PRBE002 Decision Making - Math Problem Example

PRBE002 – DECISION MAKING by: Presented to: Course/ Class: University: City and state: Due date: Question 1 a. Frequency distribution Highest = 213 Lowest = 82 Difference 131 Intervals = difference/width i.e. 131/20 Intervals = 6.55 Frequency Distribution Electricity Cost Frequency 80-100 4 100-120 7 120-140 9 140-160 13 160-180 9 180-200 5 200-220 3 More 0 Total 50 Relative frequency distribution Is obtained by dividing frequency of each class by total frequency For the first class interval, Relative Frequency=4/50 Electricity Cost Frequency Relative Frequency 80-100 4 0.08 100-120 7 0.14 120-140 9 0.18 140-160 13 0.26 160-180 9 0.18 180-200 5 0.1 200-220 3 0.06 More 0 Total 50 1 b. Histogram Frequency distribution Electricity Cost Frequency 80-100 4 100-120 7 120-140 9 140-160 13 160-180 9 180-200 5 200-220 3 More 0 Total 50 c. Cumulative relative frequency distribution Electricity Cost Frequency C.F R.C.F 80-100 4 4 0.08 100-120 7 11 0.22 120-140 9 20 0.4 140-160 13 33 0.66 160-180 9 42 0.84 180-200 5 47 0.94 200-220 3 50 1 More 0 Total 50 Data to be used in construction of Ogive are as follows: Cost C.F 80 0 100 4 120 11 140 20 160 33 180 42 200 47 220 50 d. Monthly electricity cost is concentrated around 140-160. Question 2 a. Finding mean, median, 1st quartile and 3rd quartile Grouping the data:  lower limit Higher limit Frequency Cumulative Frequency mid-point (x) fx 5.25 5.3 2 2 5.275 10.55 5.3 5.35 3 5 5.325 15.975 5.35 5.4 4 9 5.375 21.5 5.4 5.45 8 17 5.425 43.4 5.45 5.5 7 24 5.475 38.325 5.5 5.55 10 34 5.525 55.25 5.55 5.6 8 42 5.575 44.6 5.6 5.65 5 47 5.625 28.125 5.65 5.7 2 49 5.675 11.35 5.7 5.75 0 49 5.725 0 5.75 5.8 1 50 5.775 5.775 50 274.85 ∑f=50 ∑fx=274.85 Mean: Apply the formula:  Median: Apply the formula:  First quartile: Apply the formula:  Third quartile: Apply the formula:  b. Range, interquartile range, variance and standard deviation Range:  Interquatile range:  Variance:  Standard Deviation is the square root of variance and is given by:  c. On average, the sample taken weigh 5.497. The standard deviation of 0.1068 indicates that the weight varies from the average by 0.1068. The company should be concerned about the weight given that label weight is 5.5 grams whereas the real average is 5.497. It is also apparent that the weight varies by 0.1068. This should be the company’s concern. d. Constructing a box and whisker plot The data that were calculated above will be used accordingly as follows: Maximum 5.77 Max 0.2 Q3 5.57 Q3 0.065 Median 5.505 median 0.085 Q1 5.42 Q1 5.42 Minimum 5.25 min 0.17 It is skewed to the right because the tail is longer for max. e. The mean of the sample is 5.497 while the standard deviation is 0.1068. This shows that the company is not meeting the standard indicated on the label given that the weight keeps changing by 0.1068. The company should take into account the variation from the mean by taking a balance of plus or minus 0.1068. Question 3 a. Coefficient of correlation. According to Morien (2007), applicable formula for r is:  Individual Income, $ (x) Expen,diture, $ (y) x-mean y-mean (x-mean)(y-mean) (x-mean)^2 (y-mean)^2 400 350 -49.5 -49.5 2450.25 2450.25 2450.25 815 650 365.5 250.5 91557.75 133590.3 62750.25 550 525 100.5 125.5 12612.75 10100.25 15750.25 400 370 -49.5 -29.5 1460.25 2450.25 870.25 250 250 -199.5 -149.5 29825.25 39800.25 22350.25 300 295 -149.5 -104.5 15622.75 22350.25 10920.25 375 330 -74.5 -69.5 5177.75 5550.25 4830.25 380 350 -69.5 -49.5 3440.25 4830.25 2450.25 425 415 -24.5 15.5 -379.75 600.25 240.25 600 460 150.5 60.5 9105.25 22650.25 3660.25 4495 3995 0 0 170872.5 244372.5 126272.5 b. The value of r is 0.0000498. This indicates that there is no linear relationship between income and expenditure. c. The explanation is not reflected in the data since the value of r is approaching zero meaning no relationship between the income and expenditure. Question 4 Assume regression equation of the form:  a. Least Squares approach to regression The data is as follows Month Petrol Price, x Magazine sales, y xy x^2 y^2 1 85 1050 89250 7225 1102500 2 91 1112 101192 8281 1236544 3 85 982 83470 7225 964324 4 89 1000 89000 7921 1000000 5 95 950 90250 9025 902500 6 98 940 92120 9604 883600 7 105 910 95550 11025 828100 8 110 980 107800 12100 960400 9 109 900 98100 11881 810000 10 115 878 100970 13225 770884 11 118 900 106200 13924 810000 12 122 925 112850 14884 855625 Total 1222 11527 1166752 126320 11124477 Substituting into the formulas above: Regression is therefore given by:  b. The slope of the regression equation is 1344.1996. This is a positive sloping line hence indicates that petrol price rise and decline in the same direction with newspaper price. c. If the petrol price is $1.00 per liter, the magazine sales translates to: d. Y intercept shows that magazine sales will be -3.7671 when petrol price is zero. This is not useful interpretation because there cannot be a situation of negative sales. Question 5 a. Probability that from random selection, diameter is between target and the population mean of 0.503: From the table,  Hence the probability that diameter is between target and the population mean is b. Probability that from random selection, diameter is between lower specification limit and the target. Probability that diameter is between lower specification and the target limit is: c. Above the a upper specification limit: Probability that the diameter is above the upper specification limit is given by:  d. Below the lower specification limit: Probability that the diameter is below the lower specification limit e. The probability of 0.9332: Checking from the table: 0.4332 is 1.5 Hence  Question 6 a. Steps: null and alternative hypothesis: Critical values at 0.05 level of significance is 1.96 Test statistic: Decision rule: reject  The evidence is not enough hence we will not reject the statistic. b. If manufacturers standard deviation was 1.75 pounds: Test statistic is less than -1.96 hence we reject the hypothesis. Question 7 a. Testing whether population waiting time is less than 5 min Statement of null hypotheses: Test type is one tailed test, t test Level of significance:  Test statistic: Using the data, test statistic translates to: b. Manager’s assurance s not realistic. The reason is that there is no evidence to confirm that waiting time is below 5 minutes. References Morien, D 2007, Business Statistics, Thomson Learning Nelson, South Melbourne. Read More