Developing Database Query Proficiency: Assuring Compliance for

Journal of Information SystemsVol. 15, No. 1Spring 2001pp. 35–56

Developing Database Query Proficiency:Assuring Compliance for Responses to

Web Site ReferralsA. Faye BorthickDonald R. Jones

Georgia State UniversityRyan Kim

Colonial Pipeline Company

ABSTRACT: This case illustrates how database queries can be used to provide continuousassurance in a situation where two trading partners want assurance of the other’s compliancewith their agreements. In this two-sided assurance situation, a car maker wants assurancethat its dealers make timely responses to web site customers and the dealers want assurancethat the car maker is sending them all the designated customer referrals. The queries, devel-oped in Microsoft Access®, illustrate the kind of queries that accountants could prepare toperform continuous monitoring of business activities. In this situation, referrals not in com-pliance with the agreements might be sent automatically to car maker and dealer manage-ments. Query-based approaches to continuous assurance are likely to become more com-mon as trading partners devise new business relationships and want assurance that the otherparty is abiding by their mutual agreements.

Keywords: continuous assurance; database querying; query strategy; queries.

I. THE CASEThe Business Situation

When trading partners devise new ways of working with each other, they may not know what toexpect. For example, when a car maker creates a web site for potential customers to inquireabout new vehicles and refers these inquiries to dealers, the car maker will be concerned

about the timeliness of the dealers’ responses. The dealers, on the other hand, want their fair share of thereferrals. The car maker and the dealers can promise to abide by an agreement on response time and theway referrals are allocated, but neither the car maker nor their dealers seem ready to trust the other onecompletely given the history of their relationships and the potential for technology-induced changes incar sales.1

One approach to helping a car maker and its dealers work together to promote sales of vehicles is toemploy a mechanism for providing assurance to each party that the other party is abiding by their agree-ment. This kind of two-sided assurance could be achieved through the use of database queries that could

The authors are indebted to Penny Lyman and anonymous reviewers for helpful comments about this work.

1 A potential source of future conflict between car makers and their dealers is likely to be car makers’ attempts to enter retailmarkets directly through Internet sales (Wallace 1999b; Rich and Lundegaard 1999; Warner 1999). Even though dealers objectto car makers selling over the Internet (Wallace 1999a), car makers are being pressed to sell over the Internet by customers’ability to car shop on the Internet (Harris 1999).

36 Journal of Information Systems, Spring 2001

run periodically or continuously to identify noncompliant responses soon enough for corrective action tobe effective. Your assignment, explained below, is to develop database queries that give the car makerassurance about dealer compliance and give dealers assurance about car maker compliance.

The DatabaseTo make it easier for people to buy its cars, a car maker established a web site that lets potential

buyers indicate which model(s) and features interest them.2 After receiving the information from a po-tential buyer, the car maker records it in database tables like the following:

Customer: Primary key = CustomerIDFirst Last Post Phone Phone Email

Customer ID Name Name Street City Code Work Home Address

000342512 Ryan Hong 710 London Rd Atlanta 30344 404-876-4875 404-548-6625 [email protected] Daniel Lowell 225 Burbank Dr Atlanta 30314 404-567-2245 404-514-8898 [email protected] Terrel Thomas 2985 Peachtree St Atlanta 30360 770-975-6521 770-548-9658 [email protected] Cathy Allen 1827 McPherson Rd Atlanta 30303 770-988-6521 770-985-3542 [email protected]

CustomerInquiry: No primary keyCustomerID FirstName LastName ModelInInquiry InqDateTime

000342512 Ryan Hong Concorde 11/10/99000342512 Ryan Hong LHS, Sebring Coup 11/12/99000342525 Daniel Lowell Concorde 11/12/99000342525 Daniel Lowell Sebring Convertible 11/15/99000342539 Terrel Thomas 300M, LHS 11/15/99000342546 Cathy Allen Concorde 11/25/99000342546 Cathy Allen Town & Country 12/11/99

The car maker emails each inquiry to the dealer that is closest to the potential buyer, where “closest”is operationalized as the dealer whose 5-digit postal code minimizes the difference between the buyer’spostal code and the dealer’s postal code.3 Information about the inquiries that dealers receive is kept inthe ReferralToDealer table. The date/timestamp in the table is the time a referral was sent to the dealerand is automatically generated by the referral system.

ReferralToDealer: Primary key = ReferralIDReferralID CustomerID DealerID Ref DateTime

000010345 000342512 0024145 11/10/99 10:00:00 AM000010352 000342525 0016287 11/12/99 1:15:00 PM000010363 000342539 0037269 11/15/99 10:00:00 AM000010379 000342546 0405718 12/11/99 11:35:00 AM000010382 000342512 0024145 11/12/99 11:00:00 AM000010394 000342525 0016287 11/15/99 2:20:00 PM000010407 000342546 0405718 11/25/99 10:00:00 AM

2 For example, see Waltner (2000).3 The postal code scheme is used simply to illustrate the kinds of computations that could be performed in the queries. For

another allocation scheme, see Wallace (1998).

Borthick, Jones, and Kim—Developing Database Query Proficiency 37

Information about dealers is kept in the following table:

Dealer: Primary key = DealerIDDealerID Name Street City PostCode Phone

0016287 Buckhead Auto 3126 Piedment Rd Atlanta 30305 404 261-18510023718 Neal Pope Motorcar 4420 Buford Hwy Atlanta 30341 770 216-97000024145 Paul Light 4125 Piedmont Rd Atlanta 30342 404 261-18510035284 Afford Auto 3350 Cumberland Rd Atlanta 30339 404 303-14000037269 Bob Motoring 330 Forrest Rd Atlanta 30349 404 361-38320405718 Town Touring 141 Piedmont Ave Atlanta 30303 404 659-3673

After they contact customers, dealer staff enter information about the contacts, including the dateand time, into a web form. Dealers then click the Submit button on the web form, which sends this self-reported information to a web site provided by the car maker and initiates script processing that writesthe information in the following table:

DealerResponseToReferral: Primary key = ReferralIDReferralID DealerID ResDateTime PhoneResponse EmailResponse

000010345 0024145 11/12/99 11:30:00 AM Yes Yes000010352 0016287 11/12/99 4:00:00 PM No Yes000010363 0037269 11/15/99 4:30:00 PM Yes No000010379 0405718 12/11/99 4:00:00 PM No Yes000010382 0024145 11/15/99 12:20:00 PM No Yes000010394 0016287 11/16/99 9:00:00 AM Yes Yes000010407 0405718 11/25/99 1:00:00 PM No Yes

Another table in the database, EmailResponseToReferral, contains information about the dealer’semailed responses to potential buyers. Each dealer copies the car maker on its email messages to cus-tomers, and the car maker gets the dates/timestamps from the email header information that the carmaker’s email system generates automatically.

EmailResponseToReferral: Primary key = ReferralIDReferralID DealerID ResDateTime EmailAddress

000010345 0024145 11/12/99 11:30:00 AM [email protected] 0016287 11/15/99 1:36:00 PM [email protected] 0405718 12/11/99 5:00:00 PM [email protected] 0024145 11/15/99 12:20:00 PM [email protected] 0016287 11/15/99 5:00:00 PM [email protected] 0405718 11/25/99 1:00:00 PM [email protected]

Because each referral represents the potential sale of a car, dealers should be motivated to respondquickly to all the referrals they receive. Still, some potential buyers complain that a week passes beforea dealer contacts them. (That interval is long enough to change one’s mind about buying the car.) Beingexasperated with the situation,4 the car maker informed its dealers that it would cease making referrals todealers that take more than 48 hours to contact the customers referred to them. Having created thispolicy, the car maker now has to implement it, which means identifying any dealers not in compliance.

