.5

10.052631578947368

w York

20

305

15.25

17.039473684210527

th Carolina

20

279

13.95

8.681578947368411

rce of Variation

SS

df

MS

F

P

ween Groups

17.033333333333417

2

8.516666666666708

0.7142121524201889

0

hin Groups

679.7

57

11.924561403508772

va: Single Factor

MMARY

OVA

13/Case 1 - Wentworth Medical Center/BSTAT_13_Case1_WentworthMedicalCenter_Report.docx Wentworth Medical Center Sociologists and physicians at the Wentworth Medical Center in upstate New York investigated the relationship between geographic location and depression for individuals in reasonably good health and individuals with chronic health conditions such as arthritis, hypertension, and/or heart ailment.

Individuals were residents from Florida, New York, and North Carolina were sampled. This report summarizes descriptive statistics and analysis of variances results from this study.

Descriptive Statistics Medical1 – Individuals 65 or older; no chronic health condition Table 1 presents descriptive statistics for individuals 65 or older who are reasonably healthy in Florida, New York, and North Carolina. Mean responses were, respectively 5.55, 8, and 7.05. The standard deviation for each state were 2.14, 2.20, and 2.84 for Florida, New York, and North Carolina.

Medical2 – Individuals 65 or older; chronic health condition Table 2 presents descriptive statistics for individuals 65 and over with chronic health conditions in Florida, New York, and North Carolina mean responses were, respectively 14.5, 15.25, and 13.95.

The standard deviation for each state were 3.17, 4.13, and 2.95 for Florida, New York, and North Carolina.

The depression levels for the individuals who had a chronic health condition are, on average, much higher than the depression levels of individuals in reasonably good health.

Analysis of Variance In order to test statistically for the equality of depression scores in each state analyses of variance were conducted. For each health group (reasonably good health and chronic health conditions), we are interested in whether the population mean depression scores are equal. Thus, the hypotheses of interest are: H0: µFlorida = µNewYork = µNorthCarolina Ha: Not all population means are equal ANOVA: Single Factor – 65 or over; no chronic health condition Table 3 presents the results of the single factor ANOVA. The hypothesis test was conducted at a .05 level of significance. The p-value associated with this test for independence is .01. Since the p-value is less than the level of significance, the null hypothesis is rejected indicating the population mean depression scores between Florida, New York, and North Carolina are not the same.

ANOVA: Single Factor – 65 or over; chronic health condition Table 4 presents the results of the single factor ANOVA. The hypothesis test was conducted at a .05 level of significance. The p-value associated with this test for independence is .49. Since the p-value is NOT less than the level of significance, the null hypothesis is not rejected indicating there is insufficient evidence to conclude that the population mean depression scores between Florida, New York, and North Carolina are not the same.

Discussion - Inferences

Results of the analysis of variance for reasonably healthy individuals indicate that depression scores for each state are not the same. The observed differences appear to be due to the mean depression scores for residents of Florida and New York.

Results of the analysis of variance for individuals with chronic health conditions indicate that there is insufficient evidence to conclude that depression scores differ among the states. These depression scores may be the same for individuals in Florida, New York, and North Carolina.

This report summarizes findings of respondents from Florida, North Carolina, and New York regarding depression test scores of reasonably healthy individuals and individuals with chronic health conditions.

13/Case 1 - Wentworth Medical Center/BSTAT_13_Case1_WentworthMedicalCenter_Rubric.docx Wentworth Medical Center Chapter 13 (75 points total) RUBRIC

13/Case 1 - Wentworth Medical Center/BSTAT_13_Case1_Wentwo rthMedicalCenter_TeachingTips.docx Wentworth Medical Center – Teaching Tips Descriptive Statistics Make sure students use the Data Analysis Toolkit and run the Descriptive Statistics tool. There is no default automatically computed. If the student fails to choose one of the available check boxes, the following dialog box appears.

Choosing the confidence level for the mean in the Descriptive Statistics dialogue box provides the margin of error used to compute a confidence interval when sigma is unknown.

Because histograms and frequency distributions were discussed in an earlier Chapter (chapter 2), suggest reminding students that these tools can be used to further describe transmission failure incidence as part of discussion 95% Confidence Intervals Remind students to save computation time with the Descriptive Statistics – Confidence Interval option when sigma is unknown. Students should avoid use of the terms “the mean miles driven is between A and B 95% of the time”. Instead use “we are 95% confident that the mean miles driven is between A and B” Sample Size To obtain a positive value for z used to compute the margin of error, use 1-.05 (for alpha = .05) =NORM.S.INV(1-0.05/2) Planning value for sigma is the sample standard deviation (see page 395 for additional examples of planning values)

13/Case 2 - Compensation for Sales Professionals/BSTAT_13_Case2_Compensation for Sales Professionals_Report.docx Compensation for Sales Professionals A local chapter of sales professionals in the greater San Francisco area conducted a survey of its membership to study the relationship between the years of experience and salary for individuals employed in inside and outside sales positions. Year of experience were specified by level as: low (1 – 10 years), medium (11 – 20 years), and high (21 or more years). This report summarizes results of analysis conducted to study this relationship.

Descriptive Statistics Table 1 presents descriptive statistics for membership salaries summarized by experience and sales positions. Average salary for inside positions is $56,020.52. Average salary for outside positions is $73,830.43. Average salary for low experience is $59,819.63. Average salary for medium experience is $68,618.13. Average salary for high experience is $66,338.68. The largest average salary for Inside experience is for employees with 21 or more years of experience, $57,422.45. The largest average salary for outside experience is for employees with 11 to 20 years of experience, $81,628.50.

Confidence Intervals A 95% confidence interval for all salaries regardless of position or experience is: This interval indicates that 95% of the population membership salaries will be in the interval $62,966.30 to $66,884.65.

A 95% confidence interval for inside salaries is: This interval indicates that 95% of the population membership salaries of inside sales personnel will be in the interval $55,093.17 to $56,947.87.

A 95% confidence interval for outside salaries is: This interval indicates that 95% of the population membership salaries of outside sales personnel will be in the interval $71,783.72 to $75,877.15.

Analysis of Variance An analysis of variance was used to test for any significant differences due to position using a .05 level of significance, ignoring the effect of years of experience. Results of the analysis of variance indicate that there is a difference between population inside and outside salaries.

An analysis of variance was used to test for any significant differences due to experience using a .05 level of significance, ignoring the effect of position. Results of the analysis of variance indicate that there is a difference between population low, medium, and high salaries.

An analysis of variance was used to test for any significant differences due to position, level of experience, and interaction using a .05 level of significance. Results of the analysis of variance indicate that there is an interaction effect between position and level of experience. This interaction effect suggests that caution should be taken in discussing the separate effects of position and experience.

13/Case 2 - Compensation for Sales Professionals/BSTAT_13_Case2_Compensation for Sales Professionals_TeachingTips.docx Compensation for Sales Professionals – Teaching Tips Descriptive Statistics Make sure students use the Data Analysis Toolkit and run the Descriptive Statistics tool. There is no default automatically computed. If the student fails to choose one of the available check boxes, the following dialog box appears.

Choosing the confidence level for the mean in the Descriptive Statistics dialogue box provides the margin of error used to compute a confidence interval when sigma is unknown.

Because histograms and frequency distributions were discussed in an earlier Chapter (chapter 2), suggest reminding students that these tools can be used to further describe transmission failure incidence as part of discussion 95% Confidence Intervals Remind students to save computation time with the Descriptive Statistics – Confidence Interval option when sigma is unknown. Students should avoid use of the terms “the mean miles driven is between A and B 95% of the time”. Instead use “we are 95% confident that the mean miles driven is between A and B”.

Analysis of Variance Using the Data Analysis toolkit, the completely randomized design described in the text is Anova: Single Factor. The randomized block design is the Anova: Two-Factor Without Replication. The Factorial experiment is Anova: Two-Factor With Replication.

It may be worth the time to go over this distinction using the text box below.

13/Case 2 - Compensation for Sales Professionals/BSTAT_13_Ca se2_CompensationforSalesProfessionals_Rubric.docx Compensation for Sales Professionals Chapter 13 (75 points total) RUBRIC

13/Case 2 - Compensation for Sales Professionals/BSTAT_13_Case2_SalesSalary_Key.xlsx Sheet: Data

ium

High

Table 1 Descriptive Statistics

n

55607.75

Mean

57422.45

dard Error

792.627352792431

Standard Error

744.0229993438587

ian

54525

Median

57753.5

Position

e

#N/A

Mode

#N/A

Inside

dard Deviation

3544.737283339167

Standard Deviation

ple Variance

12565162.407894736

osis

wness

E

L Number Obs.

2

3327.372006712299

Average

5

Sample Variance

11071404.471052632

Std. Dev.

3

-0.11023529076944838

Kurtosis

0.02180183063500607

Number Obs.

