# Business Data Analysis

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ASSIGNMENT 1- BU1007 Question 1 (i) The following data represent the cost of electricity during July 2006 for a random sample of 50 one-bedroom apartment in a large city Electricity Charge ($)| 96| 157| 141| 95| 108| 171| 185| 149| 163| 119| 202| 90| 206| 150| 183| 178| 116| 175| 154| 151| 147| 172| 123| 130| 114| 102| 111| 128| 143| 135| 153| 148| 144| 187| 191| 197| 213| 168| 166| 137| 127| 130| 109| 139| 129| 82| 165| 167| 149| 158| | | | | | a. Form a frequency distribution and a percentage frequency distribution that have class intervals with upper class limits $99, $119 and so on. 2, 90, 95, 96, 102, 108, 109, 111, 114, 116, 119, 123, 127, 128, 129, 130, 130, 135, 137, 139, 141, 143, 144, 147, 148, 149, 149, 150, 151, 153, 154, 157, 158, 163, 165, 166, 167, 168, 171, 172, 175, 178, 183, 185, 187, 191, 197, 202, 206, 213| Limits ($)| Frequency| Percentage%| 200-219| 4| 8 (8/50 x100)| 180-199| 7| 14 (14/50 x100)| 160-179| 9| 18 (18/50 x100)| 140-159| 13| 26 (26/50 x100)| 120-139| 9| 18 (18/50 x100)| 100-119| 5| 10 (10/50 x100)| 80-99| 3| 6 (6/50 x100)| | | | b. Construct a histogram and a percentage frequency polygon (Use Excel). | | | Mid-point| Frequency| Lower Limit| Upper Limit| Class| Frequency| 200| 0| 200| 219| 200 up to 219| 4| 209. 5| 4| 180| 199| 180 up to 199| 7| 209. 5| 7| 160| 179| 160 up to 179| 9| 209. 5| 9| 140| 159| 140 up to 159| 13| 209. 5| 13| 120| 139| 120 up to 139| 9| 209. 5| 9| 100| 119| 100 up to 119| 5| 209. 5| 5| 80| 99| 80 up to 99| 3| 209. 5| 3| | | | | 209. 5| 0| c. Form a cumulative percentage frequency distribution and draw a cumulative percentage frequency curve. (4 Marks) Limits ($)| Frequency| Percentage%| Cumulative Frequency| Cumulative % frequency| | | | 0| 0/50 x 100 = 100| 00-219| 4| 8%| 4| 4/50 x 100 = 8| 180-199| 7| 14%| 11| 11/50 x 100 =22| 160-179| 9| 18%| 20| 20/50 x 100 = 40| 140-159| 13| 26%| 33| 33/50 x 100 =66| 120-139| 9| 18%| 42| 42/50 x 100 =84| 100-119| 5| 10%| 47| 47/50 x 100 = 94| 80-99| 3| 6%| 50| 50/50 x 100 = 100| | | | | Mid-point| Frequency| Lower Limit| Upper Limit| Class| Frequency| 200| 0| 200| 219| 200 up to 219| 4| 209. 5| 4| 180| 199| 180 up to 199| 7| 209. 5| 7| 160| 179| 160 up to 179| 9| 209. 5| 9| 140| 159| 140 up to 159| 13| 209. 5| 13| 120| 139| 120 up to 139| 9| 209. 5| 9| 100| 119| 100 up to 119| 5| 209. 5| 5| 80| 99| 80 up to 99| 3| 209. | 3| | | | | 209. 5| 0| d. Around what amount does the monthly electricity cost seem to be concentrated? According to the cumulative frequency table, The monthly electricity costs are concentrated at $140-$159. The reason being is that, according to the limit $140-$150, more people fall in the middle class. Thus the frequency distribution able displays that the weighted average of people who would spend $140-$150 for electricity will be at 26%. () The following data represent the fat contents, in grams per serving, for a sample of 20 chicken sandwiches from fast-food chains: Fat contents| 7| 23| 8| 30| | 25| 5| 19| 16| 29| 20| 29| 20| 30| 24| 30| 19| 40| 30| 56| | | a. Compute the sample mean, median, first quartile, and third quartile of the above data. (10 Marks) Lowest to highest 4, 5, 7, 8, 16, 19, 19, 20, 20, 23, 24, 25, 29, 29, 30, 30, 30, 30, 40, 56 | Mean = 4+5+7+8+16+19+19+20+20+23+24+25+29+29+30+30+30+30+40+56 =464/20 =23. 2 Median = (23+24)/2 = 23. 5 First Quartile =(25/100) x 20 =5 =Thus 16 is the first quartile Third Quartile = (75/100) x 20 =15 =Thus 30 is the third quartile b. Compute the sample variance, standard deviation, range, interquartile range, coefficient of variation, and z-scores.