4 See Wilson (2000).


RequiredPart 1: Assurance for the Car Maker about Dealer Compliance

Develop the following sequence of database queries and use the query results to explain the extentto which car dealers are contacting customers referred to them within 48 hours:

1. Determine response times for dealer-reported responses.2. Find the average response time by dealer for self-reported times.3. Determine response times for dealers’ emailed responses.4. Find the average response time for dealers’ emailed responses.5. Compare dealers’ self-reported response times to dealers’ emailed response times.6. Find the average time difference between dealers’ self-reported and emailed responses.

Part 2: Assurance for Dealers about Car Maker ComplianceA. Develop the following sequence of database queries and use the query results to explain the extent

to which the car maker is referring customers to the closest dealer, where “closest” is defined as thedealer that minimizes the absolute value of the difference between the customer’s and the dealer’spostal codes:1. For each referral, determine the difference between the postal codes of the customer and the

dealer receiving the referral.2. For each referral, determine the minimum postal code difference over all dealers.3. For each referral, determine whether the referred dealer corresponds to the minimum postal

code difference.B. Design a criterion for “closest” dealer that would be better than minimizing the absolute value of the

difference between the customer’s and the dealer’s postal codes for the geographic area in Figure 1.

Part 3: Objective QuestionsSelect the best response to each of the following questions.

A. Making the DealerResponseToReferral and EmailResponseToReferral tables available to a newquery will result in which of the following join results:1. No join occurs until the user connects the attributes to be joined.2. No join occurs until the user highlights the attributes to be joined and clicks the join icon.3. A join occurs automatically because the tables have identical primary keys.4. A join occurs automatically because the tables have some identical attributes.

B. Suppose the database contains the following queries:1. qDealerResponse, which has attributes DealerID, ReferralID,

DealerResponseToReferral.ResDateTime, RefDateTime, and DealerResponse whereDealerResponse = ([DealerResponseToReferral.ResDateTime] – [RefDataTime])*24

2. qEmailResponse, which has attributes DealerID, ReferralID,EmailResponseToReferral.ResDateTime, RefDateTime, and DealerResponse whereEmailResponse = ([EmailResponseToReferral.ResDateTime] – [RefDataTime])*24

Making the qDealerResponse and qEmailResponse queries available to a new query will result inwhich of the following join results:1. No join occurs until the user connects the attributes to be joined.2. No join occurs until the user highlights the attributes to be joined and clicks the join icon.3. A join occurs automatically because the tables have identical primary keys.4. A join occurs automatically because the tables have some identical attributes.

C. From the perspective of web-savvy customers, the best measure of response time to their inquirieswould be the elapsed time based on the date/time attributes in the tables:1. CustomerInquiry and DealerResponseToReferral2. ReferralToDealer and DealerResponseToReferral


3. ReferralToDealer and EmailResponseToReferral4. CustomerInquiry and EmailResponseToReferral

D. Consider a query for which the DealerResponseToReferral table and the EmailResponseToReferraltable have been made available. In the calculated expression:[DealerResponseToReferral.ResDateTime] – [EmailResponseToReferral.ResDateTime]the term “[DealerResponseToReferral]!”:1. Is necessary2. Is unnecessary3. Causes a syntax error4. Causes multiple joins

E. Suppose the CustomerInquiry table, the Dealer table, and the ReferralToDealer table are madeavailable to a new query and the user does not change the way the query manager joined the tables.After CustomerID is entered in the first field, the result for the query will have how many rows?1. The number of rows in the CustomerInquiry table2. The number of rows in the ReferralToDealerTable3. The number of rows in the CustomerInquiry table times the number of rows in the

ReferralToDealer table4. The number of rows in the CustomerInquiry table times the number of rows in the Dealer table

F. Suppose the Customer table, the Dealer table, and the ReferralToDealer table are made available toa new query and the user does not change the way the query manager joined the tables. AfterCustomerID is entered in the first field, the result for the query will have how many rows?1. The number of rows in the Customer table2. The number of rows in the ReferralToDealer table3. The number of rows in the Customer table times the number of rows in the ReferralToDealer

table4. The number of rows in the Customer table times the number of rows in the Dealer table

G. Suppose the car maker has complied with the agreement to refer customers to the closest dealer,where “closest” is defined as the dealer for which the absolute difference between the dealer’spostal code and the customer’s postal code is minimized (see the postal code map in Figure 1.) Alsosuppose that customers complain about the distances to the dealers to which they were referred. Abetter approach to allocating customers to dealers that minimizes the need for future program ordata table changes would be to:1. Preassign postal codes to dealers and use a table lookup to make the assignments2. Preassign telephone prefixes to dealers and use a table lookup to make the assignments3. Assign customers to dealers to minimize driving distance based on automatic lookup of street

addresses in a geographic database4. Assign customers to dealers such that the distance based on latitude and longitude coordinates

is less than x miles and dealers get referrals in proportion to their salesH. The potential enhancement to the car maker’s web site that would be the most attractive to custom-

ers and the least likely to prompt resistance from dealers is to:1. Give customers the ability to scan dealers’ new car inventories2. Give customers access to the prices that the car maker charges dealers3. Let customers purchase vehicles directly from the car maker4. Let customers see pricing for vehicles from other car makers without leaving the site

I. The best composition of a primary key for the Customer Inquiry table is:1. CustomerID2. InqDateTime3. CustomerID and ModelInquiry4. CustomerID and InqDateTime


J. Consider a query for which the Dealer, ReferralToDealer, and Customer tables have been madeavailable. In this query, the calculated expression:

Abs([Dealer.PostCode] – [Customer.PostCode])would be equivalent to:1. Abs([Dealer.PostCode] ) – Abs( [Customer.PostCode])2. [Customer.[PostCode] – [Dealer.[PostCode]3. Abs([Customer.PostCode] – [Dealer.PostCode])4. – ([Customer.PostCode] – [Dealer.PostCode])


II. TEACHING NOTESLearning Objectives

This case comprises a learning experience that helps students learn to: (1) decide what informationwould be relevant to solving a business problem; (2) extract the needed information; (3) put the informa-tion in a form conducive to solving the problem; and (4) analyze the information to solve the problem.The case operationalizes the learning objectives in the information-use category (Borthick 1996). Itcharacterizes a business situation for which management wants information of the kind that accountantsmight be asked to produce by querying a database (Borthick 1992).

The context for the case, assuring car dealer and car maker compliance with their agreements, waschosen because it is situated in current business practice. The context illustrates the problems that com-panies may have as they attempt to transform their interactions with each other and customers to takeadvantage of the new ways of communicating and processing information that information technologyhas enabled. The assurance need is posed as two-sided (i.e., each trading partner wants assurance aboutthe other’s behavior) because that illustrates the growing mutual dependence that trading partners haveon each other.

The case scenario was designed with the least number of tables and attributes necessary to make thesituations plausible, but the database still contains sufficient distractor tables and attributes to requirelearners to think through the implications of their querying. For example, the CustomerInquiry table, aview table with no primary key, duplicates attributes contained in the Customer table. Learners can beengaged in a discussion of whether this table ought to be included in a well-designed database.

Because business problems often come with irrelevancies intermingled with essential information,so does this case. The database tables contain very few rows, allowing learners to verify their queryresults easily by inspecting table rows. A Microsoft Access® file (Office 2000) containing the databaseis available from the first author on request.