2

0.812137393977509

Skewness

0.03831678091645043

Average

6

Outside

Sheet: Inside Inside

Inside

57718 48621

Mean

56020.51666666667

55632

Standard Error

463.4445593593819

51948

Median

56209.5

53464

Mode

53464

52833

Standard Deviation

3589.8261205786416

55959

Sample Variance

12886851.5759887

63799

Kurtosis

-0.2780415144640789

54122

Skewness

0.30676115458133524

58814

Range

15941

59276

Minimum

48621

56990

Maximum

64562

56461

Sum

3361231

49422

Count

60

57670

Confidence Level(95.0%)

927.3504212782775

52072

Inside Salaries

51119

Confidence Interval = Mean +- Margin of Error

56352

56020.51666666667

927.3504212782775

54891

55093.16624538839

to

53464 53938 52694 52031 54768 52282 51827 56588 59057 57549 63362 56080

Sheet: Outside Outside 70515

Outside

56947.86708794495

62283

Mean

73830.43333333333

63856

Standard Error

1022.8497908087188

68858

Median

76315.5

63478

Mode

#N/A

64288

Standard Deviation

7922.960410947817

71345

Sample Variance

62773301.673446395

65546

Kurtosis

-1.0856667730301193

66024

Skewness

-0.24337836995356088

61128

Range

28471

60993

Minimum

60259

61299

Maximum

88730

68560

Sum

4429826

67237

Count

60

62225

Confidence Level(95.0%)

2046.717703886799

62881

Outside Salaries

61680

Confidence Interval = Mean +- Margin of Error

69142

73830.43333333333

2046.717703886799

60561

71783.71562944654

to

60259 83846 83176 83040 88730 77683 79383 75869 76762 82794 78850 86692

Sheet: All Salary

Salary

57718 48621

Mean

64925.475

55632

Standard Error

989.4309469964535

51948

Median

61489.5

75877.15103722013

53464

Mode

53464

52833

Standard Deviation

10838.672975273119

55959

Sample Variance

117476831.86491586

63799

Kurtosis

-1.1145804751174477

54122

Skewness

0.48682834924955465

58814

Range

40109

59276

Minimum

48621

56990

Maximum

88730

56461

Sum

7791057

49422

Count

120

57670

Confidence Level(95.0%)

1959.1720959103534

52072

All Salaries

51119

Confidence Interval = Mean +- Margin of Error

56352

64925.475

1959.1720959103534

54891

62966.30290408964

to

66884.64709591035

53464 70515 62283 63856 68858 63478 64288 71345 65546 66024 61128 60993

Sheet: ANOVA-Position

Count

Sum

Average

Variance

60

3361231

56020.51666666667

12886851.5759887

60

4429826

73830.43333333333

62773301.673446395

SS

df

MS

F

P-value

9515793950.208332

1

9515793950.208332

251.54043552718787

4.96485

4463949041.716666

118

37830076.62471751

13979742991.924997

119

e

means are equal

ss than alpha, reject the null hypothesis

ulation means are not the same

Sheet: ANOVA-Experience

Count

Sum

Average

Variance

40

2392785

59819.625

36060691.163461536

40

2744725

68618.125

185541915.49679488

40

2653547

66338.675

94080544.43012805

SS

df

MS

F

P-value

1668100099.399992

2

834050049.699996

12311642892.525

117

105227717.0301282

13979742991.924992

119

7.926144111453027

0.0005912676656

e equal

pha, reject the null hypothesis

ion means are the same

Sheet: ANOVA-With Interaction

Medium

High

Total

20

20

60

1112155

1148449

3361231

55607.75

57422.45

56020.51666666667

H0:

12565162.407894736

11071404.471052632

12886851.575988699

Ha:

α= 20

20

60

1632570

1505098

4429826

F=

81628.5

75254.9

73830.43333333333

p-va

11926437.52631579

14674832.515789473

62773301.673446395

Sinc

and

40

40

2744725

2653547

68618.125

66338.675

185541915.49679488

94080544.43012805

df

MS

F

P-value

F crit

1

9515793950.208324

751.3599289266849

5.3211241752713014e-52

3.9243304846396727

2

834050049.699996

65.85596423619151

9.756202761676772e-20

3.0758526364438175

2

676033092.2333376

53.379064194786125

4.368420450159842e-17

3.0758526364438175

114

12664761.033771928

119

13/Case 3 - Touristopia Travel/BSTAT_13_Case3_TourisTopia_Key.xlsx Sheet: Data 1.

Calibri

Table 1 Descriptive Statistics

Tahoma Background

Mean

242.8

Mean

235.7

Standard Error

9.911841287124682

Standard Error

6.809585873613387

Average

Median

250.5

Median

242

Std. De

Mode

253

Mode

217

Standard Deviation

54.28939059369229

Standard Deviation

37.29763790246575

Average

Sample Variance

2947.3379310344844

Sample Variance

1391.1137931034498

Std. De

Kurtosis

0.009699110938113265

Kurtosis

-0.7840470844742868

Skewness

0.006756983001439654

Skewness

-0.08422629237951512

Average

Range

235

Range

146

Std. De

Minimum

121

Minimum

167

Total Number Obs.

Maximum

356

Maximum

313

Total Average

Sum

7284

Sum

7071

Total Std. Dev

Count

30

Count

30

Calibri

Tahoma

Mean

252.5

Mean

251.5

Standard Error

15.505733706528618

Standard Error

5.823801736552049

Median

251.5

Median

252

Mode

#N/A

Mode

245

Standard Deviation

49.033435304675294

Standard Deviation

18.41647812874836

Sample Variance

2404.277777777778

Sample Variance

339.1666666666667

Kurtosis

1.7137224055123248

Kurtosis

-0.5830552208930593

Skewness

0.7075381072642426

Skewness

-0.562136787512013

Range

183

Range

56

Minimum

173

Minimum

217

Maximum

356

Maximum

273

Sum

2525

Sum

2515

Count

10

Count

10

White

Green

Pink

Number

Number

Number

Sheet: 3-4-5 ANOVA

cation

White

Green

Pink

Total

10

10

10

30

3104

2529

2532

8165

310.4

252.9

253.2

272.1666666666667

1761.8222222222248

2439.2111111111135

2407.2888888888865

2806.971264367811

10

10

10

30

2525

2410

2349

7284

252.5

241

234.9

242.8

2404.277777777778

3104.222222222222

3810.9888888888913

2947.3379310344844

10

10

10

30

2515

2492

2064

7071

251.5

249.2

206.4

235.7

339.1666666666667

1890.6222222222195

818.9333333333359

1391.1137931034498

30

30

30

8144

7431

6945

271.46666666666664

247.7

231.5

2182.395402298855

2332.7000000000016

2567.5689655172414

SS

df

MS

F

P-value

22426.2888888889

2

11213.14444444445

5.3180577414912475

0.00675994393960

24246.2888888889

2

12123.14444444445

5.749643419240608

0.00462422223086

kground Colors

Sheet: Sheet1

13/Case 3 - Touristopia Travel/BSTAT_13_Case3_TouristopiaTravel_Report.docx Touristopia Travel TourisTopia Travel (Triple T) is an online travel agency that specializes in trips to exotic locations.

The marketing manager has been working on a revision of the website and is interested in the selection of background color (white, green, or pink) and the type of font (Arial, Calibri or Tahoma).

Visitors to the website have been randomly directed to each of several prototypes including the color and font combinations. Ten visitors to each prototype were randomly selected. This report analyzes collected data to see if time spent on the website differs by background color or font.

Additionally, analyses examine whether time spent differs by different combinations of background color and font.

Descriptive Statistics Table 1 presents descriptive statistics by background color and font. Mean responses were for background color were 271.47 for the white background, 247.7 for the green background, and 231.5 for the pink background. The standard deviations for each background respectively were 46.72, 48.30, and 50.67. The sample data suggest that visitors to the white background prototype spend more time at the Triple T website. Standard deviations for the three background colors are similar.

The sample means for the three fonts are 272.17 (Ariel), 242.8 (Calibri), and 235.7 (Tahoma).

Standard deviations were 52.98 (Arial), 54.29 (Calibri), and 37.30 (Tahoma). The sample data suggest that visitors to the Arial font prototypes spend more time at the Triple T website. The sample data also suggest that the variance for the Tahoma font prototypes is smaller than the variance for the other fonts.

For the nine combinations of background color and font, the sample data suggest that visitors to the white background and Arial font prototype spend more time (310.4) at the Triple T website. The sample data also suggest that the standard deviation for the white background and Tahoma font prototype (18.42) is smaller than the variances for the other combinations of color and font.

Since Ms. Dingle randomly assigned the prototypes featuring the three background colors and the three fonts to visitors, she is using a controlled. Experiment.

Analysis of Variance In order to test statistically for the equality of time spent by visitors for the three background colors an analysis of variance was conducted. For each background color, we are interested in whether the population mean background colors are equal. Thus, the hypotheses of interest are: H0: µWhite = µGreen = µPink Ha: Not all population means are equal The analysis includes both factors (background color and font) and their interaction. Table 2 presents the results of the ANOVA. The p-value for the hypothesis is .0046. Since the p-value is less than α = .05, background color is significant.

In order to test statistically for the equality of time spent by visitors for the three font colors, we are interested in whether the population mean font types are equal. Thus, the hypotheses of interest are: H0: µArial = µCalibri = µTahoma

Ha: Not all population means are equal Refer to Table 2 partial output from Excel’s ANOVA: Two Factor with Replication tool presents the results of the single factor ANOVA. The hypothesis test was conducted at a .05 level of significance. The p-value associated with this test is .0068. Since the p-value is less than the level of significance, the null hypothesis is rejected indicating the population means for font type are not equal. Font is significant.

