1

Chapter 2: Types, Sources and Collection of Data Case 2.1 The objective of this case is to present a questionnaire relating to live environment, for discussion relating to issues encompassing a questionnaire. There is no straightforward or unique answer to this case. The participants may be asked to offer suggestions regarding defining of objectives, comprehensiveness , clarity and unambiguity of questions and the options therein. As an illustration, a modified version of the questionnaire is given below.

Questionnaire for Consumers

What is your gender ?

□ Male □ Female

What is your age ?

□ Under 18 yrs □ 18-21yrs □ 21-50yrs □ 50+ yrs

Are you employed ?

□ Yes □ No

What is your annual household income ?

□ ≤ 2 lakhs □ 2-5 lakhs □ 5-10 lakhs □ ≥ 10 lakhs

How long have you owned a cell phone ? _____ Years ______ Months

Which Cellular Operator do you use ?

□ Vodafone □ Reliance □ AirTel □ BPL □ Others (please specify):

What type of connection do you use ?

□ Prepaid □ postpaid

2

What do you normally use the phone for? (You can choose more than one option)

□ keeping in touch with friends & family □ Business □ for emergencies

□ Others (please specify):

How much on average is your monthly expenditure on the mobile connection ?

Rs _________

Whom are the mobile charges paid by ?

□ Self □ Company □ Family

How many and how much of the following Value Added Services do you utilize ?

Kindly tick (√) those applicable

Services Not Aware Never Used Used

Occasionally

Used

Frequently

SMS

Voice Mail

Messenger Services

Ring tones / Logo

Download

Dial-in-Services

News & Cricket Updates

GPRS

Itemised Bill

STD/ISD

Roaming

3

MMS

Internet

Radio

What are the reasons for not using some services?

□ Not Aware □ Too Expensive □ Complicated □ No Utility / Time

□ other reasons (kindly specify):___________________________________________

Would you use the above services more frequently if they were cheaper? Yes / No

Kindly tick the information services that you would be interested in,

□ Movies □ Sports □ Stocks □ News □ Weather

□ Dating □ Auctions □ Jokes □ Religious Services

□ TV Listing □ Horoscopes

How much do you suggest these services should cost (per min or per message)?

□ 50p - Rs 1 □ Rs 1 – Rs 1.50 □ Rs 1.50 – Rs 2 □ > Rs 2 □ Others:

Chapter 3: Presentation of Data

Case 3.1 The Tables may be prepared for CGPA, Work Experience and Age by taking appropriate class intervals. The Tables could be

• Frequency Table • Cumulative Frequency Table

The following charts could be drawn for

• Bar charts for ‘Background’ and ‘Specialisation’ ( separately) • Histograms for CGPA, Work Experience and Age • Frequency Polygons, ‘Less Than’ and ‘Greater Than’ Ogives for CGPA, Work

Experience and Age • Pie Charts for ‘Backgrounds’ and ‘ Specialisations’

Students may be asked to suggest other Tables charts and graphs. Case 3.2 Some salient features could be as follows:

• Frequency Table and Histogram for Number of Billionaires according to wealth • Cumulative Frequency Table and Ogives for Number of Billionaires according to

wealth • Grouping of Billionaires according to Given Sectors or according to Groups of

Sectors and drawing bar chart(s) Case 3.3 Frequency table could be prepared for each of the two parameters viz. business per employee and return on assets, for each of the two years. The frequency tables will be different for each type of bank and for each year. Comparisons could be made by studying these tables. Line Graphs for each type of banks for the parameters ‘Business per Employee’ and ‘Return on Assets’ comparing figures for 2001-2 and 2005-6 Combine three charts in one showing different types of banks by different colour. Case 3.4 Line graph may be drawn by taking maturity periods on the X-axis, and expense ratios on the Y- axis Similarly multiple bar chart may be drawn by taking maturity periods on the X-axis, and expense ratios on the Y- axis For example the multiple bar chart reveals that for maturity periods 1 and 10 years, the expense ratios are minimum for Birla Sun Life Classic Life Premier and for periods 15 to 30 years HDFC Unit Linked Endowment is minimum.

1

Chapter 4: Measures of Central Tendency and Dispersion Case 4.1 All the measures of location and dispersion as also Five Number Summary can be calculated by the template, Elementary Statistics Chapter 4.xls or using various formulas given in the Chapter. The measures which can be easily combined are Arithmetic Mean, Geometric Mean and Standard Deviation ( Variance ) by using the formulas (4.3), (4.9) and (4.21). The measures derived for full data and combined data must be equal. Case 4.2 For each group of banks, Business per Employee can be worked out by dividing total business in the group by total number of employees in the group. While total number of employees can be found just by adding number of employees for each bank, the total business for all banks in a group can be found by first multiplying Number of Employees by business per Employee for each bank, and then totaling these products for all banks in a group. ROA cannot be calculated in the similar way,. For doing so, we require the knowledge of total of returns as also total assets. The total assets of each bank are not given, and hence average OA cannot be calculated. It may be noted that even though numerically it can be calculated, it will not have any physical significance with in a group however this can be used just for comparison among groups. While average business per employee can be worked out for each group for the year 2005-06 by dividing the total business by total number of employees, it cannot be worked out for the year 2001-02 because the number of employees in that year is not given. It is to be noted that we cannot calculate average percentage just by adding percentages for each bank. The productivity of an employee is defined as the business per employee. Thus, we are given the productivity of employees for each bank in the years 2001-02 and 2005- 06. We can calculate the Coefficient of Variation of productivity of banks in both the years, within each group of banks. We can also calculate variability among the groups by measuring variability among average productivity of each of the three groups. Similar analysis can be done for Return on Assets. Explain As regards the extent of skewness, it can be measured by the Bowley’s coefficient of skewness defined by the formula ( 4.23) Case 4.3 The most appropriate measure of variability is the Coefficient of Variation. It can be defined for all the parameters wherever amounts are indicated. C.V may not be much meaningful for ranks, because two sets of data having different ranks under two criteria will have the same C.V. For example, if Nidhi, Jeetendra and Sanjay have ranks 1, 2 and 3 in Economics and 2, 3 and 1 in Statistics, then C.V. will be same for both the subjects.

2

Case 4.4 The coefficient of variation may be calculated for all the four sets of data, and values compared. The ‘All India’ values may be taken as the mean for calculating C.V. If the population figures were also given then per capita values could have been calculated, and disparities among per capita could have been calculated. Caution : The data for states Bihar , Madhya Pradesh and Uttar Pradesh are not comparable as these states were divided in two states each, and could therefore be excluded from the study. Case 4.5 Mean, median, range and standard deviation could be calculated for each period(tenure like 1 year). Coefficient of variation can also be calculated for each period. The above analysis could also be done for each insurance company to study variation among periods for each company. Case 4.6 C.V. for ranks may not be much useful for the given data. However, if the data about overall combined ranks is given for countries of each continent like Asia, Europe, etc., then C.V. among countries of each continent would give an idea as to in which continent the disparity is more. Case 4.7 Average earnings per run may be found by dividing the total earnings by total of runs scored. Similarly, average earning per wicket may be worked out by dividing total earnings by total number of wickets. Consistency can be measured by C.V. which can be worked out by the formula ( 4.22). As regards improvement, discussions could be held in the class to know their ideas. Rankings of players are derived based on their performance by several authorities like ICC. The players are also ranked according to their performance, and labeled as most valuable players in a tournament. Case 4.8 The following frequency table shows that the average age in lowest group is about the same as for the highest group. Thus, age does not have any impact on the net –worth of a billionaire. Net worth ($ Billion) Average Age

1 - 5 55.92

5-10 56

10-15 60

15 - 20 55

3

Case 4.9 The variability in market capitalisation can be best measured by C.V. If the prices and earnings per share are given, comparisons could be more meaningful. Instead of fixing the ranks and arranging the companies according to their ranks, the sequence of companies could be fixed and their corresponding ranks could be indicated. Percentage increase in market capitalization can be calculated and compared for different companies.

1

Chapter 5: Probability Case 5.1 The following contingency Table (called Pivot Table in Excel terminology ) can be prepared for the given data, to answer the questions asked in the case. Background

Specialisation Commerce Economics Engineer Information Science Grand Total

Finance 12 3 2 3 20 Marketing 6 2 15 4 4 31 Resources 1 1 2 System 1 1 2 Systems 1 1 Grand Total 18 5 18 7 8 56

(i) the probability that the student is with commerce background is 18/54 (ii) the probability that the student is with commerce background given that

he/she has opted for marketing is 6/31 (iii) the probability that the student has specialization in Finance given that he/she

has engineering background 0/18 = 0 (iv) the probability that the student has engineering background as also opted for

specialization in Finance 0/56 = 0 Case 5.2 Total number of employees = 100 Experience in Years Level of Satisfaction 1 to 5 6 to 15

16 and More

Marginal Totals

Low 10 6 2 18Medium 14 20 10 44High 6 16 16 38Marginal Totals 30 42 28 100

(a) Number of Persons with Low Level of Satisfaction = 18 Number of Persons with Medium Level of Satisfaction = 44 Number of Persons with High Level of Satisfaction = 38 Number of Persons with experience from 1 to 5 years = 30 Number of Persons with experience from 6 to 15 years = 42 Number of Persons with experience of 16 years and more = 28

2

(i) The probability that an employee selected at random will have high level of satisfaction = 38 / 100.

(ii) The probability that an employee with more than 16 tears of experience is highly satisfied = 16 / 28.

(iii) The probability that an employee with medium level of satisfaction will have experience from 6 to 15 years = 20 / 44.