The case has no requirement for assessing the reliability of the database before conducting thequerying. At the instructor’s discretion, learners could be engaged in a discussion of how this would bedone. In this situation, the most likely assurors would be the car maker’s internal audit staff, who wouldlikely do no more to assure database reliability than the following:

1. Verify that the database tables are complete.2. Verify that the email system was functioning during the time the referrals were made.3. Verify that script processing for web form capture of car dealer responses about their contacts

with customers was functioning as designed.In Part 1 (assurance for the car maker about car dealer compliance), learners (1) use default table

linking to make the information they need accessible, (2) create mathematical expressions to manipulateattribute values in rows, (3) format and sort result columns, (4) apply functions to columns, and (5)combine query results to permit performance comparisons. Query Strategies 3 and 4 essentially dupli-cate Query Strategies 1 and 2 but with one of the source tables being different. This repetition yields alonger solution narrative, but many learners seem to benefit from the repetition. That is, many learnersdo not seem to be able to internalize the query strategy and the querying steps with just one encounter.The overall querying outcome identifies procedural lapses at specific car dealers.

In Part 2 (assurance for car dealers about car maker compliance), learners (1) use default and idio-syncratic table linking, (2) create mathematical expressions to manipulate attribute values in rows, (3)apply functions to columns, and (4) combine query results to permit performance comparisons. Theoverall querying outcome identifies a design flaw in the way referrals are allocated to dealers.


In Part 3 (multiple-choice questions), the learning objectives are to:Query Focus:

Problem Strategy orLearning Objective Question Querying Syntax

Use linking relationships to make the desired data(and only the desired data) accessible A, E, F Syntax

Join tables without primary keys B SyntaxAdopt a customer perspective on performance C StrategyQualify attributes to ensure uniqueness of reference D SyntaxImprove system design given constraints G, H StrategyDesignate a primary key for a view table I SyntaxSpecify precedence relations in expressions J Syntax

Case Use: Courses and Prior Learner ProficienciesThe case has been used in courses in accounting information systems (AIS) and information sys-

tems assurance. It could be used in other accounting and business courses, e.g., introductory accounting,business processes, or assurance, in which a learning objective is to have students experience databasequerying to answer business questions whose answers do not fit a pre-structured response format. In-stead, students must create a response that meets the information need. This kind of learning experienceis more likely to prepare students “for the ambiguous business world they will encounter upon gradua-tion” (Albrecht and Sack 2000, 43).

The case is usable by learners with varying degrees of proficiency in formulating query strategy andin using a database query interface. An instructor can accommodate varying levels of learner proficiencyby supplying different portions of the solution text. For example, if they are somewhat skilled in strategyformulation and in using the query interface, learners could be given the business situation, the database,and the general statements of the two questions to be answered with the suggested query sequences.(This is the way the case is posed above.) If learners are very skilled at formulating query strategy, thenthe query sequence could be withheld for the two questions. If learners need help formulating querystrategy, then some or all of the following strategy sections (identified with “Query Strategy” at thebeginning of the heading) could be supplied too. If they need help developing the database queries, thenlearners could be given some or all of the following query development sections (identified with theheading “Query Creation”). If learners have no knowledge of database systems, then it is helpful forthem to have access to a database text, e.g., Perry and Schneider (2001).

Part 1 Solution: Assurance for the Car Maker about Car Dealer ComplianceQuery Strategy 1: Determine Dealer-Reported Response Times

Join tables with referral time and dealer-reported response time. The first step in providingassurance to the car maker is deciding what data are required. Because the question is about dealerperformance in responding to customer referrals, the required tables are ReferralToDealers andDealerResponseToReferrals. Because the field of interest is the elapsed time between when the carmaker sent the referral to the dealer (field RefDateTime in table ReferralToDealer) and when the dealerresponded to it (field ResDateTime in table DealerResponseToReferral), the two tables need to be joinedon the field ReferralID. Joining the tables makes the two date/time values for each referral accessible inthe same row so that the elapsed time (the response time) can be calculated.

Query CreationLoad the database in Microsoft Access® (Office 2000). Select Queries from the Objects menu

and double-click Create query in Design view. In the Show Table screen, highlight theReferralToDealer table and select Add, which makes the table appear in the Select Query screen.(Another way to make the table appear in the Select Query screen is to double-click it in the ShowTable screen.) Similarly, make the DealerResponseToReferral table appear in the Select Queryscreen and then Close the Show Table screen.


The tables join themselves automatically on ReferralID because ReferralID is the identicalprimary key in both tables. A line connecting the two ReferralID fields signifies the join. (Foronline information about creating or deleting relationships between tables, go to the help menu,click Microsoft Access® Help, click the Index tab, type “relationship”, select Search, andselect “Create or modify relationships” and “Work with relationships”.) Linking (joining) the tablesselects only those rows with matching values in the ReferralID field in both tables.

Determine dealer-reported response times. To create a table with dealer-reported response times,select the needed fields from the ReferralToDealer and DealerResponseToReferral tables and create acalculated field for the response time, the difference between the ResDateTime field and the RefDateTimefield. Mentally calculate the response time for a sample row to verify the results of the subtraction andthe units (e.g., hours or days) in which it is presented. A result that is given in days can be converted tohours by including the appropriate multiplier in the calculation.

Query CreationTo create a table with dealer-reported response times, select the fields ReferralID, DealerID, and

RefDateTime from the ReferralToDealer table (by double-clicking them one at a time) and the ResDateTimefield from the DealerResponseToReferral table. Selecting the fields makes their values available in theresult table. Because they also appear in the DealerResponseToReferral table, the ReferralID and DealerIDfields could be selected from the DealerResponseToReferral table instead of the ReferralToDealer table.

Although it is not finished, the query needs a name, and naming it makes the field names moreaccessible during query building. To name it, select the quit icon (X) in the upper right-hand cornerof the Select Query screen, select Yes in response to the “Do you want to save changes…?” prompt,enter a meaningful name for the query, e.g., DealerResponse, and select OK. To see what the queryresult is thus far, double-click the DealerResponse query. Return to Design view (the Select Queryscreen) by selecting the design icon, the protractor symbol at the left end of the toolbar.

To finish the query, left click in the blank field to the right of the ResDateTime field in theSelect Query screen. Select the build icon (magic stick symbol) from the tool bar, which opens theExpression Builder screen. The expression builder is the tool for specifying mathematical opera-tions on the values of fields. (For online information about expressions, on the help menu, clickMicrosoft Access® Help, click the Index tab, type “expression builder” and select Search.)

To create the expression to subtract the time of the referral from the time of the dealer’s re-ported response, double-click the ResDateTime field in the middle column of the Expression Builderscreen, select the subtraction symbol in the expression builder (or type a minus sign), and double-click the RefDateTime field. The text of the expression looks like:

[ResDateTime] – [RefDateTime]Select OK in the Expression Builder screen, select the quit icon (X) in the upper right-hand

corner of the Select Query screen, and select Yes in response to the “Do you want to save changes…?”prompt. Open the query by double-clicking its name, which gives the following result:

Query: DealerResponse

ReferralID DealerID RefDateTime ResDateTime Expr1000010345 0024145 11/10/99 10:00:00 AM 11/12/99 11:30:00 AM 2.0625000010352 0016287 11/12/99 1:15:00 PM 11/12/99 4:00:00 PM 0.114583333328483000010363 0037269 11/12/99 1:15:00 PM 11/15/99 4:30:00 PM 0.270833333335759000010379 0405718 11/12/99 1:15:00 PM 12/11/99 4:00:00 PM 0.184027777773736000010382 0024145 11/12/99 11:00:00 AM 11/15/99 12:20:00 PM 3.05555555555475000010394 0016287 11/15/99 2:20:00 PM 11/16/99 9:00:00 AM 0.777777777781012000010407 0405718 11/25/99 10:00:00 AM 11/25/99 1:00:00 PM 0.125

To verify that the result is what was intended and to figure out how time is represented, men-tally subtract RefDateTime from ResDateTime for one of the rows in the result table. The result of


the subtraction is represented in days. To make the result appear as hours, the expression has toconvert days to hours.