Interaction means that the effect of one factor, i.e. background, depends on the level of the other factor. The p-value for the hypothesis test that there is no interaction between background color and font is .2269. Since the p-value is NOT less than .05, the interaction between background color and font is not significant. There appears to be no interaction between background color and font.

Discussion Results of the analysis of variance suggest that there are differences between the three background colors in the mean time spent by visitors to the Triple T website. Results also suggest that there are differences between the three fonts in the mean time spent by visitors. Descriptive statistics suggest that the white background and the Arial font will generate the greatest mean time spent by visitors to the Triple T website.

This report summarizes findings of a controlled experiment by Triple T to ascertain which background color and font on a website generate the most time visited. Triple T should adopt the white background and Arial font prototype.

13/Case 3 - Touristopia Travel/BSTAT_13_Case3_TouristopiaTravel_Rubric.docx Touristopia Travel Chapter 13 (80 points total) RUBRIC

13/Case 3 - Touristopia Travel/BSTAT_13_Case3_TouristopiaTravel_TeachingTips.docx Touristopia Travel – Teaching Tips Descriptive Statistics Make sure students use the Data Analysis Toolkit and run the Descriptive Statistics tool. There is no default automatically computed. If the student fails to choose one of the available check boxes, the following dialog box appears.

Suggest students use results of the Descriptive Statistics to create a table summarizing background color and font combinations.

ANOVA Managerial Report questions 3 through 5 may suggest that students run multiple different ANOVA. Remind students that they are are running a Factorial experiment one time that in Excel is known as ANOVA: Two Factor With Replication.

14/Case 1 - Measuring Stock Market Risk/BSTAT_14_Case1_Beta_Key.xlsx Sheet: Sheet1

Caterpillar

Johnson & Johnson

McDonald

01663722222222222

Mean

0.03009722222222222

Mean

0.005295555555555557

Mean

009223348465844635

Standard Error

0.011427208540043218

Standard Error

0.0058110000292882465

Standard

012785

Median

0.040815000000000004

Median

-0.001475

Median

N/A

Mode

#N/A

Mode

#N/A

Mode

05534009079506781

Standard Deviation

0.0685632512402593

Standard Deviation

0.03486600017572948

Standard

003062525649206349

Sample Variance

0.004700919420634919

Sample Variance

0.0012156379682539684

Sample V

584302067527922

Kurtosis

0.2857437191108141

Kurtosis

0.43449168259788795

Kurtosis

5182939115643377

Skewness

0.3305947670769223

Skewness

0.5959572119699981

Skewness

34863

Range

0.31906999999999996

Range

0.16251

Range

.11646

Minimum

-0.1006

Minimum

-0.05917

Minimum

23217

Maximum

0.21847

Maximum

0.10334

Maximum

5989399999999999

Sum

1.0835

Sum

0.19064000000000003

Sum

Count

36

Count

36

Count

14/Case 1 - Measuring Stock Market Risk/BSTAT_14_Case1_Beta_Report.docx Measuring Stock Market Risk Measures of risk or volatility of an individual stock include (1) the standard deviation of the total return over several periods of time and (2) the beta. Since the standard deviation does not take into account the extent to which the price of a given stock varies as a function of a standard market index, many financial analysts prefer the measure of risk known as the beta. Betas for individual stocks are determined by simple linear regression as the slope of the estimated regression equation. The value of beta for the stock market will always be 1. Betas greater than 1 indicate a stock more volatile than the market. Betas less than 1 indicate a stock less volatile than the market.

This report analyses the risk characteristics of a number of stocks.

Descriptive Statistics Table 1 presents selected descriptive statistics for 9 companies. Six of the 9 companies had a higher mean monthly return than the market (as measured by the S&P; 500): Exxon Mobil, Caterpillar, McDonald’s, Sandisk, Qualcomm, and Procter & Gamble. Microsoft and Johnson & Johnson had lower mean monthly returns.

Using the standard deviation as a measure of volatility, Sandisk was the most volatile stock with a standard deviation of .1954. The stocks of Johnson & Johnson and Procter & Gamble exhibit less volatility than the other individual stocks. All of the individual stocks are more volatile than the market as a whole.

Regression Analysis In order to report the beta for each stock, a simple regression was run for each company using the S&P; 500 variable as the independent variable. Table 2 reports estimated regression equations and adjusted R squares for each company. Betas for the individual stocks can be obtained from the regression output. These betas are reported in Table 3.

The beta for the market as a whole is 1. Any stock with a beta greater than 1 will move up faster than the market when the market goes up. Any stock with a beta less than 1 will not go down as fast as the market in periods when the market declines.

Conclusions Sandisk with a beta of 2.6 would benefit most from an up market. Johnson & Johnson, with a beta of .009, is least affected by the market. The effect of the market going down cannot be expected to exert downward pressure on shares of Johnson & Johnson.

Adjusted R2 values indicate that from 0% to 32% of the variability of the returns in these individual stocks is explained by the return for the market.

14/Case 1 - Measuring Stock Market Risk/BSTAT_14_Case1_Beta_Rubric.docx Measuring Stock Market Risk Chapter 14 ( 75 points total) RUBRIC

14/Case 1 - Measuring Stock Market Risk/BSTAT_14_Case1_Beta_TeachingTips.docx Measuring Stock Market Risk – Teaching Tips The report for this case involves execution of 9 simple linear regressions to obtain company betas.

In order to create a summarized table of descriptive statistics and regression results, students can make use of the copy and paste special – transpose option to create row labels and column results from summarized statistics.

Continue to remind students about Data Analysis tools for completing analysis In Excel, the Multiple R in the regression analysis is the correlation coefficient and the R squared is the coefficient of determination. Explain the reason for the adjusted R-squared measure and why it is more frequently reported.

14/Case 2 - Department of Transportation/BSTAT_14_Case2_Safety.xlsx Sheet: Sheet1

Table 3 Regression Analysis Between Percent Under 21 and Fatal Accidents

Regression Statistics Multiple R

0.8393874795142999

R Square

0.7045713407653692

Adjusted R Square

0.6971856242845035

Standard Error

0.5893502880759318

Observations

42

ANOVA df

SS

MS

Regression

1

33.13441763684027

33.134

Residual

40

13.893350482207355

0.3473

Total

41

47.02776811904762

Coefficients

Standard Error

t Stat

Intercept

-1.5974138788560592

0.37167145382869327

-4.2979

Percent Under 21

0.2870531707028242

0.029389769332818914

9.7671

14/Case 2 - Department of Transportation/BSTAT_14_Case2_Safety_Report.docx Department of Transportation As part of a study on transportation safety, the U.S. Department of Transportation collected data on the number of fatal accidents per 1000 licenses and the percentage of licensed drivers under the age of 21 in a sample of 42 cities. This report summarizes data collected over a one-year period.

Numerical and Graphical Summaries Percent Under 21 Table 1 presents descriptive statistics for the percent of licensed drivers under the age of 21. The mean is 12.26 percent with a standard deviation of 3.13. The minimum percentage is 8 and the maximum 18 percent.

Fatal Accidents per 1000 Table 2 presents descriptive statistics for the number of fatal accidents per 1000. The mean is 1.92 percent with a standard deviation of 1.07. The minimum number of fatal accidents per 1000 is .039 and the maximum is 4.1.

Figure 1 presents a scatter diagram of the relationship between the percent of licensed drivers under 21 and the number of fatal accidents per 1000. The scatter diagram suggests a linear relationship between the two variables.

Regression Analysis To further investigate the relationship between the percentage of drivers under the age of 21 and the number of fatal accidents per 1000 licenses, a regression analysis was performed. Table 3 presents the results of this analysis.

The adjusted R square indicates that approximately 70% of the variation in percentage of licensed drivers under age 21 is explained by the number of fatal accidents per 1000. The correlation coefficient of .84 indicates a strong positive relationship. The relationship is described by the following linear equation: # of Fatal Accidents = -1.597 + .287Percent Under 21 This relationship indicates that for every one percent increase in the percent under 21 there is a .287 unit increase in the number of fatal accidents.

Discussion Results of the regression analysis indicate that the number of fatal accidents per 1000 licenses is linearly related to the percentage of licensed drivers under the age of 21. The higher the percentage of drivers under 21, the larger the number of total accidents. This result suggests that the more under age 21 drivers that are on the road, the higher the incidence of fatal accidents.

One recommendation for remediating these results would be to increase training for under age 21 drivers prior to awarding drivers’ licenses. This training could include results of this study as well as reports of the types of accidents/driving conditions that lead to fatal accidents.

14/Case 2 - Department of Transportation/BSTAT_14_Case2_Safety_Rubric.docx US Department of Transportation Chapter 14 (55 points total) RUBRIC

14/Case 2 - Department of Transportation/BSTAT_14_Case2_Safety_TeachingTips.docx Department of Transportation – Teaching Tips Continue to remind students about Data Analysis tools for completing analysis In Excel, the Multiple R in the regression analysis is the correlation coefficient and the R squared is the coefficient of determination. Explain the reason for the adjusted R-squared measure and why it is more frequently reported.

14/Case 3 - Selecting a Point and Shoot Digital Camera/BSTAT_14_Case3_Cameras.xlsx Sheet: Data

Descriptive Statistics

Table 2 - Descriptive Statistics Megapixels

Table 3 - Descriptive Statist Weight (oz.)