Case 5.3 This case can be solved with the help of the Theory of Expectation discussed in Section 5.12. Probability that market will grow = 1 / ( 1 + 2 ) = 1 / 3 Probability that market will not grow = 2 / ( 1 + 2 ) = 2 / 3 If he continues with old factory Expected Profit = 3 × 1 / 3 + 0.5 × 2 / 3 = 1.33 If he sets up new factory Expected Profit = 5 × 1 / 3 − 1 × 2 / 3 = 1.00 Thus, it is advisable to continue with the old factory. If the odds are revised to 2 to 3 i.e. Probability that market will grow = 2 / ( 2 + 3 ) = 2 / 5 Probability that market will not grow = 3 / ( 2 + 3 ) = 3 / 5 If he continues with old factory Expected Profit = 3 × 2 /5 + 0.5 × 3 / 5 = 1.5 If he sets up new factory Expected Profit = 5 × 2 / 5 − 1 × 3 / 5 = 1.4 Thus, even then, it is advisable to continue with the old factory. Case 5.4 This case can be solved with the help of Bayes’ Theorem. Here: A : Event that the venture is a failure B1 : Failure is due to poor management skills B2 : Failure is due to lack of financial expertise B3 : Failure is due to lack of inadequate marketing efforts Given: P(B1) = 0.5 P(B2) = 0.3 P(B3) = 0.2

3

Also, it is given that P( A/B1) = 1 / 3 P( A/B2) = 1 / 4 P( A/B3) = 1 / 5 What is required is to find out P(B1/A)? As per Baye’s Theorem, P(B1/A) is given as:

P(B1/A) )/()()/()()/()(

)/()(

332211

11

BAPBPBAPBPBAPBPBAPBP

++=

5/12.04/13.03/15.0

3/15.0×+×+×

×=

04.0075.0167.0

167.0++

=

= 0.592

P(B2/A) )/()()/()()/()(

)/()(

332211

11

BAPBPBAPBPBAPBPBAPBP

++=

5/12.04/13.03/15.0

4/13.0×+×+×

×=

04.0075.0167.0

075.0++

=

= 0.266

P(B3/A) )/()()/()()/()(

)/()(

332211

11

BAPBPBAPBPBAPBPBAPBP

++=

5/12.04/13.03/15.0

5/12.0×+×+×

×=

04.0075.0167.0

04.0++

=

= 0.142

4

Case 5.5 This case can be solved with the help of the Theory of Expectation discussed in Section 5.12. Before applying the theory, it may be verified that the sum of probabilities for various rates of return for each of the funds is equal to 1. For Progressive Fund: Expected or Mean Rate of Return = Mean of x = E( x ) = Σ x i p ( x i ) = 5 × 1/5 + 10 × 1/6 + 15× 2/5 + 20 × 1/10 + 25 × 1/20 + 1/12 × 30 = 1 + 1.67 + 6 + 2 + 1.25+ 2.5 = 14.42 Variance of x = E [(x – E(x)]2 = E (x2) – {E(x)}2 Now, E(x2) = Σ x2 p(x) = 25 ×1/5 + 100 × 1/6 + 225 × 2/5 + 400 × 1/10 + 625 × 1/20 + 900 × 1/12 = 5 + 16.67+ 90+ 40+ 31.25 +75 = 257.92 Thus, Variance(x) = E (x2) – {E (x)}2 = 257.92 – (14.42)2 = 49.98 For Reliable Fund: Expected or Mean Rate of Return = Mean of x = E( x ) = Σ x i p ( x i ) = 5 × 1/8 + 10 × 1/6 + 15× 5/12 + 20 × 1/6 + 25 × 1/8 + 30 × 0 = 5/8 + 10 /6 + 75/12 + 20/6 + 25/8 + 0 = 15 Variance of x = E [(x – E(x)]2 = E (x2) – {E(x)}2 Now, E(x2) = Σ x2 p(x) = 25 ×1/8 + 100 × 1/6 + 225 × 25/144 + 400 × 1/36 + 625 × 1/20 + 900 × 0 = 25/8 + 100/6 + 225 × 25/144 + 400/36 + 625 /20

5

= 258.33 Thus, Variance(x) = E (x2) – {E (x)}2 = 258.33 – (15)2 = 33.33 It may be noted that for the Progressive Fund, not only the expected rate of return is marginally better ( 15 as compared to 14.42 for Reliable Fund), but the risk is also less because of lesser variance i.e. volatility. Case 5.6 This problem relates to Bayes’ Theorem as the event (Diagnosed with Mammogram ) is known , and we want to know the cause for this as to whether it was due to the woman suffering from breast cancer or not suffering from breast cancer A : Woman diagnosed positive with Mammogram B1 : Woman suffering from breast cancer B2 : Woman not suffering from breast cancer It is given that P(B1 ) = 0.005 P( B2 ) = 0.995 P(A/B1 ) = 0.8 P(A/ B2 ) = 0.1 Using Bayes’ Formula ( 5 .7 ), we get

P(B1 / A) = )()( 1

APABP =

∑ ××

)/()()/()( 11

ii BAPBPBAPBP =

1035.0004.0 = 0.037

It may be noted that the probability of a woman having breast cancer is 0.005, which gets increased to 0.037 if the woman’s mammogram indicates positive.

Chapter 6: Sampling Techniques Note: In sampling, there may not be a unique method, and one could follow more than one method. We have given only some guidelines. Case 6.1 An estimate of the work experience could be obtained by taking a representative sample of 50 students from the total strength of 500. First, one has to decide on the factors which could have impact on the work experience. These could be as follows: Sex of student – Male(400) or Female(100) Year of study – First(250) or Second year(250) Academic background – Commerce, Science, Arts, Engineering( including IT) Let us assume that there are 5 Divisions in each year. Each division contains male as well as female students. The students could go to each Division and select 5 students at random taking care of proportion of female students as well as academic background. thus, data bout 25 students from each year could be collected, and the average work experience of 50 students could be obtained. This is one of the ways of in which the answer could be provided in a couple of hours. It may be noted that each division may be considered a cluster. Case 6.2 For taking a simple random sample of, say 10 students, from the 56 students, it is better to first arrange the students in a random manner, say alphabetically which is normally available in attendance registers. Then from a the random number table given on page 6.16, select 10 3-digit numbers in any column. Divide each number by 56, and note the remainder. Select that serial number from the above list of Table, and note his/her CGPA. Like this, also note the CGPAs of nine other students. If the remainder is 0, then roll no. 56 may be selected. Average of these ten CGPAS will give an estimate of the average of 56 students; compare this with the average obtained for all the 56 students. For calculating the average work experience, it may be noted that one student has work experience of 104 months. This is an extreme value and may, therefore, may not be included in the sample. In fact, it may be eliminated and only 55 students may be considered for obtaining average work experience. Case 6.3 Since the data is arranged in descending order, it would be appropriate to adopt systematic sampling. Since two of the billionaires are having much more wealth than all others, it might be advisable to exclude these two for calculating the average. Now, 10

billionaires are to be selected out of the 100 billionaires, we may select sampling interval of 10 i.e. every 10th billionaire may be selected. To select a billionaire at random from the first 10 billionaires, we may select all the random numbers in the first column of Table, add up the last digits of those numbers and note the last digit in the sum. Since, the last digit is 7, we may select first billionaire as 7, and next 9 as 17th, 27th, 37th,………….., and 97th. The average of these 10 billionaires would provide an estimate of the average wealth of the 100 billionaires. Case 6.5 First of all, one could estimate from the past about the percentage of customers coming to different counters. If the percentages are, say: 11, 12. 11,12, 10, 12, 9, 8, 8, 7, then the number of customers to be contacted for their level of satisfaction should be the same as the percentage for that cash counter. If the store’s business hours are, say from 9 a.m. to 9 p.m., these could be divided in four slots of 3 hrs. each. The number of customers, whose opinion is recorded as they come out from the counter after paying the bill, could be roughly in the proportion of customers in various time slots. It could happen that the level of satisfaction might vary according to the rush in various periods. This may be taken in account while selecting the time period. Case 6.6 The actual value of business per employee can be found by adding business for all the banks and divided by the total of employees for all the banks. We may use stratified sampling to select banks from each of the three categories. there are 22 Public Sector, 22 Private Sector( Indian) and 5 Private Sector ( foreign banks). We may select a sample of size 13 banks – 5 from Public Sector, 5 from Private sector( Indian) and 3 from Private Sector (Foreign). Why we may select 3 out of 5 banks from foreign banks is that there is wide variation in ‘Business per Employee’ ratio, and so it may be desirable to select, the maximum, the minimum, and one out of the three others. One of the three can be selected by tossing a dice. If numbers 1 or 2 comes, select, first bank( ABN), if numbers 3 or 4 comes, select, second ( Deutsche ) and if 5 or 6 comes, select, third ( Standard). Selection of 5 banks each from Public and Private sector banks can be done either through Simple Random or Systematic Sampling. After 13 banks are selected in the sample, we may add up their business values and divide by the total number of employees in these 13 banks, to get the sample estimate of the ‘Business Per employee’.