To make the result appear in hours, switch to Design view (by selecting the design icon), clickthe field with the expression, open the expression builder, add left and right parentheses around theexpression [ResDateTime]–[RefDateTime], position the cursor after the right parenthesis, selectthe multiplication symbol (*) (or type it), and enter the digits 24. Replace “Expr1” (the default nameof the expression) with a meaningful name such as ResponseInHours. Now the expression lookslike:

ResponseInHours: ([ResDateTime] – [RefDateTime])*24

Select OK to close the expression builder. To format the ResponseInHours column to make itshow the values with two decimal digits, left click in the ResponseInHours field, right click, selectProperties, select Format, select Fixed from the pulldown menu, and close the Field Propertieswindow.

To make the result table display the result in order with the row with the largest ResponseInHoursvalue appearing first, click in the Sort row of ResponseInHours. From the pulldown menu, selectDescending. (For online information about sorting data, on the help menu, click Microsoft Access®

Help, click the Index tab, type “sort”, and select “Sort records in a table, query, or form” from thetopic list.)

Run the query (by selecting !) and verify that it gives the intended result. Once the query givesthe right result, save it by selecting the quit icon (X) in the upper right-hand corner of the SelectQuery screen and selecting Yes in response to the “Do you want to save changes…?” prompt.

With the time units in hours, the DealerResponse query gives the following result:

Query: DealerResponse

ReferralID CustomerID DealerID RefDateTime ResDateTime InHours

000010382 000342512 0024145 11/12/99 11:00:00 AM 11/15/99 12:20:00 PM 73.33000010345 000342512 0024145 11/10/99 10:00:00 AM 11/12/99 11:30:00 AM 49.50000010394 000342525 0016287 11/15/99 2:20:00 PM 11/16/99 9:00:00 AM 18.67000010363 000342539 0037269 11/15/99 10:00:00 AM 11/15/99 4:30:00 PM 6.50000010379 000342546 0405718 12/11/99 11:35:00 AM 12/11/99 4:00:00 PM 4.42000010407 000342546 0405718 11/25/99 10:00:00 AM 11/25/99 1:00:00 PM 3.00000010352 000342525 0016287 11/12/99 1:15:00 PM 11/12/99 4:00:00 PM 2.75

Query Strategy 2: Find the Average Response Time by Dealer for Self-Reported TimesExamining the average response time for each dealer (referred to as “by dealer”) is another way to

get a sense of the behavior of each dealer in responding to referrals. To find the average of self-reportedresponse times for each dealer, create a new query based on the DealerResponse query, indicate theaverage to be calculated, and group the query results with the largest average first (descending).

Query CreationTo create a query to find the average response time by dealer, select Queries in the Objects

menu, double-click Create query in Design view, select the Queries tab, select the DealerResponsequery, select Add, and select Close. (Another way to make the query appear in the Select Queryscreen is to double-click it.) Select DealerID and ResponseInHours by double-clicking them one ata time.

Select the ∑ icon from the tool bar to produce a total row in the Design view screen. The Totalrow lists Group By as the default option for each field. This option defines the groups that will entercalculations. In this query, to show average by each dealer, leave Group By in the Total row forDealerID.

Click the Total row in the ResponseInHours column and select Avg from the pulldown menu inthe field. (For online information about using Group By, on the help menu, click Microsoft Access®

Response


Help, click the Index tab, type “query group”, select Search, and select “Perform calculations ina query” from the topic list.) Run the query to see the result thus far, which shows the averageresponse in hours for the four dealers with referrals:

DealerID AvgOfResponseInHours0016287 10.70833333331390024145 61.4166666666570037269 6.500000000058210405718 3.70833333328483

To see the whole heading AvgOfResponseInHours, left-click on the right border of the heading celland drag it to the right to widen the column.

Sorting the values in descending order will arrange the rows so that the longest average responsetime appears first, which will make the results easier to analyze. To sort the rows in descending order,return to Design view, left-click the sort row in the ResponseInHours column, and choose Descendingfrom the pulldown menu. To make the values show two decimal digits, right-click in the ResponseInHoursfield, select Properties, select Format, select Fixed from the pulldown menu, and close the Field Proper-ties window. Save the query with a meaningful name, e.g., AvgResponseInHours, by selecting the quiticon (X) in the upper right-hand corner of the Select Query screen, selecting Yes in response to the “Doyou want to save changes…?” prompt, and entering the name.

The query result is:

Query: AvgResponseInHoursDealerID AvgOfResponseInHours0024145 61.420016287 10.710037269 6.500405718 3.71

From this result, based on dealers’ self-reported response times, three dealers contacted customersreferred to them within 48 hours on average and one did not. The self-reported response times may not,however, be reliable. Some factors affecting reliability of the self-reported response times are (1) timesmay not be recorded soon enough for dealer staff to remember when they contacted customers; (2)dealers may have wanted to make their responses look better than they really were; or (3) there may besome systematic flaws in the recording process. Because the process requires manual intervention, it isnot clear that the results are reliable, either to the advantage or the disadvantage of the dealers. Thesethreats to data reliability might be manageable if there were some way to assess them, which is theobjective of the next section.

Query Strategy 3: Determine Response Times for Dealers’ Emailed ResponsesBecause the timestamps on email messages were recorded automatically by the email system when

the email messages were sent, their recorded times are likely to be more reliable than the dealers’ self-reported response times. If we repeat the analysis performed above for dealers’ self-reported times onthe timestamps of the email messages, we can see if compliance differs depending on which responsemode is used in the analysis. Here are the steps: using the ReferralToDealers and EmailResponseToReferraltables, create a query that calculates the elapsed time (response time) in hours between when the carmaker sent the referral to the dealer (field RefDateTime in table ReferralToDealer) and when the dealerresponded to it (field ResDateTime in table EmailResponseToReferral).

Query CreationCalculate elapsed time. Select Queries from the Objects menu and double-click Create query

in Design view. In the Show Table screen, make the ReferralToDealer and EmailResponseToReferraltables appear in the Select Query screen and Close the Show Table screen. A line connecting the twoReferralID fields signifies that Access joined the two tables.


To create a table with email response times, select the fields DealerID, ReferralID, andRefDateTime from the ReferralToDealer table (by double-clicking them one at a time) and theResDateTime field from the EmailResponseToReferral table.

Name the query by selecting the quit icon (X) in the upper right-hand corner of the SelectQuery screen, select Yes in response to the “Do you want to save changes…?” prompt, enter thename of the query, e.g., EmailResponse, and select OK. To see the results thus far for theEmailResponse query, double-click it.

To finish the query, switch to Design view and left-click in the blank field to the right of theResDateTime field in the Select Query screen. Select the build icon from the tool bar to open theexpression builder.