Are there any outliers? Explain. (10 Marks) Sample Variance * Solving for sample variance. | 4 + 5 + 7 + 8 + 16 + 19 + 19 + 20 + 20 + 23 + 24 + 25 + 29 + 29 + 30 + 30 + 30 + 30 + 40 + 56| =23. 20| | 20| | | 368. 64 + 331. 24 + 262. 44 + 231. 04 + 51. 84 + 17. 64 + 17. 64 + 10. 24 + 10. 24 + 0. 04 + 0. 64 + 3. 24 + 33. 64 + 33. 64 + 46. 24 + 46. 24 + 46. 24 + 46. 24 + 282. 24 + 1,075. 84 = 2,915. 20| = 2,915. 20/ 20-1 = v2,915. 20 / 19 = 153. 43 Standard Deviation | | | 2,915. 20| | | 20 – 1| | | 2,915. 20| | | 19| | | 153. 43| | | S= 12. 39 Range = Largest Value – Smallest Value = 56-4 52 Interquartile range IQR = Q3- Q1 = 30 – 16 = 14 Coefficient of Variance Standard deviation / mean x 100 = 12. 39/23. 2 x 100 = 0. 534 x100 = 53. 40 Z-Scores = standard deviation – mean divided by sample size Zi = xi-x/s = 12. 39 – 23. 2 / 20 = -0. 5405 Outliner Q3 = 75/100 x 20 Q1 = 25/100 x 20 =15 =5 Q3= 30 Q1= 16 Lower = Q1 – 1. 5 (IQR) = 5+(-1. 5(14)) = -16 Upper = Q3 + 1. 5 (IQR) = 30+(+1. 5(14)) = 51 The number is above 51 therefore it can be considered as an outliner. c) Are the data skewed? If so how?

According to the mean, median and mode it can be concluded that the data is positively skewed. The reason why is, the mean is = 23. 2, and the median is 23. 5 and the mode is =30. So if you plot a graph, you will see that that the data will be leaning more to the left side, which is also known as positive skewed. And due to the outliner, the data is said to tail to the positive side. d) What can be said about the relative location of the observation 16 in the data set? In Observation 16, you will find that the data set is set at 30, which has a frequency of four (4), Therefore the mode is set by the given data.) Based on results of part (a) through part (c), what conclusions would you draw concerning the total fat of chicken sandwiches? Based on the results in part a-c, it can be concluded that the average total fat of chicken sandwich is 23. 5 when looking at the mean and median. And also based on the results of the standard deviation and range, it is also concluded that the chicken sandwich weight range is 52. Therefore, according to part C, the data is positively skewed because the data falls to the left. Question 2 (a) There are several methods for calculating fuel economy.