CHAPTER 7: Statistical Distributions Case 7.1 The mean and s.d. of CGPAs ( represented by variable x ) are 3.09 and 0.24 respectively. (i) P( x > 3.2) = P { ( x - )/ > ( 3.2 - )/ } = P { z > ) . From the Table T1, this probability can be found. Multiplying this probability by 100, we get the percentage of students obtaining marks more than as 32.34. (ii) (2.93, 3.25) (iii) 3.4 Case 7.2 The average BPE can be found by the total business of a category of banks by total number of employees of that category of banks. Total business of a bank can be obtained by multiplying BPE by the number of employees in that bank. Once the average is found, one can calculate the s.d. of BPEs for each category of banks. The standardised variable z for BPE is ( BPE – Mean of BPE) / S.D. of BPE. Incidentally, one cannot find average ROA because we do not know the assets of each bank and so the total assets for all the banks in a category are not known. Case 7.3 The daily rate of returns are as follows : Rate of Return on a Day = ( Price on the Day – Price on the previous Day) / (Price on the previous Day) Date Rate of Return (

SENSEX) Rate of Return ( NIFTY)

6-10-2006

18-10-2006 Case 7.4 P ( x ≤ 5) = P ( (x – 10) / 2 ) ≤ (( 5 – 10 ) /2) = – 2.5 ) = .0062 Thus, only 6 out of 1000 picture tubes will be required to be replaced. P ( x ≤ ? ) = 0.02 From Table T1, we find the value of z area up to which is 0.02 is 0.05. Thus, z = – 2.054 = ( x – 10 ) / 2 Therefore, x = 10 – 4.108 = 5.892 Thus, the picture tubes could be guaranteed for a period of 6 years. Case 7.5 Let x be the variable representing the diameter. Initially mean of x is m(say) = 10, and s.d. = 5 Probability of a defective disc is = 1 – P ( 9 ≤ x ≤ 11) = 1 – P ( ( 9 – 10 )/0.5 ≤ x ≤ ( 11- 10)/0. 5) = 1 – P (– 2 ≤ x ≤ 2) . = 1 – 2 x 0.0.4772 = 1- 0.9544 = 0.0456 Thus, 4.56% of discs will be defective. Probability of a disc being scrapped with this setting is = P ( x ≤ 9 ) = P ( z ≤ – 2) = 0.0.0228

Thus, 2.228% will have to be scrapped. It is desired that maximum probability of scrapping is = 0.01 From Table T1, the value of z for this area on the left side of standard normal distribution is – 2.327 We can find the mean setting ( m1) assuming to be unknown. It is desired that z = (9 – m1) / 0.5 = - 2.327 Thus, m1 = 9 + 1.1635 = 10.1635 i.e. the setting should be at 10.1635 so that 1% of output is scrapped. Probability of discs to be reworked = P ( x > 11) = P ( (x – 10.1635) / 0.5)

> (11 – 10.1635) / 0.5) = P ( z > 1.673) = 0.0472 Thus, 4.72% discs will have to be reworked. Case 7.6 Let the variable spending be represented by x with mean 20,000 and s.d 5,000 The cut off value of x z = ((x – 20000 ) / 5000) = 2.327 ( value of z, for area to the right of it = 0.01) Therefore, x = 20000 + 11635 = 31635 The reasons for more than expected % of customers qualifying for the offer could be either increase in mean or decrease in s.d. Assuming that s.d. has not changed we want that value of mean so that the probability of spending exceeding 31635 is 5%. P ( x > 31635) = 0.95 P ( ( x – m1/ 5000) > ( 31635 – m1) / 5000) = 0.95 This implies that (31635 – m1) / 5000) = = 1.645 Thus, m1 = 31635 – 1.645 x 5000 = 23410 Thus, mean spending has increased to 23,410.

Assuming that mean has not changed we want that value of s.d. , say σ1 so that the probability of spending exceeding 31635 is 5%. P ( x > 31635) = 0.95 P ( ( x – 20000/σ1) > ( 31635 – 20000) / σ1) = 0.95 This implies that (31635 – 20000) / σ1) = = 1.645 Thus, 1.645 x σ1 = 11635 or, σ1 = 7295 Thus, s.d. of spending has increased from 5000 to 7295. Case 7.7 We assume that the waiting time, say x, follows normal distribution with mean = 10. However, it is desired that m and σ should be such that P ( x ≤ 5) = 0.95 or, P ((x – m) / σ) ≤ (( 5 – m) / σ)) = 0.95 This implies that (( 5 – m) / σ)) = 1.645 or, m + 1.645 σ = 5 Thus, the system has to be designed in such a way that the above relationship is satisfied. For example, if m= 3, then σ has to 1.21. However, if m = 4, then σ has to be reduced to 0.607. Thus, if the average time increases then the variation among waiting times has to be reduced. It is due to this reason that in many systems, a single que is formed, and persons from this que go to counter whichever is free. Further, all the counters provide all services. Such system tends to reduce s.d. of waiting times.

Chapter 8: Simple Correlation and Regression Analysis Case 8.1 One may calculate correlation coefficients between CGPA and Age separately for each background . For calculating coefficient of contingency between backgrounds and CGPA, the data may be converted into a Two way Table as shown in section 8.12. One may take range of CGPA score on the horizontal side and backgrounds on the vertical side. CGPAs could be grouped in 4 class intervals. Thus, one will have a Table with five rows for each of the background viz. Commerce, Economics, Engineering, Information Technology and Science, and four columns for the four class intervals for CGPAs. Similarly, another Table with four rows one for each area of specialisation viz. Finance, Marketing, Systems and Human Resources, and four columns one for each class interval may be made to calculate coefficient of contingency. Case 8.2 For each group of banks, one may calculate Pearson’s correlation coefficients between Business per Employee in 2001-02 and Business per Employee in 2005-06, and Return on Assets in 2001-02 and Assets in 2001-02. and 2005-06. Spearman correlation coefficients may be calculated by using data on ranks in the two years. Case 8.3 Beta of each stock may be calculated by taking the stock’ values and BSE values over the given period , as illustrated on page 8.47 in Section 8.10. One could take NSE values instead of BSE values to get another set of betas. Case 8.4 This case is similar to case 8.2. Case 8.5 One may calculate correlations between various pair of parameters e.g. Average Working Funds and Net profit Net Profit and Ratio of Fee Income to Total Income Ratio of Net Profit Average to Working Funds and Cost to Income Ratio It may be noted that it is difficult to interpret relationships among various parameters as the banks of varying types concentrating on different types of business. It is advisable for such studies to take each group of banks and interpret relationships in those groups.

Case 8.6 The contribution of urban population to average life expectancy is indicated by the coefficient of determination between life expectancy and % of urban population. Case 8.7 It may be carried on the same lines ass the Case 8.5.

Chapter 9: Multiple Correlation and Regression Analysis Template for ‘Multiple Regression’ may be used for deriving multiple regression equations as per procedure explained in Section 9.19 of Chapter 9 of the book, and given in the note below. Note: The data about values of all the variables may be inputted in the template. However, the answer will not come directly. After inputting the data, one may first unprotect the worksheet for further processing by choosing the option “protect” by typing the password as “protect”. Thereafter one may chose the option “Solver” under “Tools”. This is an integral part of EXCEL.(If it is not in the tools menu, one may select tools -> add-ins and select solver option by checking the box before solver) In the template that appears on the screen, one should choose ”Solve” by pressing “OK”, and in the resulting template, select the option as “keep solver solution” by pressing “OK”. Immediately, the coefficients of the regression equation will appear under respective names of variables. It may be mentioned that fitting a multiple regression equation is only a first step in studying the relationship among the dependent and independent variables. As indicated in Section 9.14 and 9.15, one has to test the significance of the regression as a whole as also of the individual regression coefficients of the independent variables. One has also to test for multicollinearity among the independent variables. then there is the issue of selecting the best regression equation; and there are several methods to help in this process. Usually sophisticated packages like SPSS can be used to help in such detailed analysis , and are beyond the scope of this book. Case 9.1 The multiple regression equation may be obtained by using the template ‘Multiple Regression’ and the solver in Excel as indicated in the note above. The equation may be obtained as follows; CGPA = 3.195 – 0.005 Age + 0.052 Background( 0 or 1) It implies that if a student’s background is Commerce, he/she gets 0.052 more than a student who does not have the Commerce background. Incidentally, if the test of significance for regression equation as a whole or of regression coefficients is carried out using ANOVA , all will be found to be insignificant. Case 9.2 The multiple regression equation may be obtained by using the template ‘Multiple Regression’ and the solver in Excel as indicated in the note above. The equation may be obtained as follows; Net Profit = – 573.01 + 0.00965 Av. Working Fds. + 14.938 Ratio of

Cost to Income + 7.627 Ratio of Fee Income to Total Income Incidentally, if the test of significance for regression equation as a whole is carried out the regression equation is found to be significant. However, individually, only Average Working Funds is found to be significant variable, the other two variables are found to be insignificant. Regression equations may be fitted for Net Profit on one of the above three independent variables, one by one, to ascertain the extent of linear relationship between Net Profit and each of the three independent variables. With dummy Private Sector = 1 and Public Sector = 0, the multiple regression equation fitted is Net Profit = – 573.01 + 0.00965 Av. working Fds. + 14.938 Ratio of Cost to Income + 7.627 Ratio of Fee Income to Total Income + 167.7 Private/ Public Sector Case 9.3 The multiple regression equation may be obtained by using the template ‘Multiple Regression’ and the solver in Excel as indicated in the note above. The equation may be obtained as follows; P/E Ratio = 791.879 – 62.214 Current Ratio +1621.6 Debt Equity Ratio Incidentally, if the test of significance for regression equation as a whole or of regression coefficients is carried out using ANOVA , all will be found to be insignificant. Case 9.4 The multiple regression equation may be obtained by using the template ‘Multiple Regression’ and the solver in Excel as indicated in the note above. The equation may be obtained as follows; Life Expectancy = 38.14 – 0.0002 GDP + 0.476 School Enrolment + 0.0376 Percentage of Urban Population Incidentally, if the test of significance for regression equation as a whole is carried out using ANOVA , it is found to be insignificant. But among the independent variables, only School Enrolment is found to be significant. Case 9.5 The multiple regression equation may be obtained by using the template ‘Multiple Regression’ and the solver in Excel as indicated in the note above. The equation may be obtained as follows; CGPA = 3.195 – 0.005 Age + 0.052 Background( 0 or 1)