To create the expression to subtract the time of the referral from the time of the dealer’s emailresponse, double-click the ResDateTime field in the middle column of the Expression Builder screen,select the subtraction symbol in the expression builder, and double-click the RefDateTime field.The text of the expression looks like:

[ResDateTime] – [RefDateTime]

To make the result appear in hours, add left and right parentheses around the expression“[ResDateTime]–[RefDateTime]”, position the cursor after the right parenthesis, select the multi-plication symbol, and enter the digits 24. To give the calculated field a name (other than the defaultone), move the cursor to the left of the expression, enter a meaningful name, e.g., EmailResponse,and follow the name with “:”. Now the expression looks like:

EmailResponse: ([ResDateTime] – [RefDateTime])*24

Select OK to close the expression builder. Format the EmailResponse field for two decimaldigits by right-clicking the EmailResponse field, selecting Properties, selecting Fixed, and closingthe Field Properties screen. Save and run the query.The query results are:

Query: EmailResponseDealerID ReferralID RefDateTime ResDateTime EmailResponse0024145 000010345 11/10/99 10:00:00 AM 11/12/99 11:30:00 AM 49.500016287 000010352 11/12/99 1:15:00 PM 11/15/99 1:36:00 PM 72.350405718 000010379 12/11/99 11:35:00 AM 12/11/99 5:00:00 PM 5.420024145 000010382 11/12/99 11:00:00 AM 11/15/99 12:20:00 PM 73.330016287 000010394 11/15/99 2:20:00 PM 11/16/99 5:00:00 PM 26.670405718 000010407 11/25/99 10:00:00 AM 11/25/99 1:00:00 PM 3.00

Query Strategy 4: Find the Average Response Time for Dealers’ Emailed ResponsesUsing the query just developed that determines response times for dealers’ emailed responses, cal-

culate the average response by dealer based on email responses.

Query CreationTo create a query to find the average response times by dealer, select Queries in the Objects

menu, double-click Create query in Design view, select the Queries tab, double-click theEmailResponse query and select Close. Select DealerID and EmailResponse by double-clickingthem one at a time.

Select the ∑ icon from the tool bar to insert a total row in the Design view screen. The Total row listsGroup By as the default option for each field. This option defines the groups that will enter calculations.

To show average by each dealer, leave Group By in the Total row for DealerID. Left-click theTotal row of EmailResponse and select Avg from the pulldown menu in the field. (For onlineinformation about using Group By, on the help menu, click Microsoft Access® Help, select the


Index tab, type “query group” select Search, and select “Perform calculations in a query” fromthe topic list.)

To sort the rows of the EmailAverage query in descending order, click the sort row in theEmailResponse column and choose Descending from the pulldown menu in the field. (For onlineinformation about sorting data, on the help menu, click Microsoft Access Help, select the Index tab,type “sort”, and select “Sort records in a table, query, or form” from the topic list.)

Format the field for two decimal digits by right-clicking the AvgOfEmailResponse field, select-ing Properties, selecting Fixed, and closing the Field Properties screen. Save the query with a mean-ingful name, e.g., EmailAverage, and run it.The query results, with a row for each dealer that responded with email, are:

Query: EmailAverageDealerID AvgOfEmailResponse0024145 61.420016287 49.510405718 4.21

Query Strategy 5: Compare Dealers’ Self-Reported Response Times to Dealers’ Emailed ResponseTimes

How can the car maker know whether the response times are sufficiently reliable? For the self-reported times in the DealerResponseToReferral table, there is no way to know. For dealers’ emailedresponses, the automatically provided email timestamps are as reliable as the email system that pro-duced them. The email date/timestamps could be compared with the self-reported times. Some indica-tion of the reliability of dealers’ reports on their response times can be obtained by examining the differ-ence between ResDateTime fields in the EmailResponseToReferral and DealerResponseToReferralstables for the referrals for which email responses were created.

To compare matching self-report and email response times, create a query using theDealerResponseToReferrals and EmailResponseToReferral tables joined on ReferralID that includesDealerID and an expression for calculating the difference between self-reported and email responsetimes. (The query should not be based on the DealerAverage and EmailAverage queries because theaverages in these queries are for different sets of referrals. The query could be based on the DealerResponseand EmailResponse queries, which would require joining the queries on ReferralID by clicking on theReferralID field in one of the queries and dragging and dropping the cursor to the ReferralID in the otherquery.)

Query CreationSelect Queries in the Objects menu and double-click Create query in Design view. On the Show

Table screen, double-click the DealerResponseToReferral and EmailResponseToReferral tables tomake them appear in the Select Query screen and Close the Show Table screen. Double-clickReferralID, DealerID, and ResDateTime in the DealerResponseToReferral table and ResDateTimein the EmailResponseToReferral table. Select the quit icon (X), select Yes in response to the saveprompt, enter a meaningful name for the query, e.g., ResponseDifference, and select OK.

Open the query in Design view and left-click in the blank field to the right of the ResDateTimefield in the Select Query screen. Select the build icon to begin an expression to calculate the differ-ence between the two response times. In the center column of the expression builder, double-clickResDateTime from the EmailResponseToReferral table. The center column is not wide enough forall of EmailResponseToReferral.ResDateTime to be visible, but the right choice can be confirmedby examining the expression after the term has been entered. Put a minus sign (–) afterEmailResponseToReferral.ResDateTime and then double-click ResDateTime from theDealerResponseToReferral table. Modify the expression to convert it to hours (from days) and givethe expression a meaningful name, e.g., ResponseDifference. The expression should look like:


ResponseDifference: ([EmailResponseToReferral.ResDateTime] –[DealerResponseToReferral.ResDateTime])*24

Select OK to close the expression builder. Format the ResponseDifference column for twodecimal places and sort it descending. Save the query before executing it.The result from the query is:Query: ResponseDifference

Dealer Response Email Response ResponseReferral Dealer Referrals.ResDateTime Referral.ResDateTime Difference

000010352 0016287 11/12/99 4:00:00 PM 11/15/99 1:36:00 PM 69.60000010394 0016287 11/16/99 9:00:00 AM 11/16/99 5:00:00 PM 8.00000010379 0405718 12/11/99 4:00:00 PM 12/11/99 5:00:00 PM 1.00000010407 0405718 11/25/99 1:00:00 PM 11/25/99 1:00:00 PM 0.00000010382 0024145 11/15/99 12:20:00 PM 11/15/99 12:20:00 PM 0.00000010345 0024145 11/12/99 11:30:00 AM 11/12/99 11:30:00 AM 0.00

To see the whole heading for a ResDateTime column, left-click on the right border of a heading celland drag it to the right to widen the column. (Access does not permit users to manipulate column head-ings other than to change the width. The double-row presentation of the result table here was achievedoutside Access.)

Query Strategy 6: Find the Average Time Difference between Dealers’ Self-Reported Times andEmail Responses

To obtain the average difference in self-reported response times and email response times by dealer,find the average ResponseDifference by DealerID, sorted with the largest average first.

Query CreationTo create a query to find the average response differences by dealer, select Queries in the

Objects menu and double-click Create query in Design view. Select the Queries tab, double-clickthe ResponseDifference query, and select Close. Select DealerID and ResponseDifference by double-clicking them.

Select the ∑ icon from the tool bar to produce a total row. The total row lists Group By as itsdefault option for each field, which defines the groups that will enter calculations. In this query, toshow average by each dealer, leave Group By in the Total row for DealerID. Left-click the Total rowfor the ResponseDifference field, and select Avg from the pulldown menu in the field. Left-click theSort field and select Descending from the pulldown menu. (For online information about usingGroup By, on the help menu, click Microsoft Access® Help, click the Index tab, type “querygroup”, select Search, and select “Perform calculations in a query” from the topic list.) Format thefield for two decimal places. Give the query a meaningful name, e.g., AverageResponseDifference,and save it.The query gives the following results:

Query: AverageResponseDifferenceDealerID AvgOfResponseDifference0016287 38.800405718 0.500024145 0.00

This query result indicates that of the three dealers with email responses one dealer recorded itsresponse times very accurately, one dealer recorded its responses within a half-hour of the actual times(or made phone calls to customers before sending them email), and one dealer recorded its responseswith a 39 hour discrepancy with email timestamps. One of the four dealers with referrals did not respondto customers through email. If there were more referrals and dealers in the database, there would prob-ably be more different kinds of recording behavior to detect.