The following data indicate the mileage, as calculated by owners and by current government standards: Car| Owner| Government| 2005 Ford F-150| 14. 3| 16. 8| 2005 Chevrolet Silverado| 15. 0| 17. 8| 2002 Honda Accord LX| 27. 8| 26. 2| 2002 Honda Civic| 27. 9| 34. 2| 2004 Honda Civic Hybrid| 48. 8| 47. 6| 2002 Ford Explorer| 16. 8| 18. 3| 2005 Toyota Camry| 23. 7| 28. 5| 2003 Toyota Corolla| 32. 8| 33. 1| 2005 Toyota Prius| 37. 3| 56. 0| 1. Compute covariance & comment on this statistic. X| Y| 14. 3| 16. 8| 15. 0| 17. 8| 27. 8| 26. 2| 27. 9| 34. 2| 48. 8| 47. 6| 16. 8| 18. | 1) Mean X = 14. 3+15. 0+27. 8+27. 9+48. 8+16. 8+23. 7+32. 8+37. 3 = 244. 4/9 = 27. 155 Mean Y = 16. 8+17. 8+26. 2+34. 2+47. 6+18. 3+28. 5+33. 1+56. 0 =278. 5/9 = 30. 944 = (14. 3-27. 155)(16. 8-30. 944)+ (15. 0-27. 155)(17. 8-30. 944)+(27. 8-27. 155)(26. 2- 30. 944)+(27. 9-27. 155)(34. 2-30. 944)+(48. 8-27. 155)(47. 6-30. 944)+(16. 8-27. 155)(18. 3-30. 944)+(23. 7-27. 155)+(28. 5-30. 944)+(32. 8-27. 155)+(33. 1-30. 944)+(37. 3-27. 155)+(56. 0-30. 944) ……………………………………………………………………………….. 9-1 = (-12. 55)(-14. 144)+(-12. 155)(-13. 144)+(0. 645)(-4. 744)+(0. 745)(3. 256)+(21. 645)(16. 656)+(-10. 355)(-12. 644)+(-3. 455)(-2. 444)+(5. 645)(2. 156)+(10. 145)(25. 056) = 181. 82+159. 765+-3. 05+2. 425+360. 519+130. 928+8. 444+12. 17+254. 193 = 1,107. 214 =138. 40 The Covariance between the mileage of owners and government standards is 138. 40. Since the mileage is positive, the variables are positively related, they move in the same direction. 2) Compute the coefficient of correlation Correlation Co-efficient Formula: = v (14. 3-27. 155)(16. 8-30. 944)+ (15. -27. 155)(17. 8-30. 944)+(27. 8-27. 155)(26. 2- 30. 944)+(27. 9-27. 155)(34. 2-30. 944)+(48. 8-27. 155)(47. 6-30. 944)+(16. 8-27. 155)(18. 3-30. 944)+(23. 7-27. 155)+(28. 5-30. 944)+(32. 8-27. 155)+(33. 1-30. 944)+(37. 3-27. 155)+(56. 0-30. 944) =v(181. 82)2+(159. 765)2+(3. 05)2+(2. 425)2+(360. 519)2+(130. 928)2+(8. 444)2+(12. 17)2+(254. 193)2 = v253,434. 147 / 9 =55. 93 3) What conclusion would you reach about the relationship between owner and current government standards mileage? To conclude, both companies is that, A correlation coefficient of 55. 3 tells you two important things: * Because the correlation coefficient is a positive number, the owner and the current government will have a positively related standard mileage. * Because 6. 679 is relatively far from indicating no correlation, the strength of the correlation between is strong. Both covariance and correlation identified that the variables are positively related. By standardizing measures, correlation is also able to measure the degree to which the variables tend to move together. If x and y have a strong positive linear correlation, r is close to +1.

An r value of exactly +1 indicates a perfect positive fit. Therefore, positive values indicate a relationship between x and y variables such that as values for x increases, values for y also increase. Question 3(a): | | Y1| Y2| Y3| | | | $20| $20-$100| Over $100| | X1| Cash | 0. 09| 0. 05| 0. 03| 0. 17| X2| Credit| 0. 03| 0. 21| 0. 23| 0. 47| X3| Debit| 0. 04| 0. 18| 0. 14| 0. 36| | | | | | | | | 0. 16| 0. 44| 0. 4| 1| i) What proportion of purchases was paid by debit card? Proportion paid by credit card is the sum of the following joint probability; P(x1,y1) + P(x2,y2) = P(x3,y3) = 0. 4 +0. 18 +0. 14 =0. 36) Find the probability that a credit card purchase was over $100? Probability that credit purchase is over $100 is exposed by joint probability P (x2, y1) = 0. 23 i) Determine the proportion of purchases made by credit card or by debit card. Proportion paid by credit card or debit will be equal to the total proportion – the proportion paid by cash. 1-0. 17 = 0. 83 iv) What is the probability that the purchases were made in cash given that the purchase was under $100? $100 = P(x1,y1) + P(x1,y2) = 0. 09 + 0. 05 = -0. 14 b)A university found that 30% of its students withdraw without completing the introductory statistics course. Assume that 18 students registered for this course. (i) Compute the probability that two or fewer will withdraw x? binomial (n=18), (p=0. 3) p (x? 2) = p (x =0) + p(x=1) + p(x =2) from the binomial table; P (x? 2) = 0. 05995) Compute the probability that exactly 5 will withdraw P (x =5) = C P (n-p) = 0. 202 i) Compute the expected number of withdrawals E (x) = np = 18×0. 3 = 5. 4 ?6 iv) Compute the variance of number of withdrawal. Var (x) = np (n-p) = 18 x 0. 3 x 0. 7 = 3. 78