Computer Assignment

ASSIGNMENT ON QUANTITATIVE METHODS - I

Submitted to Prof. Vishnuprasad Nagadevara Indian Institute of Management, Bangalore Group 9 - Section C Manoj R Muthuraman AL Nikhil Darak Pankaj Kumar Sudheender S

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PGP2013-15 Computer Assignment 1.

Calculate a 95 percent confidence interval for the “Gross output – 2001-02 (Rs)” Solution Total number of data points in sample (n) = 10000 From the sample given, Sample mean = 1325948.88 Standard deviation of the sample (S) = 12112314.37 Total degrees of freedom = n-1=9999 As we do not know the population standard deviation, hence T distribution is to be used. Given that 95 % confidence hence α = 0.05 We compute tα/2 in this case for α = 0.05 and we get the value as 1.9602 The confidence interval for the “Gross output – 2001-02 (Rs)” is [1325948.88 ± 1.9602 × (12112314.37 ÷ 10000)] [930708.872, 1791303.32]

2.

Define two different measures that you consider most appropriate for measuring the performance of the units. This definition is up to you. These can be the variables that are already in the data or new variables defined based on the existing variables. Please explain in one paragraph why you have selected these two measures. Remaining analysis is to be carried based on these definitions. The two different measures which are most appropriate to measure performance of the units are percentage change in Gross Output and the percentage change in the Net Worth of a firm. The percentage change in Gross Output in 2001-02 is the ratio of change in Gross Output from 2000-01 and 2001-02 to the Gross Output in 2000-01. The performance of a firm can be measured by observing if the firm is growing with respect to all the other firms by measuring the percentage change in the Gross Output and then comparing it to the average percentage change of all the firms. The percentage change in Net Worth of a firm in 2001-02 is the ratio of change in Net Worth from 2000-01 and 2001-02 to the Net Worth in 2000-01.

Net worth is a important determinant of value of a company and is used to determine creditworthiness of a company. a. What is the probability that a firm selected at random is a SSSBE unit? Solution Total numbers of firms in the sample (n) = 10000 Total number of SSSBE units in the sample data = 3409 (From the data) Hence the probability that a firm selected at random is a SSSBE unit is P = 3409 / 10000 = 0.3409 b. What is the probability that a firm selected at random is GOOD in performance? (Calculate the average of the first performance measure that you had defined in question 2 above. If the firm’s performance is above this average, it considered good. If it is below average, it is considered Bad) Solution The measure used for performance of units is % change in GOP in 2001-02. In the given sample, the average % change in GOP in 2001-02 is 23.647%. The number of firms having % change in GOP in 2001-02 higher than 23.647 are 1354. Therefore, the probability that a firm selected at random from the sample will be GOOD in performance is 1354/10000 = 0.1354 c. What is the probability that a firm selected is a SSSBE Unit and ALSO GOOD in performance? Solution Total number of firms that are a SSSBE unit and ALSO GOOD in performance are 439. Therefore, the probability that a firm selected at random from the sample is a SSSBE unit and ALSO GOOD in performance is 439/10000 = 0.0439 d. What can you say about the performance of the SSSBE units in terms of GOOD or BAD based on the above?

Solution From the sample data we know that: The total number of SSSBE units = 3409. Total number of SSSBE units that are GOOD in performance = 439 Hence the probability that a SSSBE unit is GOOD in performance is given by 439/3409=0.1287 As the probability of a SSSBE unit being GOOD in performance is very small therefore we can say that the performance of SSSBE unit is below satisfaction.

3.