175.35714285714286

Mean

12.857142857142858

Mean

15.647253616191014

Standard Error

0.34776035397153426

Standard Error

160

Median

12

Median

200

Mode

12

Mode

Deviation

82.79748353919507

Standard Deviation

1.840174824912946

Standard Deviation

Variance

6855.423280423279

Sample Variance

3.386243386243391

Sample Variance

0.6634435796972946

Kurtosis

-0.6331500000000023

Kurtosis

1.056995412557892

Skewness

0.2257310608541037

Skewness

320

Range

6

Range

80

Minimum

10

Minimum

400

Maximum

16

Maximum

4910

Sum

360

Sum

28

Count

28

Count

Error

s

m

14/Case 3 - Selecting a Point and Shoot Digital Camera/BSTAT_14_Case3_Cameras_Report.docx Selecting a Point and Shoot Digital Camera Consumer Reports tested 166 different point-and-shoot digital cameras. An overall score for each camera was based on the number of megapixels, weight (oz.), image quality, ease of use, and price. The overall score ranges from 0 to 100, with higher scores indicating better overall test results. This report analyzes data collected and makes recommendations based on analysis results.

Descriptive Statistics Table 1 presents descriptive statistics for the price of point-and-shoot digital cameras. The mean is 175.36 with a standard deviation of 82.80. The minimum price is 88 and the maximum price is 400 dollars.

Table 2 presents descriptive statistics for the number of megapixels. The mean is 12.86 with a standard deviation of 1.84. The minimum number of megapixels is 10 and the maximum is 16.

Table 3 presents descriptive statistics for the weight in ounces of the camera. The mean is 5.82 with a standard deviation of .98. The minimum number of ounces is 4 and the maximum is 7.

Table 4 presents descriptive statistics for the overall score of the camera. The mean is 56.36 with a standard deviation of 6.70. The minimum score is 44 and the maximum score is 66.

An examination of correlation coefficients reveals that price, with a correlation coefficient of .683 appears to be the best predictor of the overall score.

Scatter Diagrams Figure 1 presents a scatter diagram of the linear relationship between the price of the camera and the overall score. The observation for a Nikon camera with a price of $400 appears to be an outlier observation. Consumer Reports could consider either limiting the study to cameras with a price less than $400 or consider the possibility that the relationship is in fact curvilinear.

Figure 2 presents a scatter diagram of the relationship between the megapixels and score. The number of megapixels does not appear to have much effect on the overall score. As the number of pixels increases from 10 to 14, the overall score appears to have a downward trend. This information seems to be counterintuitive in that generally speaking higher megapixel cameras are considered to have better picture quality. The overall score for the 16-megapixel cameras does increase.

Figure 3 presents a scatter diagram of the relationship between the weight and score. There appears to be a modest increase in overall score for cameras that weigh more. There is a large variability in the score for cameras with a weight of 5 and 7 ounces.

Again, price appears to be the best predictor of overall score.

Regression Analysis To further investigate the relationship between the price of the camera and the overall score, a regression analysis was performed. Table 5 presents the results of this analysis.

With a p-value = .000, price is a significant factor in predicting the overall score. The estimated regression equation explained 44.6% of the variability in the overall score. An examination of the residuals indicates that observations 17 and 28 are unusual. One recommendation would be to

remove these observations as influential and rerun the regression. The regression equation is: Score = 46.67 + .06Price This relationship indicates that for every one unit (dollar) increase in the price there is a .06 unit increase in the score.

Figure 4 presents a scatter diagram using only data for the Canon cameras. The relationship between price and overall score appears to be more curvilinear than linear. Estimated regression data are presented in Table 6. The estimated regression equation is significant and explains 65.5% (using the adjusted R-square) of the variability in the overall score. The regression equation is: Score = 47.29 + .07Price This relationship indicates that for every one unit (dollar) increase in the price there is a .07 unit increase in the score.

This report summarizes findings based on simple linear regression of the relationship between a number of variables and the overall score. Price is considered to be the best predictor of overall score. Future studies should consider eliminating potential outlier observations and considering the possibility that relationship is not linear.

14/Case 3 - Selecting a Point and Shoot Digital Camera/BSTAT_14_Case3_Cameras_Rubric.docx Selecting a Point and Shoot Digital Camera Chapter 14 (80 points total) RUBRIC

14/Case 3 - Selecting a Point and Shoot Digital Camera/BSTAT_14_Case3_Cameras_TeachingTips.docx Selecting a Point and Click Digital Camera – Teaching Tips The report for this case involves a number of tables and figures. Suggest that students create these tables around output results in Excel and use the Snipping Tool app to copy the tables/figures into the report.

Continue to remind students about Data Analysis tools for completing analysis In Excel, the Multiple R in the regression analysis is the correlation coefficient and the R squared is the coefficient of determination. Explain the reason for the adjusted R-squared measure and why it is more frequently reported.

14/Case 4 - Finding the Best Car Value/BSTAT_14_Case4_FamilySedans.xlsx Sheet: Data Car

Price ($)

Cost/Mile

Road-Test Score

Predicted Reliability

Value Score

Nissan Altima 2.5 S (4-cyl.)

23970

0.59

91

4

1.75

Kia Optima LX (2.4)

21885

0.58

81

4

1.73

Subaru Legacy 2.5i Premium

23830

0.59

83

4

1.73

Ford Fusion Hybrid

32360

0.63

84

5

1.7

Honda Accord LX-P (4-cyl.)

23730

0.56

80

4

1.62

Mazda6 i Sport (4-cyl.)

22035

0.58

73

4

1.6

Hyundai Sonata GLS (2.4)

21800

0.56

89

3

1.58

Ford Fusion SE (4-cyl.)

23625

0.57

76

4

1.55

Chevrolet Malibu LT (4-cyl.)

24115

0.57

74

3

1.48

Kia Optima SX (2.0T)

29050

0.72

84

4

1.43

Ford Fusion SEL (V6)

28400

0.67

80

4

1.42

Nissan Altima 3.5 SR (V6)

30335

0.69

93

4

1.42

Hyundai Sonata Limited (2.0T)

28090

0.66

89

3

1.39

Honda Accord EX-L (V6)

28695

0.67

90

3

1.36

Mazda6 s Grand Touring (V6)

30790

0.74

81

4

1.34

Ford Fusion SEL (V6, AWD)

30055

0.71

75

4

1.32

Subaru Legacy 3.6R Limited

30094

0.71

88

3

1.29

Chevrolet Malibu LTZ (V6)

28045

0.67

83

3

1.2

Chrysler 200 Limited (V6)

27825

0.7

52

5

1.2

Chevrolet Impala LT (3.6)

28995

0.67

63

3

1.05

14/Case 4 - Finding the Best Car Value/BSTAT_14_Case4_FamilySedans_Report.docx Finding the Best Car Value To develop car value, Consumer Reports developed a statistic referred to as a value score based upon five-year owner costs, overall road-test scores, and predicted reliability ratings. A car with a value score of 1.0 is considered to be “average-value”. A car with a value score of 2.0 is considered to be ice as good a value as a car with a value score of 1.0; a car with a value score of .5 is considered half as good as average; and so on. This report result analysis for 20 family sedans, including the price of each car tested.

Descriptive Statistics Table 1 presents descriptive statistics for the price of family sedans. The mean is 26,886 with a standard deviation of 3374.28. The minimum price is 21,800 and the maximum price is 32,360 dollars.

Table 2 presents descriptive statistics for the cost/mile (a measure of 5-year owner costs). The mean is .642 with a standard deviation of .061. The minimum cost/mile is .56 and the maximum is .74.

Table 3 presents descriptive statistics for the road-test scores of the family sedans. The mean is 80.45 with a standard deviation of 9.90. The minimum road-test score is 52 and the maximum is 93.

Table 4 presents descriptive statistics for the predicted reliability of the family sedan. The mean is 3.75 with a standard deviation of .64. The minimum reliability is 3 and the maximum reliability is 5.

Table 5 presents descriptive statistics for the value score of the family sedans. The mean is 1.46 with a standard deviation of 0.20. The minimum score is 1.05 and the maximum score is 1.75.

Regression Analysis To further investigate the relationship between the price of the car and the value score, a regression analysis was performed. Table 6 presents the results of this analysis.

With a p-value = .0083, price is a significant factor in predicting the value score. The estimated regression equation explained 29.03% (using the adjusted R square) of the variability in the value score. The regression equation is: Value Score = 2.36 - .000033Price This relationship indicates that for every one unit (dollar) increase in the price there is a .000033 unit decrease in the score. The price of the car has very little influence on the value score.

To further investigate the relationship between the cost/mile of the car and the value score, a regression analysis was performed. Table 7 presents the results of this analysis.

With a p-value = .00038, cost/mile is a significant factor in predicting the value score. The estimated regression equation explained 48.61% (using the adjusted R square) of the variability in the value score. The regression equation is: Value Score = 2.94 – 2.31Cost/Mile This relationship indicates that for every one unit (dollar) increase in the cost/mile there is a 2.31 unit decrease in the score. Increased cost/mile results in a decreasing value score, a moderate negative relationship.

To further investigate the relationship between the road test of the car and the value score, a regression analysis was performed. Table 8 presents the results of this analysis.

With a p-value = .07, road test is NOT a significant factor in predicting the value score. The estimated regression equation explained only 12.32% (using the adjusted R square) of the variability in the value score. The road test score is not of value in predicting the value score.