Incidentally, if the test of significance for regression equation as a whole or of regression coefficients is carried out using ANOVA template, all will be found to be insignificant. Case 9.6 The multiple regression equation may be obtained by using the template ‘Multiple Regression’ and the solver in Excel as indicated in the note above. The equation giving dependence of SENSEX on RIL and INF may be obtained as follows; SENSEX = 3364.2 + 6.672 RIL + 0.688 INF Incidentally, if the test of significance for regression equation as a whole or of regression coefficients is carried out using ANOVA template, all will be found to be insignificant. The equation giving dependence of NIFTY on RIL and INF may be obtained as follows; NIFTY = 964.3 + 1.930 RIL + 0.198 INF Incidentally, if the test of significance for regression equation as a whole or of regression coefficients is carried out using ANOVA template, all will be found to be insignificant. Case 9.7 The multiple regression equation may be obtained by using the template ‘Multiple Regression’ and the solver in Excel as indicated in the note above. The equation may be obtained as follows; Final Grade Score = – 24.207 – 0.730 Graduate Level Score + 2.226 Entrance Exam. Score For a student with Graduate level Score as 80 and Entrance Level Score as 70, the predicted score for Final Grade is 73. Incidentally, if the test of significance for regression equation as a whole or of regression coefficients is carried out using ANOVA, all are found to be insignificant. *****

Chapter 10: Statistical Inference

Case 10.1 Calculation of correlation can be made through template on ‘Simple Correlation’ given on page 8.60. It can be tested for significance by using ‘t’ test illustrated on page 10.88. If it is significant, it is a significant factor explaining the variation in the CGPA of students. Its squared value indicates the proportion of variance of CGPAs explained by age. The following tests for equality of means(CGPAs) of students with various backgrounds may be carried out by using Formula 10.11 or Template 7 on page 10.103: Mean for Commerce Students = Mean for Mean for Science Students Mean for Commerce Students = Mean for Engineering Students Mean for Commerce Students = Mean for Information technology Students Similar analysis may be carried out for various specilasations. CGPA scores of students with various backgrounds may be grouped in suitable number of class intervals along with the frequencies in each class interval. Assuming normal distribution with the calculated means and standard deviations, expected frequencies may be worked out. Then Chi –Square test for goodness of fit given on page 10.72 may be used to judge goodness of fitting normal distribution. Case 10.2 Simple Correlation coefficients may be calculated as desired, and their significance may be tested by using the ‘t’ test illustrated on page 10.88. If it is significant, use of relationships is justified. Use of normal distribution is justified through the test of goodness of fit, mentioned in the Case 10.1. Case 10.3 The simple correlations between various pairs of the four indicators may be calculated, and tested for significance, as indicated in Case 10.1, above. Case 10.4 This case is similar to the Case 10.3, above. Case 10.5 Here, the template 11 on page 10.107 for ‘Chi Square Test for Independence’ can be used to test the association between the areas of specialisation and size of pay packages.

Case 10.6 This case is just like the Case 10.5, above. Case 10.7 The first part of the case is similar to Cases 10.5 and 10.6, above. For the second part, the Template 10 on page 106 for ‘ Chi Square Test for Goodness of Fit’ may be used. While the observed preferences are given, the expected preferences are 120 for each colour, on the assumption that there is no preference for any colour. Case 10.8 Here, again the first part of the Case is similar to Cases 10.5 and 10.6, above. As regards the second part, the conclusion would not change as the test only assumes three areas of specialisations without any name label.

Chapter 11: Analysis of Variance and Design of Experiments Case 11.1 It is a case relating to One Way ANOVA. (a) Data about CGPAs of students having five different backgrounds viz Commerce, Economics, Engineering, Information Technology and Science are given. The backgrounds may be termed as ‘Treatments’ in ANOVA terminology. The template ‘ONE WAY’ on page 11.27 may be used to draw conclusion about significant differences( variation) among CGPAs of students with different backgrounds. (b) Data about CGPAs of students having four different areas of specialisation viz. Finance, Marketing, Human Resources and Systems are given. The areas of specialisation may be termed as ‘Treatments’ in ANOVA terminology. The template ‘ONE WAY’ on page 11.27 may be used to draw conclusion about significant differences( variation) among CGPAs of students with different areas of specialisations. Case 11.2 It is a case relating to Two Way ANOVA. One factor is the ‘Indices’ and the other is ‘Months’. The template ‘ TWO WAY’ given on page 11.28 may be used to generate ANOVA Table. The table will indicate the significant differences among the four indices, if any. The table will also indicate the significant differences among the four months, if any but that is not required in the case. Case 11.3 It is a case relating to Two Way ANOVA with interaction. One factor is the ‘Area’ and the other is ‘Location’. The template ‘TWO WAY WITH INTERACTION’ given on page 11.29 may be used to draw conclusions about significant differences among areas and locations as among interactions. Also, refer to Illustration 11.3 in Ch.11. Case 11.4 It is a case relating to Two Way ANOVA. One factor is ‘Series’ and the other is ‘Days’. The template ‘ TWO WAY’ given on page 11.28 may be used to generate ANOVA Table. The table will indicate the significant differences among the ratings of three series, if any. The table will also indicate the significant differences among the ratings on eight days, if any..

STATISTICS FOR MANAGEMENT ( Some Changes) (Reference: Pages 11.15, 11.16 7 11.17 – Please replace printed values by values given in red ink. No change in text.) 11.4 Two – Way or Two – Factor ANOVA with Interaction It has been observed that there are variations in the pay packages offered to the MBA students. These variations could be either due to specialisation in a field or due to the institute wherein they study. The variation could also occur due to interaction between the institute and the field of specialisation. For example it could happen that the marketing specialisation at one institute might fetch better pay package rather than marketing at the other institute. These presumptions could be tested by collecting following type of data for a number of students with different specialisations and different institutes. However, for the sake of simplicity of calculations and illustration, we have taken only two students each for each interaction between institute and field of specialisation. Illustration 11.3: The data is presented below in a tabular format.

Institute A

Institute B

Institute C

Marketing 8 10 8 10 11 7

Finance 9 11 5 11 12 6

HRD 9 9 8 7 7 5

Here, the test of hypotheses will be For Institute: H0 : Average pay packages for all the three institutes are equal

H1 : Average pay packages for all the three institutes are not equal For Specialisation: H0 : Average pay packages for all the specialisations are equal

H1 : Average pay packages for all the three specialisations are not equal For Interaction: H0: Average pay packages for all the nine interactions are equal

H1: Average pay packages for all the nine interactions are not equal

Now we present the analysis of data and the resultant ANOVA Table. Incidentally, this is an example of two – way or two – factor (Institute and Specialisation) with interaction. First of all, a Table is reconstructed by working out marginal totals for Institutes and Specialisations as follows:

Institute A

Institute B

Institute C

Total

Marketing 8 10 8 54 10 11 7

Finance 9 11 5 54 11 12 6

HRD 9 9 8 45 7 7 5

Total 54 60 39 153 From the above Table, we work out the requisite sums of squares for preparing the ANOVA Table, as follows:

nsObservatioofnumber

ns)Observatio All of (Sum (CF) 2

TotalFactorCorrection =

= 18

)153( 2

= 1300.5 Total Sum of Squares = (Sum of Squares of All Observations) – CF (TSS) = 1375 – 1300.5 = 74.5

Sum of Squares Between or = 5.13006

456

546

54 222

−++

due to Fields of Specialisation (Row: SSR) = 9

Sum of squares due to Institutes Column:SSC = 5.13006

396

606

54 222

−++

= 39

2.... )( xxxxnSSI jiij +−−= ∑∑ ………….(11.5)

where, n is the number of observations for each interaction. In this example it is equal to 2.

ijx is the mean of the observations of ith row and jth column.

.ix is the mean of the observations in the ith row.

jx. is the mean of the observations in the jth column.

..x is the grand mean of all the observations. These terms can be calculated by first calculating the means of all the interactions as also the means of corresponding rows and columns, as given below,

Institute A

Institute B

Institute C

Row Mean

Marketing 9 9.5 8.5 9 Finance 10 10.5 6.5 9

HRD 8 7 6 7 Column Mean

9 9 7

Grand Mean 8.33

and then calculating the sum of squares for each interaction by the formula (11.5 ) as follows.

Institute A

Institute B

Institute C

Marketing 0.25 0 0.25 Finance 0.25 1 2.25

HRD 0 1 1 Total 6

Thus, Interaction SS (SSI) = 2 × 6 = 12 Sum of Squares due to Error = Total Sum of Squares – SS Due to Specialisation (Row: SSR) (column: SS – SS Due to Interaction – SS due to Institute = TSS – SSR – SSC – SSI = 74.5 – 9 – 39 – 12 = 14.5

Now, the ANOVA Table can be prepared as follows.

ANOVA Table

Source of Variation

Sum of Squares

(SS)

Degrees of Freedom

(d.f.)

Mean Sum of Squares(MSS) = SS÷ d.f.

‘F’= MSST÷ MSSE

α = 5% F(Tabulated

) Decision

Specialisation 9 2 4.5 2.7931 4.2565 Institute 39 2 19.5 12.103 4.2565 Reject

Interaction 12 4 3 1.8621 3.6331 Error 14.5 9 1.61111 Total 74.5 17

From the above Table, we conclude that the while the pay packages among the institutes are significantly different, there is no significant difference among the pay packages for fields of specialisations as also among interactions between the institutes and the fields of specialisations.

Chapter 12 : Non – Parametric Tests Case 12.1 Using the template on “Simple Correlation and Regression Analysis”, we can find the daily rate of returns on SENSEX as well as NIFTY. Thereafter, one can carry out the test as indicated in Illustration 12.4 in Section 12.5. Case 12.2 This case can be solved just like the above Case 12.1. Case 12.3 This Case can be solved by first calculating the daily returns of all the four variables viz. RIL, Infosys, NIFTY and Dollar, with the help of the template on “Simple Correlation and Regression Analysis”, then calculating medians of respective rate of returns, and finally carrying out the run test as described in Illustration 12.1 in Section 12.2. Case 12.4 As indicated as a hint for the Case, first of all, the ranks for all the four cities have to be converted to be uniformly varying from 1 to 50 in all the four cities. Thus, the rank ‘79’ of Suresh Krishna in Delhi ( lowest rank in Delhi) may be converted to 50. Similarly, rank’71’ of Renuka Ramnath in Delhi (second lowest in Delhi) may be converted to 49. Similarly, the rank ‘66’ of Sanjay Nair in Chennai (lowest in Chennai), may be converted to 50. Then, the F test, as per formulas 12.7 to 12.10 in the illustration 12.6 in Section 12.11, may be carried out.