Assessment of Dealer ComplianceBased on dealers’ self-reported response times, the result in the DealerAverage query indicates that

one of the four dealers exceeded an average 48-hour response time. This result is consistent with cus-tomers complaining about tardy responses.

Based on timestamps from emailed dealer responses, the results in the EmailAverage query indicatethat two of the three dealers making email responses exceeded an average 48-hour response time. One ofthese dealers appeared to be responding timely according to self-reported response times. Comparingthe self-reported response times with the automatically timestamped email responses (in theAverageResponseDifference query) reveals discrepancies for two dealers. For one dealer, the discrep-ancy is large; for the other one, small.

These results illustrate the importance of being able to assess the reliability of data used for analyz-ing performance. Without this kind of detailed analysis, it is easy to mischaracterize performance andthe behavior associated with it. As this example demonstrates, performance may look different depend-ing on which data are analyzed, e.g., self-reported versus automatically recorded timestamps.

Another data reliability issue concerns access to the queries. Because it has exclusive control overthe web site, the car maker has control over the recording of queries and how they are made available tothe dealers. Thus, dealers are able to view only the customer inquiries that the car maker sends themthrough email. Statements in the case that make this apparent are: (1) “the car maker records it [thecustomer inquiry] in database tables” and (2) “The car maker emails each inquiry to the dealer that isclosest to the potential buyer.”

Part 2 Solution: Assurance for Dealers about Car Maker Compliance

A. Database QueriesOver time, several dealers wondered whether they were getting all the referrals that were due them.

Their doubts increased as they heard potential customers say they had been referred to dealers that werefarther away from them. The queries developed in this section are designed to give dealers assurancethat the car maker is allocating referrals as they agreed.

Query Strategy 1: For Each Referral, Determine the Difference between the Customer’s andDealer’s Postal Codes

Make postal code fields accessible. The first step is to decide what data are required. Because thequestion is about whether each customer is referred to the nearest dealer (the definition of which in-volves customers’ and dealers’ postal codes), the appropriate tables are Dealer, ReferralToDealer andCustomer. The Dealer and Customer tables contain postal codes, and the ReferralToDealer table desig-nates the dealers to which customers were referred. The postal codes in the Customer and Dealer tablescan be brought together by joining the Customer and ReferralToDealer tables on CustomerID and join-ing the ReferralToDealer and Dealer tables on DealerID. The absolute value of the difference betweencustomer and dealer postal codes can be obtained after joining the three tables.

Query CreationLoad the database in Access. Select Queries from the Objects menu and double-click Create

query in Design view. In the Show Table screen, double-click the Dealer, ReferralToDealer, andCustomer tables to make them appear in the Select Query screen. Close the Show Table screen.Determine postal code differences for referrals. The differences between customers’ postal codes

and the postal codes of the dealers to which customers were referred can be determined by comparingpostal code values. For each referral, this means calculating the absolute value of the difference betweenthe customer’s postal code and the postal code of the dealer receiving the referral. Specifying the abso-lute value in the calculation (with the “Abs” function) causes the result to be expressed as a positivenumber regardless of which sign (plus or minus) the result actually had.

Query CreationTo start the query, double-click the following fields to make them appear in the query: ReferralID

from the ReferralToDealer table; CustomerID and PostCode from the Customer table; and DealerID


and PostCode from the Dealer table. (The CustomerID and DealerID fields could also be selectedfrom the ReferralToDealer table.) Save the query with a meaningful name e.g., PostDiffActual.

Open the query in Design view. To finish the query, create an expression that calculates theabsolute value of the difference between the customer postal code and the dealer postal code. To dothis, left-click in the first blank field in the Select Query screen. Select the build icon (magic sticksymbol) from the tool bar to open the Expression Builder screen. (For online information aboutexpressions, on the help menu, Microsoft Access® Help, select the Index tab, type “expressionbuilder”, and select Search.)

In the middle column of the expression builder, double-click Customer.PostCode, select the mi-nus sign (–), and double-click Dealer.PostCode. To make the result of the calculation an absolutevalue, put parentheses around the whole expression and type “Abs” in front of the left parenthesis.Give the field a name by moving the cursor to the left of “Abs” and entering “AbsDiffActual: ”.Now the expression looks like:

AbsDiffActual: Abs([Customer.PostCode] – [Dealer.PostCode])Select OK and save the query.

Running the query gives the following result:

Query: PostDiffActualCustomer Dealer Abs

ReferralID CustomerID .PostCode DealerID .PostCode DiffActual000010345 000342512 30344 0024145 30342 2000010352 000342525 30314 0016287 30305 9000010363 000342539 30360 0037269 30349 11000010379 000342546 30303 0405718 30303 0000010382 000342512 30344 0024145 30342 2000010394 000342525 30314 0016287 30305 9000010407 000342546 30303 0405718 30303 0

Query Strategy 2: For Each Referral, Determine the Minimum Postal Code Difference over AllDealers

To find the minimum difference for postal code (closest dealer) for each referral, it is necessary tocompare the postal code of each customer for each referral with the postal codes of all dealers. (Al-though minimizing the absolute value of postal code differences was the approach given in this assign-ment, many other allocation schemes are possible.) Comparing all pairs of postal codes requires (1)joining the ReferralToDealer and Customer tables so that customers’ postal codes are linked to referralsand (2) not joining the tables on DealerID so that the query produces every combination of rows from thejoined ReferralToDealer and Customer tables and the DealerID table. This arrangement is necessary inorder to put the two postal code values in the same row for each referral so that an expression can becreated to calculate the difference between them.

Query CreationTo create the comparison table, select Queries from the Objects menu and double-click Create

query in Design view. On the Show Table screen, double-click the Dealer, ReferralToDealer, andCustomer tables and Close the Show Table screen.

The query manager joins the three tables (Customer and ReferralToDealer on CustomerID;Dealer and ReferraltoDealer on DealerID). The default link between ReferralToDealer and Dealertables is not, however, needed for this query. To pair every referral with every dealer in order to beable to find the difference in each pair of postal codes, highlight the link between the ReferralToDealerand Dealer tables and delete it by pressing [Delete]. (For online information about creating ordeleting relationships between tables, on the help menu, click Microsoft Access® Help, click theIndex tab, type “relationship”, select Search, and select “Create or modify relationships” and“Work with relationships”.)


Double-click the following fields to make them appear in the query: ReferralID fromReferralToDealer; CustomerID and PostCode from Customer; and DealerID and PostCode fromDealer. Give the query a meaningful name, e.g., AllCombinations, and save it. Run the query andexamine the result to verify that every referral has been paired with every dealer.

Open the query in Design view. To calculate the difference between the pairs of postcodes, left-click in the first blank field (to the right of the PostCode field from the Dealer table). Select the buildicon (magic stick symbol) from the tool bar to open the Expression Builder screen. (For onlineinformation about expressions, on the help menu, select Microsoft Access® Help, select the Indextab, type “expression builder”, and select Search).

Double-click Customer.PostCode from the middle column of the Expression Builder, select theminus sign (–), and Dealer.PostCode. To express the result of the calculation as an absolute value,enclose the expression in parentheses and type “Abs” in front of the left parenthesis. Give theexpression a meaningful name by typing “AbsDifference:” in front of “Abs”. The completeexpression is:

AbsDifference: Abs([Customer.PostCode] – [Dealer.PostCode])Select OK. Save the query and run it.The query result is:

Query: AllCombinationsCustomer. Dealer.