To further investigate the relationship between the predicted reliability of the car and the value score, a regression analysis was performed. Table 9 presents the results of this analysis.

With a p-value = .13, predicted reliability is NOT a significant factor in predicting the value score.

The estimated regression equation explained only 7.4%% (using the adjusted R square) of the variability in the value score. Predicted reliability is not of value in predicting the value score.

This report summarizes findings based on simple linear regression of the relationship between a number of variables and the value score associated with family sedans. Consumer Reports did not include the price as one of the components of value score; however, there is a significant statistical relationship between price and value score. A review of remaining regression output indicates that cost/mile is the best single predictor of value score. To further investigate the relationship among these variables requires the use of multiple regression analysis.

14/Case 4 - Finding the Best Car Value/BSTAT_14_Case4_FamilySedans_Rubric.docx Finding the Best Car Value Chapter 14 (90 points total) RUBRIC

14/Case 4 - Finding the Best Car Value/BSTAT_14_Case4_FamilySedans_TeachingTips.docx Finding the Best Car Value – Teaching Tips The report for this case involves a number of tables and figures. Suggest that students create these tables around output results in Excel and use the Snipping Tool app to copy the tables/figures into the report.

Continue to remind students about Data Analysis tools for completing analysis In Excel, the Multiple R in the regression analysis is the correlation coefficient and the R squared is the coefficient of determination. Explain the reason for the adjusted R-squared measure and why it is more frequently reported.

14/Case 5 - Buckeye Creek Amusement Park/BSTAT_14_Case5_BuckeyeCreek.xlsx Sheet: Data

opulation

Season Pass Holders

4171

224

7576

42

3437

15

731

78

952

19

913

28

3349

83

5713

75

1353

69

5105

83

411

9

233

8

566

36

193

63

1043

207

1596

133

Mean

15738.218543046358

Mean

128.264

1683

102

Standard Error

1028.5914706961541

Standard Error

11.8306

642

36

Median

12675

Median

59

603

42

Mode

#N/A

Mode

11

028

55

Standard Deviation

12639.543601409052

Standard Deviation

145.377

2356

207

Sample Variance

159758062.4519205

Sample Variance

21134.6

5849

193

Kurtosis

1.612577218914491

Kurtosis

1.45454

913

41

Skewness

1.306203371020306

Skewness

1.42891

1137

86

Range

61076

Range

652

8236

424

Minimum

1227

Minimum

5

3968

342

Maximum

62303

Maximum

657

6485

269

Sum

2376471

Sum

19368

5868

236

Count

151

Count

151

941

111

029

84

2737

564

Table 1 - Descriptive Statistics Population

Table 2 - Descriptive Statistics Season Pass Holders

ders

Sheet: Regression

SS

MS

F

Significance F

2020777.2545030734

2020777.2545030734

261.954982231383

1.217548793303831e-34

1149418.149470439

7714.215768257979

3170195.4039735124

Standard Error

t Stat

P-value

Lower 95%

Upper 95%

11.43776639748799

-1.4214634557843062

0.157271725387927

-38.85954407459384

6.3428101749400

0.0005673732668196426

16.18502339298225

1.2175487933039003e-34

0.008061812544124492

0.0103040866479

Residuals

Standard Residuals

110.12678822450457

1.258054854234706

-103.14115514997309

-1.178253111706803

-92.13292677201039

-1.05249846680435

41.63088281498656

0.4755785132378555

-56.13034742985046

-0.6412159765346831

-19.223363607519325

-0.2196018452100579

-23.324827207559878

-0.266455714960432

-53.03332005257167

-0.6058364799698644

14/Case 5 - Buckeye Creek Amusement Park/BSTAT_14_Case5_BuckeyeCreek_Report.docx Buckeye Creek Amusement Park Buckeye Creek Amusement Park, open from the beginning of May to the end of October, relies heavily on the sale of season passes. The director of marketing at Buckeye Creek has been asked to develop a targeted marketing campaign to increase season pass sales. This report analyses last season data of season pass holders for each zip code within 50 miles of Buckeye Creek using total population data of each zip code. If the number of season pass holders in a zip code can be predict, a direct mail campaign could target zip codes with fewer than expected season pass holders.

Descriptive Statistics Table 1 presents descriptive statistics for the population for zip codes within 50 miles of Buckeye Creek. The mean is 15,738 with a standard deviation of 12639.5. The minimum population is 1227 and the maximum population is 62303 dollars.

Table 2 presents descriptive statistics for the number of season pass holders in reported zip codes.

The mean is 128.3 with a standard deviation of 145.4. The minimum number is 5 and the maximum is 657.

Figure 1 presents a scatter diagram of the relationship between the population of a zip code and the number of season pass holders. There appears to be a positive linear relationship between the population of a zip code and the number of season pass holders.

Regression Analysis To further investigate the relationship between the population of a zip code and the number of season pass holders, a regression analysis was performed. Table 3 presents the results of this analysis.

With a p-value = 1.21755E-34, population is a significant factor in predicting the number of season pass holders. The estimated regression equation explained 63.50% (using the adjusted R square) of the variability in the number of season pass holders; a reasonably good fit. The regression equation is: Value Score = -16.3 + .009183Population This relationship indicates that for every one unit (person) increase in the population there is a .009 unit increase in the score. This translates to approximately 9 new season passholders for every 1000 person increase in population.

Residual Analysis In order to check the model assumptions for the validity of this inference, we examine the residuals.

There are a number of unusual observations and observations with large residuals. Figure 2 presents a plot of the residuals against the predicted number of season pass holders. The assumption of constant variances appears to be violated; larger predicted values exhibit greater variability. This information suggests consideration of a curvilinear or multiple regression model.

Conclusions The model provides a reasonable fit (63.50% adjusted r-squared) and could be used to guide the marketing campaign. The scatter diagram in Figure 1 shows the estimated regression line and the data. Any data below the estimated regression line has fewer observed number of season pass holders than is estimated by the regression line. These zip codes are good targets for the direct

mail campaign.

Other independent variables could be considered for a multiple regression model. These variables include the distance of the zip code from the park, the average household income of the zip code, and the average number of children per household in the zip code.

14/Case 5 - Buckeye Creek Amusement Park/BSTAT_14_Case5_BuckeyeCreek_Rubric.docx Buckeye Creek Amusement Park Chapter 14 (65 points total) RUBRIC

14/Case 5 - Buckeye Creek Amusement Park/BSTAT_14_Case5_BuckeyeCreek_TeachingTips.docx Buckeye Creek Amusement Park – Teaching Tips The report for this case involves a number of tables and figures. Suggest that students create these tables around output results in Excel and use the Snipping Tool app to copy the tables/figures into the report.

Continue to remind students about Data Analysis tools for completing analysis In Excel, the Multiple R in the regression analysis is the correlation coefficient and the R squared is the coefficient of determination. Explain the reason for the adjusted R-squared measure and why it is more frequently reported.

In the residual analysis, call attention to the standardized residuals with absolute values greater than 2 as possible influential observations.

15/Case 1 - Consumer Research Inc/BSTAT_15_Case1_ConsumerResearchInc_Report.docx Consumer Research Inc. Consumer Research, Inc. is an independent agency that conducts research on consumer attitudes and behaviors for a variety of firms. This report investigates consumer characteristics that can be used to predict the amount charged by credit card users. Data included are annual income, household size, and annual credit card charges for a sample of 50 consumers.

Descriptive Statistics Table 1 presents descriptive statistics for the income independent variable. Mean income is $43,480. The standard deviation is 14.55. Minimum income is 21,000 and maximum income is 67,000.

Table 2 presents descriptive statistics for the household size independent variable. Mean household size is 3.42. The standard deviation is 1.74. Minimum household size is 1 and maximum household size is 7.

Table 3 presents descriptive statistics for the amount charged dependent variable. Mean amount charged is $3,964.06. The standard deviation is 933.49. Minimum amount charged is 1864 and maximum amount charged is 5678.

Figure 1 presents a scatter diagram of the relationship between income and amount charged.

There appears to be a positive relationship between income and amount charged. The R2 is .398 indicating a small positive relationship between income and amount charged.

Figure 2 presents a scatter diagram of the relationship between household size and amount charged. There appears to be a positive relationship between household size and amount charged.

The R2 is .567 indicating a moderate positive relationship between income and amount charged.

Income Regression Analysis In order to test statistically for the relationship between income, household size, and amount charged two simple linear regressions were performed and a multiple regression.

Simple Regression – Income Table 4 presents results of a simple linear regression using income as the independent variable and amount charged as the dependent variable.

The adjusted R square indicates that approximately 38.6% of the variation in amount charged is explained by the income. The correlation coefficient of .63 indicates a moderate strong positive relationship. The relationship is described by the following linear equation: Amount Charged = 2204 + 40.48Income This relationship indicates that for every thousand dollar increase in income amount charged increases by $40.48.

Simple Regression – Household Size Table 5 presents results of a simple linear regression using household size as the independent variable and amount charged as the dependent variable.

The adjusted R square indicates that approximately 55.8% of the variation in amount charged is explained by household size. The correlation coefficient of .75 indicates a moderate strong positive relationship, a relationship stronger than that of income. The relationship is described by the following linear equation: Amount Charged = 2581.94 + 404.13Household Size This relationship indicates that for every additional person in the family (a one unit increase in household size) the amount charged is increased by $404.13.