Chapter 13 : Time Series Analysis Case 13.1 The first step in analysing a time series data is to plot the data. Doing so, we get the following graph:

Demand Deposits

0

500

1000

1500

2000

2500

3000

3500

4000

4500

0 5 10 15 20 25 30 35 40

Demand Deposits

The graph indicates seasonality, and therefore the template ‘ Trend and Seasonal time Series’ described on page 13.17 may be used to forecast demand deposits for the months in the year 2007-08. Case 13.2 Plotting the data over the five years we have the following graph:

Returns

-40

-20

0

20

40

60

80

100

0 5 10 15 20 25

Returns

It may be noted that neither there is some visible trend and nor there is any seasonality, and therefore time series analysis can not be used to analyse and forecast. Case 13.3 Plotting the data over the three years we have the following graph:

Web hits

0

100

200

300

400

500

600

700

800

900

1000

0 5 10 15 20 25 30 35 40

Web hits

The graph shows that there are different trends during the three years. On this basis, it is difficult to project the trend in the next year. In such a situation, one can use only trend method on the monthly values for the year 2004-05, and determine the trend equation. With this equation, one can project monthly values for the next year i.e. 2005-06. Case 13.4 Plotting the data over the five years we have the following graph:

HPI

170

175

180

185

190

195

200

205

0 10 20 30 40 50 60 70

HPI

It may be noted that there is secular increasing trend as well as seasonality, and hence time series analysis template may be used to analyse and forecast.

Chapter 14: ABC ANALYSIS Case 14.1 First, we may tabulate the cumulative wealth of billionaires starting from the first wealthiest one. From this cumulative value of wealth, we may note that 7 billionaires ( out of 102 i.e. about 7%) account for about 60% of the wealth. These may be clubbed together to form ‘A’ category. Then next 30% ( 30 out of 102) billionaires account for about 25% of wealth. These may be clubbed together to form ‘B’ category billionaires. The rest 63% ( 65 out 0f 102) accounting for only 15% wealth, may constitute ‘C’ category of billionaires. Name Wealth Sector Cumulative

Wealth Cumulative Percentage

Ratan Tata 125674.12 Large diversified 125674.1 16.8476 P. R. S. Oberoi 83739.00 Hospitality 209,413 28.0735 Azim H. Premji 64855.27 Software 274,268 36.7678 Mukesh Ambani 56414.35 Petrochemicals 330,683 44.3306 Sunil Mittal & Family 35558.22 Telecom 366,241 49.0975 Anil Ambani 34993.98 Large diversified 401,235 53.7880 Tulsi R. Tanti & Family Promoters 26139.69 Wind energy 427,375 57.2929 Anil Agarwal 18108.75 Metals 445,483 59.7205 Shiv Nadar 16698.47 Software & hardware 462,182 61.9591 Kumarmangalam Birla 16643.04 Large diversified 478,825 64.1902 Rahul Bajaj 12455.99 Auto 491,281 65.8600 Dilip S. Shanghvi 10584.49 Pharmaceuticals 501,865 67.2790 Baba Kalyani 7857.83 Auto components 509,723 68.3324 Savitri Jindal 7238.64 Steel 516,962 69.3028 Naresh Goyal (through Tail Winds) 6874.30 Aviation 523,836 70.2243 V. N. Dhoot 6706.88 Durables 530,543 71.1234 Vijay Mallya 6504.73 Spirits 537,048 71.9954 V. C. Burman 5815.94 FMCG 542,864 72.7751 Malvindar Singh & Shivinder Singh 5680.21 Pharmaceuticals 548,544 73.5366 Adi Godrej 5560.76 FMCG 554,105 74.2820 Subhash Chandra 5423.62 Media 559,528 75.0091 Brijmohanlal Munjal 5309.41 Auto 564,838 75.7209 S.P.Hinduja 5071.27 Large diversified 569,909 76.4007 Jaiprakash Gaur 5038.50 Construction 574,947 77.0762 Uday Kotak 5033.98 Banking & Fin.

Services 579,981 77.7510 Ajay G. Piramal 4923.32 Diversified 584,905 78.4110 M.V. Subbiah & Family 4784.08 Diversified 589,689 79.0524 The Kirloskar Family 4744.98 Engineering 594,434 79.6885 Keshub Mahindra 4506.42 Diversified 598,940 80.2926

The Murthy Family 4310.44 Software 603,251 80.8705 Habil F. Khorakiwala 4186.75 Pharmaceuticals 607,437 81.4317 B. Ramalinga Raju, Rama Raju & Family

3862.18 Software 611,300 81.9495

Venu Srinivasan 3629.69 Auto 614,929 82.4361 K. K. Birla 3533.71 Diversified 618,463 82.9098 Sashi Rula 3527.39 Diversified 621,990 83.3827 Jignesh Shah 3463.89 Software 625,454 83.8470 Yusuf K. Hamied & Family 3250.83 Pharmaceuticals 628,705 84.2828 The Gopalakrishnan Family 3156.28 Software 631,861 84.7059 Karsanbhai K. Patel & Family 3143.78 FMCG 635,005 85.1274 The Punj Family 3097.85 Construction 638,103 85.5427 Gautam Thapar 3070.21 Diversified 641,173 85.9543 Zydus family Trust 3054.94 Pharmaceuticals 644,228 86.3638 K. Anji Reddy & Family 3000.50 Pharmaceuticals 647,229 86.7660 The Dani family & the Vakil Family 2917.68 Paints 650,146 87.1580 G. K. Patni 2864.79 Software 653,011 87.5412 Sushir Mehta 2796.48 Power 655,808 87.9161 The Nilekani Family 2795.89 Software 658,604 88.2909 Kiran Mazumdar-Shaw 2716.55 Biotech 661,320 88.6551 L.K. Jain 2676.48 Diversified 663,997 89.0139 The Abraham Family 2534.10 Diversified 666,531 89.3536 Kishore Biyani Associates 2332.83 Retail 668,863 89.6664 Harsh Goenka 2324.81 Diversified 671,188 89.9780 The Sawhney Family 2287.71 Sugar & engineering 673,476 90.2847 B. K. Birla 2270.13 Diversified 675,746 90.5890 Desh Bandhu Gupta & Family 2144.61 Pharmaceuticals 677,891 90.8765 Ramesh Chandra & Family 2096.74 Construction 679,987 91.1576 Harsh C. Mariwala & Family 2085.63 FMCG 682,073 91.4372 Ajit Gulabchand 2082.21 Construction 684,155 91.7163 Arvind Dham 2054.75 Auto components 686,210 91.9918 Saldanha Family trust 2047.04 Pharmaceuticals 688,257 92.2662 P. V. Ramaprsad Reddy & Family 2028.61 Pharmaceuticals 690,286 92.5382 B. G. Bangur 1983.38 Diversified 692,269 92.8041 B. K. Parekh 1918.99 Chemicals 694,188 93.0613 R. S. Lodha 1888.03 Cement 696,076 93.3144 The Patni Family 1886.82 Software 697,963 93.5674 Rajan Raheja 1827.90 Diversified 699,791 93.8124 P. R. Ramasubrahmaneya Rajha 1789.10 Diversified 701,580 94.0523 Vidya Murkumbi & Narendra Murkumbi

1698.04 Sugar 703,278 94.2799

Soshil Kumar Jain & Family 1679.98 Pharmaceuticals 704,958 94.5051

R. D. Shroff & Family 1568.50 Chemicals 706,526 94.7154 Rupen Patel & family 1529.41 Construction 708,056 94.9204 The Saraogi family 1486.71 Sugar 709,543 95.1197 T. Venkattram Reddy & T. Vinayak Ravi Reddy

1463.80 Media 711,006 95.3159

S. P. Oswal 1403.92 Textiles 712,410 95.5041 Nusli Wadia 1390.14 Diversified 713,800 95.6905 Kamlesh Kumar Agarwal 1324.74 Engineering 715,125 95.8681 Abhijit Rajan & Associates 1306.00 Construction 716,431 96.0432 Vijaypat Singhania 1305.69 Textile 717,737 96.2182 SanjayDalmia 1303.43 Chemicals. Cement 719,040 96.3930 Murali K. Divi & Family 1295.30 Pharmaceuticals 720,336 96.5666 K. Raheja & Family 1285.40 Retail 721,621 96.7389 Sheth family 1278.47 Shipping 722,899 96.9103 Narrotam Sekhsaria & Suresh Neotia 1260.00 Cement 724,159 97.0792 G. V. Krishna Reddy 1253.52 Diversified 725,413 97.2473 K. S. Dhingra & Family 1252.21 Paints 726,665 97.4151 D. Jayavarthanavelu 1246.86 Textile machinery 727,912 97.5823 Capt. C. P. Krishnan Nair 1242.36 Hospitality 729,154 97.7488 Analjit Singh 1211.09 Healthcare 730,366 97.9112 B. M. Khaitan 1208.98 Diversified 731,574 98.0733 Asha Dinesh 1199.18 Software 732,774 98.2340 Dharaprasad R. Poddar 1195.44 Tyres 733,969 98.3943 The Ajmera Family 1175.44 Steel 735,145 98.5519 L. M. Thapar 1151.40 Paper 736,296 98.7062 Jamuna Raghavan 1117.00 Software 737,413 98.8559 Nitin Sandesara 1114.00 Software 738,527 99.0053 K. K. Modi 1097.29 Cigarettes 739,624 99.1524 Sushil Ansal 1081.28 Construction 740,706 99.2974 Pollachi Mahalingam 1073.74 Sugar, auto

components 741,779 99.4413 AC Muthiah 1048.92 Diversified 742,828 99.5819 The Sarin family 1040.73 Real estate 743,869 99.7214 Dinesh Hinduja,Rajendra Hinduja & Family