ReferralID CustomerID PostCode DealerID PostCode AbsDifference

000010345 000342512 30344 0016287 30305 39000010345 000342512 30344 0023718 30341 3000010345 000342512 30344 0024145 30342 2000010345 000342512 30344 0035284 30339 5000010345 000342512 30344 0037269 30349 5000010345 000342512 30344 0405718 30303 41000010382 000342512 30344 0016287 30305 39000010382 000342512 30344 0023718 30341 3000010382 000342512 30344 0024145 30342 2000010382 000342512 30344 0035284 30339 5000010382 000342512 30344 0037269 30349 5000010382 000342512 30344 0405718 30303 41000010352 000342525 30314 0016287 30305 9000010352 000342525 30314 0023718 30341 27000010352 000342525 30314 0024145 30342 28000010352 000342525 30314 0035284 30339 25000010352 000342525 30314 0037269 30349 35000010352 000342525 30314 0405718 30303 11000010394 000342525 30314 0016287 30305 9000010394 000342525 30314 0023718 30341 27000010394 000342525 30314 0024145 30342 28000010394 000342525 30314 0035284 30339 25000010394 000342525 30314 0037269 30349 35000010394 000342525 30314 0405718 30303 11000010363 000342539 30360 0016287 30305 55000010363 000342539 30360 0023718 30341 19000010363 000342539 30360 0024145 30342 18000010363 000342539 30360 0035284 30339 21000010363 000342539 30360 0037269 30349 11000010363 000342539 30360 0405718 30303 57000010379 000342546 30303 0016287 30305 2000010379 000342546 30303 0023718 30341 38


Customer. Dealer.ReferralID CustomerID PostCode DealerID PostCode AbsDifference000010379 000342546 30303 0024145 30342 39000010379 000342546 30303 0035284 30339 36000010379 000342546 30303 0037269 30349 46000010379 000342546 30303 0405718 30303 0000010407 000342546 30303 0016287 30305 2000010407 000342546 30303 0023718 30341 38000010407 000342546 30303 0024145 30342 39000010407 000342546 30303 0035284 30339 36000010407 000342546 30303 0037269 30349 46000010407 000342546 30303 0405718 30303 0

Query Strategy 3: For Each Referral, Determine Whether the Referred Dealer Corresponds to theMinimum Postal Code Difference

Find the minimum postal code difference for each referral. Based on the query created above,another query needs to be created to find the minimum postal code difference for each ReferralID. Thisminimum value identifies the “closest” dealer to the customer for each ReferralID.

Query CreationTo create a query to find the minimum postal code difference for each ReferralID, Select Que-

ries from the Objects menu and double-click Create query in Design view. Select the Queries tab,double-click the AllCombinations query, and Close the Show Table screen. Double-click ReferralIDand AbsDifference from the AllCombinations query to add them to the query.

Select the ∑ icon from the tool bar to produce a total row. The total row lists Group By as thedefault option for each field. This option defines the groups for calculations. Leave Group By in theReferralID column. Left-click in the Total row of the AbsDifference column and choose Min fromthe pulldown menu to show the minimum value of AbsDifference for each ReferralID. (For onlineinformation about calculations in queries, on the help menu, Microsoft Access Help, select theIndex tab, type “calculate”, select Search, and select “Perform calculations in a query” in thetopic list.) Save the query with a meaningful name, e.g., MinDifferences. Access supplies a defaultname for the expression, i.e., MinOfAbsDifferences.The result shows the minimum value of the postal code difference for each ReferralID.

Query: MinDifferencesReferralID MinOfAbsDifference000010345 2000010352 9000010363 11000010379 0000010382 2000010394 9000010407 0

Determine whether the referred dealer corresponds to the minimum postal code difference. Adealer is defined to be “closest” if the absolute value of the difference between its postal code and thecustomer’s postal code is no greater than the difference for any other dealer. Thus, comparing theMinOfAbsDifference field in the MinDifferences query with the DiffAsReferred field in the PostDiffActualquery identifies referrals for which the minimum was not attained. Such referrals would not comply withthe car maker and dealers’ agreement about how to allocate referrals..

Query CreationSelect Queries from the Objects menu and double-click Create query in Design view. Select the

Queries tab and double-click the queries PostDiffActual and MinDifferences to make them appear


FIGURE 1Postal Code Map

30349

Legend:

D = Dealer location

Scale: 1.75 inches = 5 miles


in the Show Table screen. Close the Show Table screen. Join the queries by clicking on ReferralIDin one query and dragging to ReferralID in the other query. Select the needed fields: ReferralID,CustomerID, Customer.PostCode, DealerID, Dealer.PostCode, and AbsDiffActual from thePostDiffActual query and MinOfAbsDifferences from the MinDifferences query. Save the tablewith a meaningful name, e.g., PostcodeDiscrepancies.

Open the query in Design mode and left-click in the blank field to the right of theMinOfAbsDifference field. Select the build icon (magic stick symbol) from the tool bar to open theExpression Builder screen. In the middle column, double-click AbsDiffActual, enter a minus sign(–), and double-click MinOfAbsDifferences. Take the absolute value of the expression by enclosingthe expression in parentheses and typing “Abs” to the left of the left parenthesis. Give the expres-sion a name by typing “Discrepancy:” to the left of “Abs”. The expression is:

Discrepancy: Abs([AbsDiffActual] – [MinOfAbsDifferences])Select OK and save the query.

The following result of the query shows the postal codes of the actual dealer and the closest dealerfor the postal code of each customer. The Discrepancy field of the table shows whether postal codes ofthe referred and closest dealers are identical. The number 0 indicates that they are identical. In otherwords, all customers were referred to a “closest” dealer in terms of the minimum postal code difference.This analysis does not identify referrals for which there might be more than one “closest” dealer; it onlydetermines whether customers were referred to a dealer that was as “close” as any other dealer.

Query: PostcodeDiscrepancies

Customer Dealer AbsDiff MinOfAbsReferralID CustomerID .PostCode DealerID .PostCode Actual Difference Discrepancy

000010345 000342512 30344 0024145 30342 2 2 0000010352 000342525 30314 0016287 30305 9 9 0000010363 000342539 30360 0037269 30349 11 11 0000010379 000342546 30303 0405718 30303 0 0 0000010382 000342512 30344 0024145 30342 2 2 0000010394 000342525 30314 0016287 30305 9 9 0000010407 000342546 30303 0405718 30303 0 0 0

Assessment of Car Maker ComplianceThe zero values in the Discrepancy column mean that the car maker referred all customers to deal-

ers that were as “close” to the customers as any other dealers. Thus, the car maker complied with theagreement it had with the dealers to refer customers to the “closest” dealer where closest is defined as thedealer such that the absolute value of the difference between the dealer’s postal code and the customer’spostal code is minimized.

So why would customers complain? Figure 1 illustrates why: postal codes that are adjacent to eachother on the map may have larger absolute numerical differences than postal codes that are not adjacentto each other on the map. For example, the referral scheme refers a customer in postal code 30360 to adealer in postal code 30349, which minimizes the postal code difference between the customer and thedealer, but dealers in postal codes 30305 and 30342 are physically closer to the customer. The reasonthat customers complain is not because the car maker did not implement the agreement correctly butbecause the design of the system, i.e., using this approach to allocating referrals, is flawed. This designflaw is relatively easy to detect with just a little thought about the process and what following it impliesfor customers. The nature of the flaw is apparent from the postal code map.