Multiple Regression Table 6 presents results of a multiple regression using income and household size as the independent variables and amount charged as the dependent variable.

The adjusted R square indicates that approximately 81.8% of the variation in amount charged is explained by income and household size. The correlation coefficient of .91 indicates a strong positive relationship. The relationship is described by the following linear equation: Amount Charged = 1304.90 + 33.13Income + 356.30Household Size This relationship indicates that for every additional person in the family if income is held constant the amount charged is increased by $356.30 and for every additional thousand dollars in income if household size is held constant amount charged is increased by 33.13.

The multiple regression explains a high percentage of the variability in the dependent variable (amount charged). Using the regression equation to predict annual credit card charge for a household of size 3 with an income of 40,000 yields a predicted charge of $3,699.11. Residual analysis output identifies three observations as having a large standardized residual (value greater than the absolute value of 2). These observations should be treated as possible outliers.

Discussion This report summarizes findings of an investigation of consumer characteristics that can be used to predict the amount charged by credit card users including annual income, household size, and annual credit card. Further analysis could include additional independent variables including: amount of savings, debt to equity ratio, number of pets, and average age of household members.

15/Case 1 - Consumer Research Inc/BSTAT_15_Case1_ConsumerResearchInc_Rubric.docx Wentworth Medical Center Chapter 13 (85 points total) RUBRIC

15/Case 1 - Consumer Research Inc/BSTAT_15_Case1_ConsumerResearchInc_TeachingTips.docx Compensation for Sales Professionals – Teaching Tips The report for this case involves a number of tables and figures. Suggest that students create these tables around output results in Excel and use the Snipping Tool app to copy the tables/figures into the report.

Descriptive Statistics Make sure students use the Data Analysis Toolkit and run the Descriptive Statistics tool. There is no default automatically computed. If the student fails to choose one of the available check boxes, the following dialog box appears.

Scatter Diagrams Remind students to include axis titles in scatter diagrams.

Scatter diagrams can be used to quickly ascertain degree of linear relationships by adding the trend line and R-squared value. Suggest that students use these in their future analyses to determine if they want to continue with regression based on these initial results.

Regression Analysis In Excel, the Multiple R in the regression analysis is the correlation coefficient and the R squared is the coefficient of determination. Explain the reason for the adjusted R-squared measure and why it is more frequently reported.

15/Case 1 - Consumer Research Inc/BSTAT_15_Case1_Consumer_Key.xlsx Sheet: Data

Table 1 - Descriptive Statistics Income ($1000s)

Table 2 - Descriptive Statistics Household Size

Table 3 - Descr Amount Charge

Mean

43.48

Mean

3.42

Mean

Standard Error

2.0577856138706623

Standard Error

0.2459301377486322

Standard Error

Median

42

Median

3

Median

Mode

54

Mode

2

Mode

Standard Deviation

14.550741617960679

Standard Deviation

1.7389886810019957

Standard Devia

Sample Variance

211.72408163265297

Sample Variance

3.0240816326530604

Sample Varianc

Kurtosis

-1.247719422087707

Kurtosis

-0.722808552447856

Kurtosis

Skewness

0.09585563857075645

Skewness

0.5278959770849904

Skewness

Range

46

Range

6

Range

Minimum

21

Minimum

1

Minimum

Maximum

67

Maximum

7

Maximum

Sum

2174

Sum

171

Sum

Count

50

Count

50

Count

Confidence Level(95.0%)

4.1352750129552724

Confidence Level(95.0%)

0.49421511488343395

Confidence Lev

Sheet: Simple Regression - Income

4737

8059

67687

435

SS

MS

F

Significance F

16999744.785958763

16999744.785958763

31.751232388314136

9.012482330632816e-07

25699404.034041222

535404.2507091921

42699148.819999985

Standard Error

t Stat

P-value

Lower 95%

Upper 95%

1222

329.04892814487334

6.698090860590032

2.1358810835413103e-08

1542.4024064070418

2865.596830181202

165

7.183857988059566

5.634823190510431

9.012482330632797e-07

26.035655023822606

54.9238841530807

Sheet: Simple Regression - Size

53589

6992

75543

7835

SS

MS

F

Significance F

24200717.481357794

24200717.481357794

62.79637542447097

2.864949751488129e-10

18498431.33864219

385383.9862217123

42699148.819999985

Standard Error

t Stat

P-value

Lower 95%

Upper 95%

1198

195.26257949093934

13.222917695814862

1.2795954740305425e-17

2189.339288493351

2974.542746869045

1585

50.99786994284894

7.924416409078404

2.8649497514880767e-10

301.5902675312663

506.6664472750506

Sheet: Multiple Regression

Amount Charged = 1304.90 + 33.13I Predicted charge =

F

Significance F

111.21764679933152

1.508756981100993e-18

P-value

Lower 95%

Upper 95%

Lower 95.0%

36785247879

3.286637325190513e-08

907.2745236165953

1702.5350346146279

907.2745236165953

50854139272

7.682061406164437e-11

25.15061221111238

41.1154060863031

25.15061221111238

51643710967

3.123421167496439e-14

289.5043250993967

423.08747788222655

289.5043250993967

377.83624059

6.44994720863

Sheet: Residual Analysis

SS

MS

F

Significance F

35250755.67248118

17625377.83624059

111.21764679933152

1.508756981100993e-18

7448393.147518806

158476.44994720863

42699148.819999985

Standard Error

t Stat

P-value

Lower 95%

Upper 95%

197.65484305021513

6.601936785247879

3.286637325190513e-08

907.2745236165953

1702.5350346146279

3.9679058422877795

8.350250854139272

7.682061406164437e-11

25.15061221111238

41.1154060863031

33.200890440677846

10.73151643710967

3.123421167496439e-14

289.5043250993967

423.08747788222655

Residuals

Standard Residuals

-146.97497761826435

-0.37697291408755046

147.51314344153343

0.37835324386849445

1309.655322162494

3.3591063679434994

-1.0347440050563819

-0.0026539923274140637

-1180.6198657071745

-3.028146140367086

230.18791472383964

0.590403961322999

-156.1220191086095

-0.40043396127920955

15/Case 2 - Predicting Winnings for NASCAR Drivers/BSTAT_15_Case2_NASCAR_Key.xlsx Sheet: Simple Regression - Poles

26

687

77

04

55

SS

MS

F

Significance F

9315165014055.352

9315165014055.352

6.516542743240377

0.015498222284203007

47172320903241.78

1429464269795.2056

56487485917297.13

Standard Error

t Stat

P-value

Lower 95%

Upper 95%

L

266797.0953847961

15.970453530500576

4.3485899053363053e-17

3718067.842039997

4803673.385790834

3

184736.9159992138

2.5527519940723598

0.015498222284202784

95737.44909175689

847437.612299785

9

Sheet: Simple Regression - Wins

2

1

44

SS

MS

F

Significance F

24722535670037.87

24722535670037.87

25.683770028307922

1.5100803165027288e-05

31764950247259.26

962574249916.9473

3

56487485917297.13

Standard Error

t Stat

P-value

Lower 95%

Upper 95%

205150.13955216663

19.95357019445653

5.493599270906838e-20

3676096.6127639594

4510858.807149459

120766.13073239927

5.0679157479488515

1.5100803165027149e-05

366332.0353687429

857733.1161464092

Sheet: Simple Regression - Top Five

766

28

319

79

SS

MS

F

Significance F

41891672484386.555

41891672484386.555

94.7138162829397

3.181514222684686e-11

14595813432910.58

442297376754.8661

56487485917297.13

Standard Error

t Stat

P-value

Lower 95%

Upper 95%

L

38

164420.73130367548

21.516600814414176

5.420202614580496e-21

3203258.748020103

3872291.734130373

3

36

23461.331681825046

9.732102356784978

3.181514222684663e-11

180595.6431487977

276060.5195592095

1

481

22

256

57

Sheet: Simple Regression - Top Ten

SS

MS

F

Significance F

45526971887647.836

45526971887647.836

137.0729573657095

2.7120190492125313e-13

10960514029649.295

332136788777.25134

56487485917297.13

Standard Error

t Stat

P-value

Lower 95%

Upper 95%

2

171768.9285623709

17.751503059847888

1.8913282318778907e-18

2699690.148175094

3398623.173748346

79

13831.274123635718

11.707816080111177

2.7120190492124444e-13

133794.07480537583

190073.9523808799

Sheet: Multiple Regression

896

692

718

1

SS

MS

F

Significance F

46347328154801.81

11586832038700.453

34.280034818262415

8.619415056129453e-11

10140157762495.324

338005258749.8441

56487485917297.14

Standard Error

t Stat

P-value

Lower 95%

Upper 95%

493

184229.02428764632

17.0459953259994

5.594538579989538e-17

2764121.2249667374

3516612.948874561

8348

107205.07514534681

-0.12069317412814626

0.9047388015650049

-231880.89294698453

206003.0513431078

7

111226.21629834718

0.12177715954222364

0.9038875697943726

-213609.42527727655

240699.0506521599

6

50666.86770704488

1.4137324157003694

0.16773416250212375

-31846.155085704813

175104.941644608

633

33432.88381531903

3.501659547861806

0.001470314344780265

48791.51907372085

185349.6345752118

Sheet: Data Top 10

Top2-5

Top6-10

Winnings ($)