Test the null hypothesis that the population average of the variable “Value of Exports for 2001-02” = 87,300. Carry out a one sided test. Clearly state your null and alternate hypotheses. Solution H0 : Average of Value of Exports for 2001-02 ≤ 87300 (Null Hypothesis) H1 : Average of Value of Exports for 2001-02 > 87300 (Alternate Hypothesis) Total Number of Samples = 10000 Sample mean for Value of Exports for 2001-02 = 133717.5 Sample Standard deviation = 5129673.88 Total degrees of freedom = n-1=9999 As we do not know the population standard deviation, hence T distribution is to be used. Assume that 95 % confidence hence α = 0.05 So, we compute tα in this case for α = 0.05 and find that the corresponding tvalue is 1.645 For carrying out a right tailed t test, the value of t= 1.645 Hence Tcritical = 1.645. The T test statistic is given by T calculated = (133717.5 – 87300) × 10000/ 5129673.88 T calculated = 0.9051 As Tcritical > T calculated hence we do not reject the null hypothesis. Hence, the average value of exports for 2001-02 is probably greater than 87300.

4.

What are the variables that could be used for determining the two measures of performance selected by you? Explain your methodology and justify. Interpret and explain the results. Solution The Two measures selected for measuring the performance of a firm are % change in Gross Output and % change in Net Worth. The variables used to determine the % change in Gross Output for 2001-02 are Gross output for the year 2000-01 and Gross output for the year 2001-02. The % change in Gross output is calculated as : % change in GOP for 2001-02 = (GOP2001-02 – GOP2000-01 )/GOP 2000-01 × 100% After calculating the % change in GOP for the firms individually we can calculate the average % change in GOP for all the firms. In the given sample, the average % change in GOP in 2001-02 is 23.647%. Hence for a firm if the % change in GOP in 2001-02 is greater than 23.647% we can say that the firm is growing at a faster rate than the market. After the analysis it was found that out of the 10000 firms only 1354 had a higher % change in GOP in 2001-02 than 23.647%. Hence we can interpret that some of the firms are growing at a very high rate as compared to the others and that the distribution is left skewed. The variables used to determine the % change in Net Worth for 2001-02 are Net Worth for the year 2000-01 and Net Worth for the year 2001-02. The % change in Net Worth is calculated as: % change in NW for 2001-02 = (NW2001-02 – NW2000-01 )/NW 2000-01 × 100% In the given sample, the average % change in NW in 2001-02 is 23.846%. After the analysis it was found that out of the 10000 firms only 1205 had a higher % change in NW in 2001-02 than 23.846%.Hence we can interpret that the net worth of only few of the firms increased above the average rate.

5.

Some male chauvinists like to think that the productivity of women employees is significantly less than that of the male employees. Do you agree or disagree with this? Justify your answer with appropriate analysis. Solution Let µ1 = mean of male employees productivity µ2 = mean of female employees productivity Let us divide the sample data by the number of firms managed by male and female in order to get the productivity of male and female. Total number of firms managed by male (n1) = 9192

Total number of firms managed by female (n2) = 808 The average productivity is calculated by taking the average of the Gross Output for the year 2001-02. Average productivity of male employees(X1) = 1408040.81 Average productivity of female employees(X2) = 392051.65 Standard deviation of male employees productivity (S1) = 12604851.76 Standard deviation of female employees productivity (S2) = 2699562.94 In order to check whether the population standard deviation is same we need to perform a F test. The null and alternate hypothesis for the F test are : H0: σ12 = σ22 H1: σ12 ≠ σ22 F (n1-1, n2-1) = S12/S22 Fcalculated (9191, 807) = 21.80161 F critical (9197, 801) = 1.091 (From table with α = 0.05) As Fcalculated >> Fcritical , we reject the Null Hypothesis. Hence we have σ1 ≠ σ2 Null hypothesis Alternative hypothesis

H0: µ1 ≥ µ2 H1: µ1 < µ2

We assume α=5%, i.e, 0.05 for doing the left-test. Degrees of freedom (df) = [(S12/n1 + S22/n2)2/ {(S12/n1)2/ (n1-1) + (S22/n2)2/ (n2-1)}] = 5190 Tcritical (0.1, 5190) = 1.645 The T test statistic is given as Tcalculated = ((X1-X2)-(µ1-µ2)) / (S12/n1 + S22/n2)) = 6.2643 As Tcalculated > Tcritical , it lies to the right of Tcritical , so we do not reject the Null Hypothesis. Hence, we agree with the statement that the women employees’ productivity is considerably less than the men employees’ productivity.