1040.73 Real Estate 744,910 99.8609

Dinesh Hinduja, Rajendra Hinduja & family

1037.28 Textiles 745,947 100

745,947 Case 14.2 First, we arrange the banks in descending order as per their business, and then we calculate cumulative business of banks starting from the first bank. We may note that only 4 out of 22 (

about 18%) conduct about 48% of business. These may be termed as ‘A’ category banks. Next 6 out of 22 ( 27%) account for about 32% of business, and these may be termed ‘B’ category banks. The rest 12 out of 22 ( i.e. about 55%) account for only 20 % of business, and may be termed as ‘C’ category banks in term of business parameter.. Name of bank Business

State Bank of India 59479144.02 59479144.02 24.8223

Canara Bank 20706542.01 80185686.03 33.4637

Panjab National Bank 19208913.24 99394599.27 41.4801

Bank of India 15928848.00 115323447.3 48.1276

Bank of Baroda 15339852.00 130663299.3 54.5294

Union Bank of India 11095503.87 141758803.1 59.1599

UCO Bank 9485370.00 151244173.1 63.1184

Central Bank of India 8954970.86 160199144 66.85552

Syndicate Bank 8584911.36 168784055.4 70.43824

Indian Overseas Bank 8576661.94 177360717.3 74.01752

Oriental Bank of Com 8532230.12 185892947.4 77.57825

Allahabad Bank 6297312 192190259.4 80.20629

Indian Bank 6284090 198474349.4 82.82882

Corporation Bank 5667358 204141707.4 85.19396

Andhra Bank 5619870.75 209761578.2 87.53929

State Bank of Hyderabad 5431168.72 215192746.9 89.80586

State Bank of Travancore 4437813.98 219630560.9 91.65789

United Bank of India 4399026 224029586.9 93.49372

Bank of Maharashtra 4302441.36 228332028.2 95.28925

Vijay Bank of India 4244274.44 232576302.7 97.0605

Dena Bank 3696784 236273086.7 98.60327

State Bank of Bikaner & Jaipur 3346839.65 239619926.3 100

Case 14.3 First, we arrange the banks in descending order as per the number of their employee and then we calculate cumulative number of employees of banks starting from the first bank. We may note that only 4 out of 22 ( about 18%) have about 50% of employees. These may be termed as ‘A’ category banks. Next 6 out of 22 ( 27%) account for about 25 % of employees, and these may be termed ‘B’ category banks. The rest 12 out of 22 ( i.e. about 55%) have only 25 % of employees, and may be termed as ‘C’ category banks in term of number of employees.

Bank Number of Employees

Cumulative Number of Employees

Cumulative % of Employees

State Bank of India 198774 198774 28.8487

Punjab National Bank 58047 256821 37.2734

Canara Bank 46893 303714 44.0790

Bank of India 41808 345522 50.1467

Bank of Baroda 38737 384259 55.7688

Central Bank of India 37241 421500 61.1737

Union Bank of India 25421 446921 64.8631

Syndicate Bank 24624 471545 68.4369

UCO Bank 24510 496055 71.9940

Indian Overseas Bank 24178 520233 75.5031

Indian Bank 21302 541535 78.5947

Allahabad Bank 18742 560277 81.3148

United Bank of India 17319 577596 83.8284

Oriental Bank of Com 14962 592558 85.9999

Bank of Maharashtra 14052 606610 88.03928

Andhra Bank 13169 619779 89.95054

State Bank of Hyderabad 13108 632887 91.85295

State Bank of Bikaner & Jaipur 12089 644976 93.60746

State Bank of Travancore 11642 656618 95.2971

Vijaya Bank 11494 668112 96.96526

Corporation Bank 10754 678866 98.52603

Dena Bank 10156 689022 100 Case 14.4 For this type of situation, there may be more than one solution/approach. One such solution/approach is indicated below. ABC analysis may be carried out for each type of loan i.e. auto, home, etc. For monitoring at the beginning of a loan, it may be classified according to the amount, as also the type of customer. The customers themselves may be classified in three categories, say ‘A’, ‘B’ and ‘C’, depending on the credibility of such customers in the past. For example, a senior executive in a Government/Reputed organisation may require less monitoring as compared to a self employed professional/businessman/trader or a junior/middle level employee. For defaulting loans, there are two parameters of importance. One is the amount, and the other is the period or number of installments for which default is reported. For monitoring purpose, the product of the amount and the number of monthly

defaults may be taken, and ‘ABC’ analysis may be carried out on this product. For loans relating to carry out business/trade activities, the borrowers may be arranged as per amount sanctioned/utilized, and ‘ABC’ analysis may be carried out. The outstanding balance may be monitored daily/weekly for top ‘A’ category of customers, and weekly/monthly for customers under ‘B’ category. For this type of situation, there may be more than one solution/approach. One such solution/approach is indicated below. ABC analysis may be carried out for each type of loan i.e. auto, home, etc. For monitoring at the beginning of a loan, it may be classified according to the amount, as also the type of customer. The customers themselves may be classified in three categories, say ‘A’, ‘B’ and ‘C’, depending on the credibility of such customers in the past. For example, a senior executive in a Government/Reputed organisation may require less monitoring as compared to a self employed professional/businessman/trader or a junior/middle level employee. For defaulting loans, there are two parameters of importance. One is the amount, and the other is the period or number of installments for which default is reported. For monitoring purpose, the product of the amount and the number of monthly defaults may be taken, and ‘ABC’ analysis may be carried out on this product. For loans relating to carry out business/trade activities, the borrowers may be arranged as per amount sanctioned/utilized, and ‘ABC’ analysis may be carried out. The outstanding balance may be monitored daily/weekly for top ‘A’ category of customers, and weekly/monthly for customers under ‘B’ category.

Chapter 15: Forecasting Methods Case: 15.1

:

Linear

y = 3.4336x + 56.348R2 = 0.8237

0

20

40

60

80

100

120

0 2 4 6 8 10 12 14

t

Y

Exponential

y = 57.762e0.0453x

R2 = 0.8436

0

20

40

60

80

100

120

0 2 4 6 8 10 12 14

t

Y

Quadratic

y = -0.2333x2 + 6.466x + 49.273R2 = 0.8592

0

20

40

60

80

100

120

0 2 4 6 8 10 12 14

t

Y

Black Best equation : Quadratic y = -0.2333x2 + 6.466x + 49.273

R2 = 0.8592

y = -0.2333x2 + 6.466x + 49.273 X2 x Const -0.2333 6.466 49.273 Forecasts

13 93.9033 14 94.0702 15 93.7705 16 93.0042

1 55.5057 58 6.2215322 61.2718 62 0.5302753 66.5713 65 2.4689844 71.4042 68 11.588585 75.7705 72 14.216676 79.6702 75 21.810777 83.1033 98 221.91178 86.0698 84 4.2840729 88.5697 87 2.463958

10 90.603 90 0.36360911 92.1697 93 0.68939812 93.2698 92 1.612392

MSE 24.01349

Black Best α MSE(Best α)

0.780006672 68.98t X Deviation^2 1 58 58.00 2 62 58.00 16.003 65 61.12 15.054 68 64.15 14.855 72 67.15 23.506 75 70.93 16.547 98 74.11 570.958 84 92.74 76.459 87 85.92 1.16

10 90 86.76 10.4811 93 89.29 13.7812 92 92.18 0.0313 92.04

Colour

Best α MSE(best α) 0.298435453 10.88

t X Deviation^2 1 40 40.00 2 38 40.00 4.003 41 39.40 2.554 46 39.88 37.465 42 41.71 0.096 41 41.79 0.63

Quadratic

y = 0.008x2 + 0.3996x + 39.636R2 = 0.3344

0

10

20

30

40

50

60

0 2 4 6 8 10 12 14

t

Y

Exponential

y = 39.444e0.0117x

R2 = 0.3425

0

10

20

30

40

50

60

0 2 4 6 8 10 12 14

t

Y

7 41 41.56 0.318 47 41.39 31.469 42 43.06 1.1310 43 42.75 0.0611 42 42.82 0.6812 49 42.58 41.2513 Forecast 44.49

Case 15.2 By plotting the graph of the given data, and fitting varios curves by using the templates on Curve Fitting described on page 13.8, we observe that the best fitted curve is exponential. In fact, it provides an excellent fit with R2 as about 0.99. In such a case, it is advisable to use this curve itself to project the future values of credit for the year 2007. However, a closer look at the graph shows that the first four observations one trend, the next 9 observations follow different trend, and the last 4 observations follow yet another trend. Thus, it may be more realistic to fit a curve to the last four observations and project the fitted curve to forecast credit for 2007.