19

4

10

6529870

26

18

7

8485990

19

5

10

6197140

20

9

8

6183580

14

7

4

5087740

21

12

7

6296360

12

4

8

4163690

18

10

5

5912830

14

4

9

5401190

17

8

8

5303020

16

6

8

5936470

18

10

4

6161020

16

3

12

5633950

15

7

7

4775160

10

1

9

4825560

10

3

7

4318050

8

3

4

3853690

12

3

9

3955560

Poles

1

12

4

7

4750390

Wins

0.13313526567544892

1

5

2

3

3807780

Top 5

0.4373121802104965

0.7251702533340862

1

8

2

6

5020780

Top 10

0.45781177278353896

0.6971780940326693

0.90174

10

2

8

3830910

Winnings ($)

0.40608706224995345

0.6615617434550297

0.86116

8

3

4

4203660

6

4

2

3856010

7

3

4

4301880

5

1

3

4579860

4

2

2

4794770

3

1

2

4374770

2

1

1

4505650

2

1

1

3878390

0

0

0

2838320

Table 1 Correlation Coefficients Poles

Sheet: Final Regression - wo Poles

Wins

Top 5

4

5

6

SS

MS

F

Significance F

46342404485507.414

15447468161835.805

47.20233309477055

1.1467678762393771e-11

10145081431789.717

327260691348.05536

56487485917297.13

Standard Error

t Stat

P-value

Lower 95%

Upper 95%

2

180327.47547272596

17.402195507981027

1.4925077415348823e-17

2770313.672655687

3505874.2946183733

2

86426.80872966659

2.368888720394348

0.02425383823703342

28466.853806101426

381004.1308726724

3

30210.466904432138

6.182560554456236

7.306683583166837e-07

125163.38754154471

248392.69448855074

2

32102.07395461106

3.6193541122065622

0.0010389509133533407

50716.16189033954

181661.3848656219

Sheet: Data2

5 and Top6-10

0.9058081593252896 0.8204884215002692 0.7965535443669718

581382.1967947109 35

df

SS

MS

F

Significance

4

46347328154801.81

11586832038700.453

34.280034818262436

8.61941505

30

10140157762495.318

338005258749.84393

34

56487485917297.13

Coefficients

Standard Error

t Stat

P-value

Lower 95%

3140367.086920651

184229.0242876463

17.045995325999414

5.594538579989418e-17

2764121.22

-12938.920801938435

107205.07514534674

-0.12069317412814716

0.9047388015650042

-231880.89

202244.78279135964

90225.86833906137

2.2415387794478234

0.03254153484745904

17978.9770

188699.97010391767

34586.32225625556

5.455913141206793

6.430280756385385e-06

118065.276

117070.57682446629

33432.88381531903

3.501659547861805

0.0014703143447802674

48791.5190

Predicted Winnings ($)

Residuals

6064157.728735844

465712.2712641563

7519888.606947978

966101.393052022

6063551.836850341

133588.1631496586

6343149.018419905

-159569.01841990463

5523344.6125180805

-435604.6125180805

15/Case 2 - Predicting Winnings for NASCAR Drivers/BSTAT_15_Case2_NASCAR_Report.docx Predicting Winnings for NASCAR Drivers Can NASCAR data be used to predict winnings? Data for the 2011 NASCAR season for the top 35 drivers was collected. NASCAR results for the 2011 season include points, poles, wins, top 5, top 10, and winnings. Points are awarded based on finishing position, leading a lap, leading the most laps, and bonus points for winning. This report uses regression to predict winnings based on number of poles won, number of wins, number of top five finishes, and the number of top ten finishes.

Best Single Predictor of Winnings Table 1 presents correlation coefficients for the NASCAR variables. The variable most highly correlated with Winnings ($) is the number of Top 10 finishes (.8977), indicating a very strong positive relationship.

Figure 1 presents a scatter diagram of the relationship between Winnings and the number of Top 10 finishes. There appears to be a positive relationship between winnings and top 10. The R2 is .806 indicating that approximately 80% of the variability in winnings can be explained by the linear relationship between winnings and top 10.

Table 2 presents results of a simple linear regression using Top 10 as the independent variable and Winnings ($) as the dependent variable. The adjusted R square reiterates that approximately 80% of the variability in winnings can be explained by the linear relationship. The relationship is described by the following linear equation: Winnings ($) = 3,049,156.67 + 161,934.01Top 10 This relationship indicates that for every additional top 10 finish, winnings increase by 161,934.01.

Regression Analysis – Given Data A multiple regression was executed to estimate a regression equation that can be used to predict Winnings ($). Independent variables included were the number of poles won, number of wins, number of top five finishes, and the number of top ten finishes. Table 3 presents the results of this analysis.

The regression equation is Winnings ($) = 3140367 – 12939Poles + 13545Wins + 71629Top 5 + 117071Top 10 Examination of the p-values corresponding to the t values for each of the independent variables reveals that the only significant variable is Top 10 with a p-value of .001. Comparison of the adjusted r-square from the simple regression (.80) and the adjusted r-square for the multiple regression (.797) indicate that adding Poles, Wins, and Top 5 to the model as independent variables added little to the model’s ability to explain variation in Winnings.

Regression Analysis – Newly Created Variables (Top 2-5, Top 6-10) A multiple regression was executed to estimate a regression equation that can be used to predict Winnings ($). Two new independent variables replaced the Top 5 and Top 10 variables. Top 2-5 represents the number of times the driver finished between second and fifth place. Top 6-10 represents the number of times the driver finished between sixth and tenth place. Independent variables included were Poles, Wins, Top 2-5, and Top 6-10. Table 4 presents the results of this analysis.

The regression equation is Winnings ($) = 3140367 – 12939Poles + 202245Wins + 188700Top 5 + 117071Top 10 Examination of the p-values corresponding to the t values for each of the independent variables reveals that the only variable that is not significant variable is Poles with a p-value of .905.

Replacing the Top 5 with Top2-5 and Top 10 with Top6-10 greatly reduced multicollinearity in the model since Top 5 and Top 10 both included Win values as well. The adjusted r-square remains approximately the same (.80).

Having removed issues of multicollinearity, a final model was estimated removing Poles from the regression analysis. Results are presented in Table 5.

The regression equation is Winnings ($) = 3138094 + 204735Wins + 186778Top 5 + 116189Top 10 Examination of the p-values corresponding to the t values for each of the independent variables reveals that all variables are significant.

Discussion This report establishes the best linear model for predicting Winnings ($) using 2011 NASCAR results. Interpretation of the variable coefficients is summarized below.

Wins: An estimate of the expected increase in total winnings corresponding to having one more win, when all other independent variables are held constant is $204, 735.

Top 2-5: An estimate of the expected increase in total winnings corresponding to having one more Top 2-5 finish, when all other independent variables are held constant is $186,778.

Top 6-10: An estimate of the expected increase in total winnings corresponding to having one more Top 6-10 finish, when all other independent variables are held constant is $116,189.

15/Case 2 - Predicting Winnings for NASCAR Drivers/BSTAT_15_Case2_NASCAR_Rubric.docx Predicting Winnings for NASCAR Drivers Chapter 15 ( points total) RUBRIC

15/Case 2 - Predicting Winnings for NASCAR Drivers/BSTAT_15_Case2_NASCAR_TeachingTips.docx Predicting Winnings for NASCAR Drivers – Teaching Tips The report for this case involves a number of tables and figures. Suggest that students create these tables around output results in Excel and use the Snipping Tool app to copy the tables/figures into the report.

Regression Analysis It might be worth calling attention to the comparison of adjusted r-squares for models where statistically insignificant variables have been removed. It should make since to the student that since the variable was not significant it had little effect on the coefficient of determination.

In Excel, the Multiple R in the regression analysis is the correlation coefficient and the R squared is the coefficient of determination. Explain the reason for the adjusted R-squared measure and why it is more frequently reported.

Multicollinearity Ask students if they see where the potential of multicollinearity lies in this regression equation.

Winnings ($) = 3140367 – 12939Poles + 13545Wins + 71629Top 5 + 117071Top 10 Alternately, use the regression coefficients table and reference the rule of thumb on page 705.

15/Case 3 - Finding the Best Car Value/BSTAT_15_Case3_CarValues_Key.xlsx Sheet: Data Size

Family Sedan

Upscale Sedan

Price ($)

Cost/Mile

Road-Test Score

Predicted Reliability

Value

a Corolla (base, manual)

Small Sedan

0

0

16419

0.44

70

4

1.99

3 i Touring (manual)

Small Sedan

0

0

18895

0.5

74

5

1.94

a Corolla LE

Small Sedan

0

0

18404

0.47

71

4

1.89

3 i Touring

Small Sedan

0

0

19745

0.52

70

5

1.82

ai Elantra GLS

Small Sedan

0

0

18445

0.53

80

3

1.64

n Sentra 2.0 SL

Small Sedan

0

0

20150

0.57

74

4

1.51

rte Sedan EX

Small Sedan

0

0

19040

0.57

71

3

1.32

ocus SE

Small Sedan

0

0

20280

0.52

68

2

1.3

iesta SE

Small Sedan

0

0

16595

0.47

61

2

1.25

wagen Jetta SE (2.5)

Small Sedan

0

0

20300

0.54

60

3

1.24

wagen Jetta TDI

Small Sedan

0

0

25100

0.5

68

2

1.18

olet Cruze LS (1.8)

Small Sedan

0

0

18375

0.57

67

1

0.96

olet Cruze 1LT (1.4T)

Small Sedan

0

0

20530

0.6

69

1

0.91

n Altima 2.5 S (4-cyl.)

