Organizational_Learning_and_Knowledge_Management_in_Mcdonalds.

APPENDIX A: ASSIGNMENT COVER SHEET DATE RECEIVED :______________ DATE RETURNED :______________ Programme | Master of Business Administration | Module Name | MANAGERIAL STATISTICS | Surname | ZONDI | Full Names | GABRIEL JABULANI | Student Number | MBA 113 1080 | Date Submitted | 17 APRIL 2013 | Postal Address | P.O. BOX 36273 | | EMPANGENI | | 3880 | | KWAZULU-NATAL., SOUTH AFRICA | e-mail, my regent address | | e-mail (alternate e-mail address | gjzondi@webmail.co.zazondig@unizulu.ac.za | Contact Numbers | Cell: 083 392 1595 | | Home: 076 728 5200 | | Work: 035-902 6946 | Alternate Contact | Mrs Nontobeko Patricia Zondi | | 076 728 5200 | I hereby confirm that the assignment submitted herein is my own original work. Signature of Student: GJZondi Date: 17 April 2013 MASTER OF BUSINESS ADMINISTRATION (MBA) Managerial Statistics Module Assignment Name: Gabriel JabulaniZondi Student Number: MBA1131080 Table of Contents Pages Question 1 1.1 Frequency distribution 5 1.2 Histogram 5 1.3 Mean, the median and the mode 6 1.4 The standard deviation 6 Question 2 2.1 The probability that an employee who is under 30 has a salary under R25 000 7 2.2 A probability that an employee who has a salary under R25,000 is under 30 7 2.3.1 A probability that an employee is under 30 or has a salary under R25,000. 7 2.3.2 The probability that the employee’s salary is at least R25,000 given that the Employee is at least 30 years old. 8 2.4 Are being under 30 and having a salary under R25,000 independent events/' Explanation. 8 Question 3 3.1.1 Probability that in any single day worker A will produce more units than worker B 9 3.1.2 Probability that during 1 week, worker A will produce more units than worker B On average. 10 3.2.1 Test to determine at the 5% significance level whether there is enough evidence to infer that mean amount of television watched daily by all young adult men is greater than 53 munites. 11 3.2.2 Test to determine at the 1% significance level whether there is enough evidence to infer that the mean amount of television watched daily by all young adult men is less than 53 minutes. 12 Question 4 4.1 The dependent and the independent variable 13 4.2 Scatter diagram of the data 13 4.3 The least square regression equation 14 4.4 Person’s Correlation coefficient for the data. 15 4.5 Predicting the amount of sale for a broker who brought in 35 new clients last year.15 Question 5 5.1 Calculating the IRR for each project 16 5.2 Calculating the NPV for each project if the discount rate is 10%. 16 5.3 The project that is more profitable on basis of the NPV. 17 5.4 The project that is more profitable on basis of the IRR 17 Question 6 6.1 The Laspeyres Price. 18 6.2 Quantity Indices. 18 QUESTION 1 1.1 FREQUENCY DISTRIBUTION USING ’50 BUT LESS THAN 60’ AS THE FIRST CLASS. Response to customer complaint 1.1 A frequency distribution Classes | Frequency (f) | Variance | Relative frequency (f) | | | | | 50≤X<60 | 8 | 2366,72 | 8/50 | 60≤X<70 | 15 | 777,60 | 15/50 | 50≤X<60 | 16 | 125,44 | 16/50 | 50≤X<60 | 5 | 819,20 | 5/50 | 50≤X<60 | 6 | 3119 | 50/50 | Total | 50 | 7208 | 50/50 | 1.2 THE HISTOGRAM OF THE FREQUENCY 1.3 MEANS, THE MEAN AND THE MODE OF THE DATA Measures of central tendency (mean, median and mode) Mean = i=15f.Mn = 361050 = 72, 2 Median Me =Sum of all observationsTotal number of observations, =L+ n2- c 16 x i = 70 + 502- 2316 x 10% = 71, 25 Mode Mo = L + d1d2+d3 x i = 70 + 16-1512 x 10 = 70, 83 1.4 THE VALUE OF THE STANDARD The value of standard deviation S = Varience = fx2-fx2n =149x 7208 = 12, 13 QUESTION 2 2.1 A PROBABILITY THAT AN EMPLOYEE WHO IS UNDER 30 HAS A SALARY UNDER 25,000. Probability Company employees classified according to age and salaries Classes | Frequency (f) | Relative frequency (f) | Variance | | | | | 50≤X<60 | 8 | 8/50 | 2366,72 | 60≤X<70 | 15 | 15/50 | 777,60 | 70≤X<80 | 16 | 16/50 | 125,44 | 80≤X<90 | 5 | 5/50 | 819,20 | 90≤X<99 | 6 | 50/50 | 3119 | Total | 50 | 50/50 | 7208 | Classical approach | Salaries | | Age | X < 25 000 | R25000R45000 | Total | x<30 | 32 | 3 | 0 | 35 | 30≤x <45 | 10 | 18 | 21 | 49 | X > 45 | 1 | 10 | 5 | 16 | TOTAL | 45 | 31 | 26 | 100 | 2.1 Event A = an employee is under 30 Probability P = (X <30 and S< R25000-00) =32/100 = 0, 32 2.2 Event B = salary under R25000-00 P = R25000-00< S < R30 000-00 = 32/100 = 0, 32 2.3.1 Probability that an employee is under 30 or has the salary under R25 000 Probability P =(X <30 or SR25 000 , x ≤30 or S≥ R25000 and x ≥30 P1 = 32/100 P2 = 100-32100 = 65100 = 0, 65 P (s is greater than R25000-00 given greater or equal to 30) P (s is greater or equal to 25000-00 and x is greater or equal to 30) all divide by p(greater or equal to 30) P 1 – 32/100 = 100 – 32 divide by 100 divide by 65/100 2.4 A and B are mutually exclusive event, If A and B are independent then the probability of A divide by B equal to the probability of A i.e. P (A) = 35100 = 0, 35 P (B) = 43100 = 0,43 P (A) x P (B) =35100 X 43100 = 0,1 5 PA ƞ B → 32100 = 0, 32 ≠ 0, 15 Therefore A and B are not independent events QUESTION 3 3.1.1 THE PROBABILTY THAT IN ANY Let A denote productivity of worker A per day and B denote productivity of worker B per day. Therefore A˜ N (µ a I ) = (84; 24) And B˜ N (µB ) = (72,252) ------------------------------------------------- Required probability = P(A>B) = P (A-B>0) Let D = A-B (=difference between the productivity of A and that of B) The corresponding standard normal variable is : Z = ZD = D- µD Z = x-μ ϬD Ϭ Where μ D = 84-74 = 10; and ϬD = √242+252 = √1201 = 34,65 Therefore the required probability = P (A-B>0) = P (Z> - 10 √1201 = P (Z> -0,2885 = P (Z < 0,2885) = P (Z < 0,29) = 0,614 3.1.2 A PROBABILITY THAT DURING ONE WEEK (5 WORKING DAYS), WORKER ‘A’ WILL PRODUCE MORE UNITS THAN WORKER ‘B’ ON AVERAGE. Required Probability = P (A-B>0) Where A = 1/5 ∑ A and Note: n =5 B = 1/5 ∑B The corresponding standard normal variable is Z = ZD = D - µ D where Ϭ D The corresponding standard normal variable is: Z = ZD = D- µD ϬD Where µD = 84-74 = 10 ; and ϬD = 242+252 = 576+625 5 5 5 = 2402 = 15,498 Therefore Z = -10 = - 0,645 ≠ 0,65 15,4984 Therefore the required probability = P (Z> - 0,65) = P (Z <0,65 = 0,742 3.2.1 TEST TO DETERMINE AT THE 5% SIGNIFICANCE LEVEL WHETHER THERE IS ENOUGH EVIDENCE TO INFER THAT THE MEAN AMOUNT OF TELEVISION WATCHED DAILY BY ALL YOUNG ADULT MEN IS GREATER THAN 53 MINUTES. 1. given sample size, n = 18 Standard deviation, = 12; And X = 5 x/n 60,11 The sample is from a normally distributed 2. Hypothesis Ho: µ = 53 H1: µ >b 53 = 0,05 Now, =0,05 for a one-tailed test Means that Z critical = ± 1,645 3. Calculations/ Analysis From the sample, Z calculated = 60,11-53 12/ 18 = 2,514 Therefore Z calculated > Z critical 4. Decision: Ho is rejected. 5. Conclusion: There is enough evidence to infer that the mean amount of television watched daily is greater than 53 minutes. 3.2.2 TEST TO DETERMINE AT THE 1% SIGNIFICANCE LEVEL WHETHER THERE IS ENOUGH EVIDENCE TO INFER THAT THE MEAN AMOUNT OF TELEVISION WATCHED DAILY BY ALL YOUNG ADULT MEN IS LESS THAN 53 MINUTES. 