Linear

y = 25065x - 32715R2 = 0.8427

-100000

0

100000

200000

300000

400000

500000

600000

0 2 4 6 8 10 12 14 16 18

t

Y

Exponential

y = 44923e0.1368x

R2 = 0.9875

0

100000

200000

300000

400000

500000

600000

0 2 4 6 8 10 12 14 16 18

t

Y

Quadratic

y = 2099.6x2 - 12727x + 86960R2 = 0.955

0

100000

200000

300000

400000

500000

600000

0 2 4 6 8 10 12 14 16 18

t

Y

Y = 44923e0.1368x

R2 = 0.9875 A b 44923 0.1368 Year Forecast

18 527078.7 Fore Actual

1 51508.66 53805 2 59059.77 61689 3 67717.86 65240 4 77645.21 78662 5 89027.91 80482 6 102079.3 102310 7 117044 124937 8 134202.5 138548 9 153876.4 161038

10 176434.5 178999 11 202299.6 200133 12 231956.5 218839 13 265961.1 229523 14 304950.6 295562 15 349656 313065 16 400915.2 426892 17 459688.9 549057

Y = 26166x2 - 43396x + 308393

R2 = 0.9906 X2 x const 26166 -43396 308393

1 291163 295562 2 326265 313065 3 413699 426892 4 553465 549057

5 745563

Case 15.3 Linear

y = 46018x + 80295R2 = 0.9662

0

50000

100000

150000

200000

250000

300000

350000

0 1 2 3 4 5 6

t

Y

Exponential

y = 110498e0.2119x

R2 = 0.9963

0

50000

100000

150000

200000

250000

300000

350000

0 1 2 3 4 5 6

t

Y

Quadraticy = 7026.2x2 + 3861.5x + 129478

R2 = 0.9977

0

50000

100000

150000

200000

250000

300000

350000

0 1 2 3 4 5 6

t

Y

y = 7026.2x2 + 3861.5x + 129478

R2 = 0.9977 x2 x const

Total 7026.2 3861.5 129478139032.2 1 140365.7 167003.8 2 165305.8 207203.3 3 204298.3 251770.3 4 257343.2 326741.2 5 324440.5 Forecast 6 405590.2

Case 15.4

Linear

y = 1.2857x + 89.143R2 = 0.27

86

88

90

92

94

96

98

100

102

104

0 1 2 3 4 5 6 7 8

t

Y

Exponential

y = 89.173e0.0136x

R2 = 0.2716

86

88

90

92

94

96

98

100

102

104

0 1 2 3 4 5 6 7 8

t

Y

Quadratic

y = 0.0952x2 + 0.5238x + 90.286R2 = 0.2744

86

88

90

92

94

96

98

100

102

104

0 1 2 3 4 5 6 7 8

t

Y

Best α MSE(best α) 0.289006766 38.39

t X Deviation^2

1 90 90.00

2 94 90.00 16.00

3 88 91.16 9.96

4 102 90.24 138.21

5 90 93.64 13.26

6 96 92.59 11.63

7 100 93.57 41.288 95.43

Chapter 16: Decision Theory

Case 16.1 Entering the data in the template “Decision Analysis.xls” (Available on CD) , following output would be generated. 1 2 3 4

Alternatives

Take Away Budget Speciality Hotel

Sr no

States of nature

1 LP, LI 8 -20 -65 -710 2 HP , LI 20 92 10 -560 3 LP , HI 15 110 375 405 4 HP , HI 7 30 85 1200

Maximum Maximum 20 110 375 1200 1200 MaxiMax Hotel

Minimum 7 -20 -65 -710 7

Alpha MaxiMin Take

Away 0.5 Hurwicz 13.5 45 155 245 245

Hurwicz Hotel Laplace 12.5 53 101.25 83.75 101.25

Laplace Speciality Above are the choices for different criteria. Depending on the priorities of a decision maker, different approaches could be selected. For example, an optimistic person may select Hotel as the best choice, A pessimistic one may opt for Take Away as the best choice, and a balanced person( Alpha =0.5) may opt for Hotel as the best choice.

Case 16.2 Sunsoft Information Services is a small IT company based in Mumbai. It primarily deals

with accounting software which is quite popular with Small and Medium Enterprises. It

employs around 50 employees, and has a annual turnover of about Rs. 22 Crores. It has

come up with a new software, CostCalc which is a cost accounting software. This is a

unique software and has features to carry out cost accounting in a very easy way. The

price of this software is around Rs. l1 lakhs. The fixed cost which is the cost incurred to

make the software has been Rs. 4 Crores. The variable cost associated with shipping and

onsite support, is around Rs. l lakh per copy of the software. The average sale of this

software is dependant on the following factors:

• Awareness

• Economic condition in the country

• Market conditions

• Competition

Based on these factors, the probabilities of the approximate sales that Sunsoft expects to

make in the year are as follows: Sunsoft’s Expected Profit

( Rs. in Lakhs) Sales for the Year ( Number of Copies)

Profit = (Sales * Cost) - (Fixed Cost) + Sales * Variable Cost

Probability of Occurrence

50 Rs. 1 Crore 0.4 60 Rs. 2 Crore 0.3 80 Rs. 4 Crore 0.3

As regards the competition, AcctCalc is the market leader in accounting software, and

also has a big software R&D department. It also has an annual revenue of Rs. 96

Crores. The solutions given by this company are in the domain of accounting as well

of Mobile Software. AcctCalc has also been working on a similar project for cost

accounting software which could ease the way of cost accounting. But due to some

internal problems, they are not in a stage to ship the software till the end of the year.

Due to this AcctCalc faces an opportunity cost because the entry of the new product

viz. in the market. AcctCalc solution, stands to lose substantial amount of Rs.5.4

Crores that it has invested in the startup of the project.

Now, AcctCalc Solutions wants to buy out the software CostCalc from the Sunsoft for

an amount of Rs. 15 Crores. The management at AcctCalc decides that if this is not

possible then they can attempt to enter into an agreement with Sunsoft where they can

use the superior reach and network of AcctCalc to garner more sales for that company.

The revenue sharing agreement with AcctCalc envisages that the Product Costcalc

would use AcctCalc's brand name, and would retail for Rs.12 lakhs. Out of this,

AcctCalc would get Rs.l lakh for every copy sold as also subsequently in the next year

even though the price will come down to ll lakhs, because of competition. Out of the

balance of Rs 10 Lakhs per copy of software,. Sunsoft would get Rs. 5 lakhs and

AcctCalc would get Rs.5 lakhs. The approximate sales that could be garnered on this

basis are as follows: Sunsoft Alliance with AcctCalc: Expected Profit

Sales for theYear

Profit for AcctCalc

Profit for Sunsoft

Probability of Occurrence

80 Rs. 0.8 crore Rs. 4 crore 0.25 100 Rs. 1.0 crore Rs. 6 crore 0.42 120 Rs. 1.2 crore Rs. 8 crore 0.28 140 Rs. 1.4 crore Rs. 10 crore 0.05

Incidentally, both the above Tables relate to the first year. Decision Situation by SunSoft:

Now, Sunsoft has to decide as to how it can maximize its profits. Normally, a software

like this, has a shelf life of around 2 years after which competition enters the markets and

the product is unable to sustain due to declining operating margins.

The following assumptions are to be considered when we solve the problem for Sunsoft

• The sales are can be calculated over two years because this is the shelf life of

the product after which competitors will make it unviable to operate in the market

• The probabilities of the sales in each category remain same. The sales increase by

20% in the second year.

What decision Sunsoft should take? Solution There are three options available for the SunSoft namely

• Market the product self • Market the product with the agreement with AcctCalc • Sell the product to AcctCalc for Rs 15 Crores

Following is the table calculating EMV for SunSoft if they market the product themselves in 1st year. Sales Profit Probability

50 1 0.4 0.460 2 0.3 0.680 4 0.3 1.2

EMV= 2.2 Following is the table calculating EMV for SunSoft if they market the product themselves in 2nd year. Sales IN 1st year

Sales in 2nd year Profit Probability

50 60 6 0.4 2.460 72 7.2 0.3 2.1680 96 9.6 0.3 2.88

EMV= 7.44Profit in the 2nd year = sales ×0.11 – (sales × 0.01) Following is the table calculating EMV for 1st year for SunSoft if they have agreement with AcctCalc .

Sale Profit Sunsoft Probability

80 4 0.25 1100 6 0.42 2.52120 8 0.28 2.24140 10 0.05 0.5

EMV = 6.26Following is the table calculating EMV for 2nd year for SunSoft if they have agreement with AcctCalc . Sales IN 1st year

Sales in 2nd year

Profit Sunsoft Probability

80 96 4.8 0.25 1.2100 120 6 0.42 2.52120 144 7.2 0.28 2.016

140 168 8.4 0.05 0.42 EMV = 6.156

Profit in the 2nd year = (sales ×0.09)/2 Note that in the 2nd year the Sunsoft’s profit = (10 × Sales) ÷2 It may be noted that the sales in the 2nd year =1.2×Sales in the 1st year. It may be also noted that since the fixed cost of the software is taken into consideration in the first year, it is not taken into account in the second year. Total expected profit if Sunsoft markets itself = 2.2+7.44 =9.64 Crores Total expected profit if Sunsoft markets with the help of agreement with AcctCalc = 6.26 + 6.156 =12.416 Crores Total Profit if Sunsoft buyout the software to AcctCalc = 15 – 4 = 11 Crores Hence the best choice for SunSoft is to market with the help of agreement with AcctCalc. The expected profit for the two years =12.416 Crores Case 16.3 Haridas has specialised in making Lord Ganesh idols which are installed during Ganesh Chaturthi festival, in homes as well at public places. He is based in Mumbai, and earns his livelihood for the year from this profession which keeps him busy for about two months in a year. He makes two types of idols viz. big and small. The small ones ( of about 2 feet height) cost him Rs. 250, and sell for Rs. 400. The big ones ( of about 5 feet) cost him Rs.1,000, and sell for Rs. 1500. Last year, because of the supply of idols in Mumbai by another idol maker Jagdip, based in Pune, Haridas had suffered a big loss, so this time he wanted to consider the impact of the competition before deciding on the number of idols to be made. Due to this activity being the only source of livelihood, he was averse to take undue risk. He came to know that his last year’s competitor was not certain to supply the idols in Mumbai as he has the habit of supplying the idols to different cities in different years. From some knowledgeable sources, Haridas has predicted 40% probability of the idols being supplied, again this year, by Jagdip. If it is so, the demand for idols, made by him, will reduce by 30%, in case of small idols and 20% in case of big idols. The following Table gives the demand for Ganpati idols under the two conditions.

In case Jagdip does not supply idols in Mumbai:

Demand for Demand

For Big Idols

For Small Idols

Idols(x1000) Probability big idols Probability5 0.3 250 0.05

10 0.3 500 0.120 0.2 1000 0.3

30 0.1 1500 0.340 0.05 2000 0.250 0.05 2500 0.05

In case, Jagdip supplies idols in Mumbai:

Demand for Big Idols

Probability Demand for Small

Probability

4 0.3 200 0.057 0.3 400 0.1

14 0.2 800 0.321 0.1 1200 0.328 0.05 1600 0.235 0.05 2000 0.05

Assuming that Haridas knows in advance about Jagdip’s plan of supplying idols in Mumbai, before starting his work on the Idols, help Mr. Haridas regarding which option he should choose?