The lesson in this example is not that postal code differences are a poor basis for allocating referralsbut that information systems may have design flaws. Even systems created by the most well-intentioneddevelopers exhibit flaws. Systems that begin life with no flaws may exhibit some as the world whoserealities they represent changes and people use them in unanticipated ways. Some of the flaws are easyto detect with a little thought, but others are hard to detect even with a great deal of thought. To make


matters worse, developers often introduce new flaws during attempts to correct existing ones. Further-more, it may be difficult to distinguish between system deficiencies due to lack of compliance withsystem procedures (e.g., car dealers not complying with the agreement to contact customers within 48hours) and system deficiencies that are a function of a flawed design (e.g., allocating referrals on aminimize postal code difference basis).

B. New Design for “Closest” Dealer CriterionAlthough it has the advantage of being easy to implement because all the required data (postal

codes) are available in the tables, the “minimize postal code difference” approach to referring customersto dealers is sure to lead to unhappy customers because the allocation to dealers is unrelated to custom-ers’ physical proximity to dealers. While appealing because the data are available in the tables, allocat-ing referrals based on telephone number area codes or prefixes would suffer from the same deficiencybecause area codes and prefixes no longer correspond to compact locales in high-growth areas.

A better design for the “closest” dealer criterion might be to minimize the physical distance betweenthe customer and the referred dealer but to do so in a way that minimizes the processing effort. Forexample, if a geographic database were available, the computer program that refers customers couldlook up customers’ and dealers’ addresses in units that could be manipulated to give physical distancesexpressed in miles or driving time between the customer’s and dealer’s postal codes.

Part 3 Solution: Objective QuestionsA. Solution: 3. When tables have identical primary keys, the query manager joins them automatically

on their primary keys.B. Solution: 1. Queries do not inherit the primary key designations from their source tables so the user

has to join the tables by connecting the attributes to be joined.C. Solution: 4. Customers are likely to think of response time as beginning when they inquired rather

than when the car maker sent the referrals to dealers, which eliminates b and c. Web-savvy custom-ers are more likely to be interested in the elapsed time to the dealer’s email responses, which elimi-nates a.

D. Solution: 1. The term “[DealerResponseToReferral]!” is necessary in order to distinguish whichResDateTime attribute is being referred to in the calculation.

E. Solution: 3. Because the CustomerInquiry table has no primary key and the Dealer table and theReferralToDealer table joined themselves on DealerID, the query manager matches every row inthe CustomerInquiry table with every row in the join of Dealer and ReferralToDealer (which has arow for every row in the ReferralToDealer table), the result has the number of rows equal to thenumber of rows in the CustomerInquiry table times the number of rows in the ReferralToDealertable.

F. Solution: 2. The Customer table and the ReferralToDealer table joined themselves on CustomerID,and the Dealer table and the ReferralToDealer table joined themselves on DealerID. This gives onerow in the results for each row in the ReferralToDealer table provided the Customer and Dealertables contained rows corresponding to the customers and dealers appearing in the ReferralToDealertable.

G. Solution: 3. As dealer locations and postal code areas change, changes in the lookup table would berequired, which eliminates choice 1. In high-growth areas, telephone prefixes (xxx in yyy xxx-dddd) do not represent compact areas, and they may not be unique in areas with multiple telephonearea codes, which eliminates choice 2. Choice 3 remedies the defect of the postal code scheme andis least likely to require future modifications. All dealers except the one with the most sales are sureto insist that they receive more referrals, which would necessitate changes, which eliminates choice4.

H. Solution: 1. Choice 1 would be the most attractive one because it would help customers find the dealerthat has the specific configuration the customer wants in stock. Dealers are likely to resist choice 2,even though the information is already available on the web, because they prefer to negotiate bystarting from the higher sticker price, which eliminates choice 2. Dealers would resist choice 3


because it would likely decrease their sales, which eliminates choice 3. In the short run, many stateshave laws prohibiting car makers from selling vehicles directly. Customers can get the informationin choice 4 from web sites independent of the car maker, which customers might deem to be morereliable, which eliminates choice 4.

I. Solution: 4. Primary keys must uniquely identify the row in the table. Choice 1 does not en-sure uniqueness because there may be more than one inquiry per customer. Choice 2 does not ensureuniqueness because there may be more than one inquiry with the same timestamp across manyinquiries from many persons. Choice 3 does not ensure uniqueness because someone may inquireabout more than one configuration per model. Choice 4 is the one most likely to be unique. Al-though someone may inquire about a model in more than one configuration, he or she is unlikely tobe able to do so at exactly the same time when the timestamp includes seconds.

J. Solution: 3. In choice 1, subtracting an absolute value from an absolute value may give a result witha minus sign. In choice 2, the expression will give a result with a minus sign when Dealer.PostCodeis greater than Customer.PostCode. Choice 3 is equivalent to the calculated expression becauserearranging the order of the subtraction within the absolute value function does not change theresult. In choice 4, the expression will give a result with a minus sign when Customer.PostCode isgreater than Dealer.Postcode.

III. SUMMARYThis case illustrates database querying as a way to provide continuous assurance. It develops data-

base queries for giving a car maker assurance that dealers respond to web site customers on a timelybasis and for giving car dealers assurance that the car maker sends them all the appropriate customerreferrals. The queries are developed in Microsoft Access, but any database system supporting QBE orSQL could be used. The execution of the queries could be automated so that they run continuously (or ona periodic basis), and referrals not in compliance with the agreements might be referred automatically tocar maker and dealer managements. As trading partners evolve their ways of relating to each other totake advantage of new communication capabilities (especially web-based features) and to respond tonew entrants (especially web-based ones), this approach to continuous assurance is likely to becomemore common. Accountants should be able to satisfy this need by developing continuous assuranceapplications based on database queries.

REFERENCESAlbrecht, W. S., and R. J. Sack. 2000. Accounting Education: Charting the Course through a Perilous Future.

Accounting Education Series, Volume 16. Sarasota, FL: AAA. Available at: http://aaahq.org/pubs/AESv16/toc.htm.

Borthick, A. F. 1992. Helping users get the information they want, when they want it, in the form they want it:Integrating the choice and use of information. Journal of Information Systems 6 (2): v–ix.

———. 1996. Helping accountants learn to get the information managers want: The role of the accountinginformation systems course. Journal of Information Systems 10 (2): 75–85.

Harris, E. 1999. Web car-shopping puts buyers in driver’s seat. Wall Street Journal (April 15): B10.Perry, J. T., and G. P. Schneider. 2001. Building Accounting Systems Using Access2000. Cincinnati, OH: South-

Western.Rich, M., and K. Lundegaard. 1999. Car dealers fight forays by makers. Wall Street Journal (April 28): S1.Wallace, B. 1998. Car dealers yield sales to Internet. Computerworld (August 31): 1, 17. Available at: http://

www.computerworld.com/cwi/story/0,1199,NAV47_STO43338,00.html.———. 1999a. Car-buying site could help put GM on the road again. Computerworld (March 8): 43. Available at:

http://www.computerworld.com/cwi/story/0,1199,NAV47_STO34814,00.html.———. 1999b. Carmakers resist selling direct over the Internet. Computerworld (May 31): 6. Available at: http://

www.computerworld.com/cwi/story/0,1199,NAV47_STO35857,00.html.Waltner, C. 2000. Auto dealers come to terms with the Web. Informationweek (April 24): 104–110. Available at:

http://www.informationweek.com/783/dealer.htm.Warner, F. 1999. Auto makers and dealers wage battle. Wall Street Journal (June 16): A2.Wilson, T. 2000. Auto dealers take web offensive. Internetweek (February 7): 1, 51. Available at: http://

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Developing Database Query Proficiency: Assuring Compliance for