1. Using the data in 3.2.1: Hypothesis: Ho: µ = 53 H1: µ < 53 Ϭ = 0.01 2. For a one tailed test Ϭ = 0.01 means that Z critical = ± 2,33 3. Calculations/Analysis From 3.2.1, Z calculated = 2.514 i.e. Z calculated > Z critical 4. Decision: Ho is rejected. 5. Conclusion: There is enough evidence to infer that the mean amount of television watched daily is less than 53 minutes. QUESTION 4 4.1 A DEPENDENT AND THE INDEPENDENT VARIABLE. 1.1 A frequency distribution Classes | Frequency (f) | Midpoint(M) | f.M | M – x bar all squared | Variance | Relative frequency (f) | | | | | | | | 50≤X<60 | 8 | 55 | 440 | (55 – 72,2) = 295,84 | 2366,72 | 8/50 | 60≤X<70 | 15 | 65 | 975 | (65 -72,2) = 51,84 | 777,60 | 15/50 | 50≤X<60 | 16 | 75 | 1200 | (75 – 72,2) = 7,84 | 125,44 | 16/50 | 50≤X<60 | 5 | 85 | 425 | (85 – 72,2) = 163,84 | 819,20 | 5/50 | 50≤X<60 | 6 | 95 | 570 | (95 – 72,2) = 519,84 | 3119 | 50/50 | Total | 50 | | 3610 | | 7208 | 50/50 | 4.2 A SCATTER DIAGRAM OF DATA Dependent variables – Clients – X 4.3 THE LEAST SQUARE REGRESSION EQUATION. ŷ= a + bx | Clients | Sales | yi2 | xi2 | xiyi | 1 | 27 | 52 | 2074 | 792 | 1404 | 2 | 11 | 37 | 1369 | 121 | 407 | 3 | 42 | 64 | 4096 | 1764 | 2688 | 4 | 33 | 55 | 3025 | 1089 | 1815 | 5 | 15 | 29 | 841 | 225 | 435 | 6 | 15 | 34 | 1156 | 225 | 510 | 7 | 25 | 58 | 3364 | 625 | 1450 | 8 | 36 | 59 | 3481 | 1296 | 2124 | 9 | 28 | 44 | 1936 | 784 | 1232 | 10 | 30 | 48 | 2304 | 900 | 1440 | 11 | 17 | 31 | 961 | 289 | 527 | 12 | 22 | 38 | 1444 | 484 | 836 | | | | y2=26681 | | | | i=lnxi=301 | =lnyi=549 | i=lnx2=8531 | i=lnxiyi=14868 | x=i=lnxi y=i=lnyi Sxy=i=lnxiyi-nxy N n = 301 549 = 14868-12(25.08)(45.75) 12 12 = 14868-13768.92 = 25.083 45.75 = 1099.08 .Sxx=i=lnx2-nx2 = 8531-12(2508)2 b = δxyδxx a=y-bx = 8531-7548.08 = 1099.08982.92 = 45.75-1.12(25.08) = 982092 = 1.12 = 45.75-28.09 = 17.66 Therefore the least square regression line is now defined as: (y=17.66+1.12x) 4.4 DETERMINING AND COMMENTS ON THE PERSON’S CORRELATION COEFFICIENT FOR THE DATA GIVEN. n∑xy-∑x∑y (n∑x2-(x)2(n∑y2-(y)2 = 1214868-301(549)128531-(301)2(1226681-(549)2 = 178 416-165 249102 372-90601(320 172-301 401) = 1316711771×18771 = 1316714864.5 = 0.9 -1 ≤ r ≤+ 1 in this case r < 1 therefore – Strong Positive 4.5 PREDICTING THE AMOUNT OF SALE FOR A BROKER WHO BROUGHT IN 35 NEW CLIENTS LAST YEAR. ŷ = a + bx y=35 y = a+b (35) y=17,6919+1,1186 x 35 y= 57 QUESTION 5 5.1 CALCULATING THE IRR FOR EACH PROJECT Project | Cost | Cash | Term | A | 15,000 | 20,000 | 1 | | C | F | 1 | B | 28,000 | 35,000 | 1 | = 1+r = 20,000 15,000 A 20,000-1 15,000 ------------------------------------------------- ------------------------------------------------- = 0.333 = 33⅓% 35,000=28,000 (1+r)b r b= 35,000 - 1 28,000 = 0,25 =25% ------------------------------------------------- 5.2 CALCULATING THE NPV FOR EACH PROJECT IF THE DISCOUNT RATE IS 10% A = 20,000 = 18181,82 (1+10%) ------------------------------------------------- NPV A = 18181,82 – 15,000= 3181,82 5.3 THE PROJECT THAT IS MORE PROFITABLE ON BASIS OF THE NPV Project B 5.4 THE PROJECT THAT IS MORE PROFITABLE ON THE BASIS OF THE IRR Project A QUESTION 6 THE INFORMATION ON A BASKET OF TOILETRIES FOR YEAR 2008 AND 2011 | Price | Quantities | Product | 2008 | 2011 | 2008 | 2011 | Soap (125g) | 2,65 | 3,05 | 5 | 7 | Deodorant (50 ml) | 24,65 | 25,35 | 6 | 8 | Toothpaste (100ml) | 5,25 | 6,15 | 10 | 9 | THE LASPEYRES PRICE = (3,05x5)+(25,35x6)+(6,15x10) x 100% (2,65x5)+(24,65x6)+(5,25x10) = 228,85 x100% 213,65 = 107,11 QUANTITY INDICES (7x2,65) + (8 x 24,65) + (9 x 5,25) x100% (2,65x5)+(24,65x6)+(5,25x10) = 263___ x100% 213,65 = 123,1