Entering the data in the template “Decision Analysis.xls” (Available on CD) , following output would be generated.

Small Idols Marginal Profit 150

Marginal Loss 250 Opportunity Loss 0

When Jagdip does not supply 1 2 3 4 5 6

Produce 250 500 1000 1500 2000 2500

Sr no Probability

Demand 1 0.05 250 37,500.00 -25,000.00 -150,000.00 -275,000.00 -400,000.00 -525,000.00 2 0.1 500 37,500.00 75,000.00 -50,000.00 -175,000.00 -300,000.00 -425,000.00 3 0.3 1000 37,500.00 75,000.00 150,000.00 25,000.00 -100,000.00 -225,000.00 4 0.3 1500 37,500.00 75,000.00 150,000.00 225,000.00 100,000.00 -25,000.00 5 0.2 2000 37,500.00 75,000.00 150,000.00 225,000.00 300,000.00 175,000.00 6 0.05 2500 37,500.00 75,000.00 150,000.00 225,000.00 300,000.00 375,000.00

Produce 1000 Total Expected Profit 37,500.00 70,000.00 115,000.00 100,000.00 25,000.00 -90,000.00

When Jagdip supplies:

1 2 3 4 5 6

Produce 200 400 800 1200 1600 2000

Sr no Probability

Demand 1 0.05 200 30,000.00 -20,000.00 -120,000.00 -220,000.00 -320,000.00 -420,000.00

2 0.1 400 30,000.00 60,000.00 -40,000.00 -140,000.00 -240,000.00 -340,000.00 3 0.3 800 30,000.00 60,000.00 120,000.00 20,000.00 -80,000.00 -180,000.00 4 0.3 1200 30,000.00 60,000.00 120,000.00 180,000.00 80,000.00 -20,000.00 5 0.2 1600 30,000.00 60,000.00 120,000.00 180,000.00 240,000.00 140,000.00 6 0.05 2000 30,000.00 60,000.00 120,000.00 180,000.00 240,000.00 300,000.00

Produce 800 Total Expected Profit 30,000.00 56,000.00 92,000.00 80,000.00 20,000.00 -72,000.00

Big Idols

Marginal Profit 500

Marginal Loss 1000 Opportunity Loss 0

When Jagdip does not supply

1 2 3 4 5 6

Produce 5 10 20 30 40 50

Sr no Probability

Demand 1 0.3 5 2,500.00 -2,500.00 -12,500.00 -22,500.00 -32,500.00 -42,500.00 2 0.3 10 2,500.00 5,000.00 -5,000.00 -15,000.00 -25,000.00 -35,000.00 3 0.2 20 2,500.00 5,000.00 10,000.00 0.00 -10,000.00 -20,000.00 4 0.1 30 2,500.00 5,000.00 10,000.00 15,000.00 5,000.00 -5,000.00 5 0.05 40 2,500.00 5,000.00 10,000.00 15,000.00 20,000.00 10,000.00 6 0.05 50 2,500.00 5,000.00 10,000.00 15,000.00 20,000.00 25,000.00

1

Produce 10 Total Expected Profit 2,500.00 2,750.00 -1,250.00 -8,250.00 -16,750.00 -26,000.00

When Jagdip supplies

1 2 3 4 5 6

Produce 4 7 14 21 28 35

Sr no Probability

Demand 1 0.3 4 2,000.00 -1,000.00 -8,000.00 -15,000.00 -22,000.00 -29,000.00 2 0.3 7 2,000.00 3,500.00 -3,500.00 -10,500.00 -17,500.00 -24,500.00 3 0.2 14 2,000.00 3,500.00 7,000.00 0.00 -7,000.00 -14,000.00 4 0.1 21 2,000.00 3,500.00 7,000.00 10,500.00 3,500.00 -3,500.00 5 0.05 28 2,000.00 3,500.00 7,000.00 10,500.00 14,000.00 7,000.00 6 0.05 35 2,000.00 3,500.00 7,000.00 10,500.00 14,000.00 17,500.00

Produce 7 Total Expected Profit 2,000.00 2,150.00 -650.00 -5,550.00 -11,500.00 -17,975.00

0.4Jagdip supplies Make Small 800 and Big 7

1 82150Supply Idols 0 82150 0 82150

0 103510 0.6Jagdip do not supply Make Small 1000 and big 10

1 1177501 0 117750 0 117750

103510

Do not supply Idols0

0 0 It may be noted from above tables and the decision tree that the maximum EMV for Haridas, If Jagdip Supplies, is for 800 small and 7 Big idols. If Jagdip does not supply, the maximum EMV for Haridas is for 1000 small and 10 Big idols. His expected profit is 103510/- Case 16.4 Entering the data in the template “Decision Analysis.xls” (Available on CD) , following output would be generated.

Enter Details Marginal Profit 10

Marginal Loss 20

Opportunity Loss

1 2 3 4 5 6 7

prepare 180 190 200 210 220 230 240

Sr no Probability

Demand

1 0.14 180 1,800.00 1,600.00 1,400.00 1,200.00 1,000.00 800.00 600.00

2 0.15 190 1,800.00 1,900.00 1,700.00 1,500.00 1,300.00 1,100.00 900.00

3 0.18 200 1,800.00 1,900.00 2,000.00 1,800.00 1,600.00 1,400.00 1,200.00

4 0.16 210 1,800.00 1,900.00 2,000.00 2,100.00 1,900.00 1,700.00 1,500.00

5 0.15 220 1,800.00 1,900.00 2,000.00 2,100.00 2,200.00 2,000.00 1,800.00

6 0.12 230 1,800.00 1,900.00 2,000.00 2,100.00 2,200.00 2,300.00 2,100.00

7 0.1 240 1,800.00 1,900.00 2,000.00 2,100.00 2,200.00 2,300.00 2,400.00

prepare 200 Total Expected Profit 1,800.00 1,858.00 1,871.00 1,830.00 1,741.00 1,607.00 1,437.00

It may be noted from the above table that the maximum EMV is for preparing 200 Meals everyday. It may be added that the above case could also be considered with some amount of opportunity loss for the hostel mess. Case 16.5 Entering the data in the template “Decision Analysis.xls” (Available on CD) , following output would be generated.

Enter Details Marginal Profit 500

Marginal Loss 2000 Opportunity Loss

1 2 3 4 5 6 7 8 9 10 11

0 1 2 3 4 5 6 7 8 9 10 Sr no Proba

1 0.05 0 0.00 -2,000.00 -4,000.00 -6,000.00 -8,000.00 -10,000.00 -12,000.00 -14,000.00 -16,000.00 -18,000.00 -20,000.00

2 0.08 1 0.00 500.00 -1,500.00 -3,500.00 -5,500.00 -7,500.00 -9,500.00 -11,500.00 -13,500.00 -15,500.00 -17,500.00

3 0.1 2 0.00 500.00 1,000.00 -1,000.00 -3,000.00 -5,000.00 -7,000.00 -9,000.00 -11,000.00 -13,000.00 -15,000.00

4 0.15 3 0.00 500.00 1,000.00 1,500.00 -500.00 -2,500.00 -4,500.00 -6,500.00 -8,500.00 -10,500.00 -12,500.00

5 0.2 4 0.00 500.00 1,000.00 1,500.00 2,000.00 0.00 -2,000.00 -4,000.00 -6,000.00 -8,000.00 -10,000.00

6 0.15 5 0.00 500.00 1,000.00 1,500.00 2,000.00 2,500.00 500.00 -1,500.00 -3,500.00 -5,500.00 -7,500.00

7 0.11 6 0.00 500.00 1,000.00 1,500.00 2,000.00 2,500.00 3,000.00 1,000.00 -1,000.00 -3,000.00 -5,000.00

8 0.06 7 0.00 500.00 1,000.00 1,500.00 2,000.00 2,500.00 3,000.00 3,500.00 1,500.00 -500.00 -2,500.00

9 0.05 8 0.00 500.00 1,000.00 1,500.00 2,000.00 2,500.00 3,000.00 3,500.00 4,000.00 2,000.00 0.00

10 0.04 9 0.00 500.00 1,000.00 1,500.00 2,000.00 2,500.00 3,000.00 3,500.00 4,000.00 4,500.00 2,500.00

11 0.01 10 0.00 500.00 1,000.00 1,500.00 2,000.00 2,500.00 3,000.00 3,500.00 4,000.00 4,500.00 5,000.00

1

2

0.00 375.00 550.00 475.00 25.00 -925.00 -2,250.00 -3,850.00 -5,600.00 -7,475.00 -9,450.00 It may be noted that the maximum EMV in the above table is for overbooking 2 extra rooms.

Statistics for Management

Hint for Solving Exercises

Chapter 9 on

Multiple Correlation and Regression Analysis ( Please e-mail any query to the authors : dr_tns@rediffmail.com as also to

shailarego@gmail.com )

Template for ‘Multiple Regression’ may be used for deriving multiple regression equations as per procedure explained in Section 9.19 of Chapter 9 of the book, and given in the note below. Note: The data about values of all the variables may be inputted in the template. However, the answer will not come directly. After inputting the data, one may first unprotect the worksheet for further processing by choosing the option “protect” by typing the password as “protect”. Thereafter one may chose the option “Solver” under “Tools”. This is an integral part of EXCEL.(If it is not in the tools menu, one may select tools -> add-ins and select solver option by checking the box before solver) In the template that appears on the screen, one should choose ”Solve” by pressing “OK”, and in the resulting template, select the option as “keep solver solution” by pressing “OK”. Immediately, the coefficients of the regression equation will appear under respective names of variables.