Family Sedan

1

0

23970

0.59

91

4

1.75

ptima LX (2.4)

Family Sedan

1

0

21885

0.58

81

4

1.73

u Legacy 2.5i Premium

Family Sedan

1

0

23830

0.59

83

4

1.73

usion Hybrid

Family Sedan

1

0

32360

0.63

84

5

1.7

Accord LX-P (4-cyl.)

Family Sedan

1

0

23730

0.56

80

4

1.62

6 i Sport (4-cyl.)

Family Sedan

1

0

22035

0.58

73

4

1.6

ai Sonata GLS (2.4)

Family Sedan

1

0

21800

0.56

89

3

1.58

usion SE (4-cyl.)

Family Sedan

1

0

23625

0.57

76

4

1.55

olet Malibu LT (4-cyl.)

Family Sedan

1

0

24115

0.57

74

3

1.48

ptima SX (2.0T)

Family Sedan

1

0

29050

0.72

84

4

1.43

usion SEL (V6)

Family Sedan

1

0

28400

0.67

80

4

1.42

n Altima 3.5 SR (V6)

Family Sedan

1

0

30335

0.69

93

4

1.42

ai Sonata Limited (2.0T)

Family Sedan

1

0

28090

0.66

89

3

1.39

Accord EX-L (V6)

Family Sedan

1

0

28695

0.67

90

3

1.36

6 s Grand Touring (V6)

Family Sedan

1

0

30790

0.74

81

4

1.34

usion SEL (V6, AWD)

Family Sedan

1

0

30055

0.71

75

4

1.32

u Legacy 3.6R Limited

Family Sedan

1

0

30094

0.71

88

3

1.29

olet Malibu LTZ (V6)

Family Sedan

1

0

28045

0.67

83

3

1.2

Sheet: Costpermile-Dependent (1)

43304711

28643714

19963075

9864042296

SS

MS

F

Significance F

0.43253641025641026

0.21626820512820513

79.8883940510505

1.9300054810608024e-16

0.13806358974358962

0.00270712921065862

0.5705999999999999

Standard Error

t Stat

P-value

Lower 95%

Upper 9

30769233

0.014430547755099296

36.247891067896894

4.4860921495272285e-38

0.49410638960701836

0.55204

692307683

0.01853637421517844

6.415660125468179

4.55556827421033e-08

0.08170975288818366

0.15613

025641022

0.018361689010150156

12.540045206578045

3.337176053086406e-17

0.19339378140404614

0.26711

Sheet: Value Score - Dependent (2,3,4)

Mile, Road-Test, Predicted Reliability, Type of Sedan 0.9671162522810983 0.9353138454262371 0.9285757043248034 0.07198758214875287 54

df

SS

MS

5

3.5966853062675566

0.7193370612535113

48

0.24874617521392528

0.005182211983623443

53

3.845431481481482

Coefficients

Standard Error

t Stat

1.3709586723452776

0.1396617601004339

9.816278066089032

0.022778289445574976

0.03798910853951438

0.5996005255527945

0.06811050163916937

0.05370603653252463

1.2682094236822212

-2.2658805533579875

0.19384057991765594

-11.689402468361065

0.011133265173551921

0.0013132890038119313

8.477391603247028

0.16620991206739313

0.01043340599916072

15.930551545752493

Mile, Road-Test Score, Predicted Reliability 0.9655874609243954 0.9323591446944207 0.9283006933760859 0.07212603890779093 54

Sheet: Value Score - Road Test (5)

949622487

883760355

5930634383

750224963

SS

MS

F

Significance F

0.14471807945491744

0.14471807945491744

2.033483632516563

0.15984362452676995

3.7007134020265644

0.07116756542358778

3.845431481481482

Standard Error

t Stat

P-value

Lower 95%

Upper 95%

88278792

0.31868052264249935

2.830538595042404

0.00659109611783543

0.2625582662319196

1.5415167

62768725156

0.004055200157338251

1.42600267619544

0.1598436245267705

-0.0023546281627359325

0.0139200

Sheet: Value Score - Pred Rel (6)

6886440576893

3350704888266

0722833828425

11121091035673

SS

MS

F

Significance F

1.5932971239395828

1.5932971239395828

36.787969673038006

1.51818680561319e-07

2.252134357541899

0.04331027610657498

3.845431481481482

cients

Standard Error

t Stat

P-value

Lower 95%

Upper

9329608938549

0.10140594713669207

7.5235524388470525

7.250260588559289e-10

0.5594470410577574

0.9664

32402234636868

0.028576290310871916

6.0653087038532485

1.518186805613209e-07

0.11598150069354232

0.2306

15/Case 3 - Finding the Best Car Value/BSTAT_15_Case3_CarValues_Report.docx Finding the Best Car Value To measure car value, Consumer Reports developed a statistic referred to as a value score. The value score is based upon five-year owner costs overall road-test sores, and predicted-reliability ratings. A car with a value score of 1.0 is considered to be an “average-value” car. A car with a value score of 2.0 is considered to be twice as good a value as a car with a value score of 1.0. Data for three sizes of cars (small sedans, family sedans, and upscale sedans) is included along with the car price. This report considers a number of regression analyses designed to determine the best combination for predicting car value.

Regression Results – Relationship Between Type of Sedan and Cost/Mile Table 1 presents results of a linear regression using Family-Sedan and Upscale-Sedan dummy variables as the independent variable and cost/mile as the dependent variable. Both independent variables are significant at a .05 level of significance.

The adjusted R square indicates that approximately 74.9% of the variation in cost/mile is explained by the type of sedan. The relationship is described by the following linear equation: Cost/mile = .52 + .12FamilySedan + .23UpscaleSedan This relationship indicates that a family sedan increases the cost/mile by .12 and an upscaled sedan increases the cost/mile by .23. A small sedan adds no additional cost to the cost/mile variable.

Regression Results – Relationship Between Value Score and Type of Sedan, Cost/Mile, Road-Test Score, and Predicted Reliability Table 2 presents results of a linear regression using type of sedan, cost/mile, road-test score, and predicted reliability as the independent variables and value score as the dependent variable.

The adjusted R square indicates that approximately 92.9% of the variation in value score is explained by the independent variables. Examination of the p-values for each independent variable indicate that the Family Sedan and Upscale Sedan variables are not significant in the presence of the other independent variables and should be removed from the regression.

Table 3 presents results of a linear regression using cost/mile, road-test score, and predicted reliability as the independent variables and value score as the dependent variable.

The adjusted R square indicates that approximately 92.8% of the variation in value score is explained by these independent variables. All independent variables are significant to the relationship. The relationship is described by the following linear equation: Value-score = 1.244 - 2.04Cost/Mile + .011Road-Test Score + .17Predicted Reliability This equation indicates that cost/mile has a -2.04 per unit effect on value-score when road-test score and predicted reliability are unchanged; road-test score has a .011 per unit effect on value-score when cost/mile and predicted reliability are unchanged; predicted reliability has a .17 per unit effect on value-score when cost/mile and road-test score are unchanged.

Table 4 presents results of a simple linear regression using road-test score as the independent variable and value score as the dependent variable.

The adjusted R square indicates that only 1.9% of the variation in value score is explained by road-test score. While significant, the road-test variable alone does not explain the variability in value score.

Table 5 presents results of a simple linear regression using predicted reliability as the independent variable and value score as the dependent variable.

The adjusted R square indicates that only 40.3% of the variation in value score is explained by predicted liability. While significant, the predicted reliability variable alone does not explain the variability in value score.

Discussion Regression results in this analysis indicate that type of sedan explains 74.9% of the variability in cost/mile. When the cost/mile, predicted reliability, and road-test scores are included with sedan values in predicting value score, 92.9% of the variability is explained but the sedan variables are not significant. Removal of the sedan variables only alters variability explained by .1% to 92.8%.

Clearly, the cost/mile value takes in to account the type of sedan. When tested in a simple linear regression to predict value score, neither the predicted reliability or road-test score alone explain a significant amount of variability.

As a result of analyses of these regression models, the regression equation that best describes the relationship is Value-score = 1.244 - 2.04Cost/Mile + .011Road-Test Score + .17Predicted Reliability

15/Case 3 - Finding the Best Car Value/BSTAT_15_Case3_CarValues_Rubric.docx Finding the Best Car Values Chapter 15 (85 points total) RUBRIC

15/Case 3 - Finding the Best Car Value/BSTAT_15_Case3_CarValues_TeachingTips.docx Finding the Best Car Value – Teaching Tips The report for this case involves a number of tables and figures. Suggest that students create these tables around output results in Excel and use the Snipping Tool app to copy the tables/figures into the report.

Regression Analysis In Excel, the Multiple R in the simple linear regression analysis is the correlation coefficient. The multiple R value does not have the same meaning in a multiple regression. The R squared is the coefficient of determination. Explain the reason for the adjusted R-squared measure and why it is more frequently reported.