Diploma Part 2 Quantitative Methods Examiners Suggested Answers Q1 (a) A frequency distribution is a table or graph (i.e. a histogram) that shows the total number of measurements that fall in each of a number of class intervals. A probability distribution shows the probabilities associated with all the possible values of a random variable. (b) Using the binomial distribution with p = 0.25, q = 0.75 and n = 5, P(2) = 5 C 2 (0.25) 2 (0.75) 3 = 0.2637 (c) Using the Poisson distribution with µ = 3.5, P(more than 3) = 1 [P(0) + P(1) + P(2) + P(3)] = 0.4634 (d) (i) z = (500 505)/4 = 1.25. Required proportion = 0.1057 (ii) z 1 = (498 505)/4 = 1.75 z 2 = (503 505)/4 = 0.5 Required proportion = 0.2685
Q2 (a) Mean = 97.33 Median = 100 Mode = 105 (b) Standard deviation (n) = 11.95 Standard deviation (n 1) = 12.16 (c) The standard error of the mean is the standard deviation of the sampling distribution of the mean. σ 12. 16 SE = = 222. n 30 (d) 95% confidence interval = 97.33 ± (1.96 2.22) 95% confidence interval = 97.33 ± 4.35 99% confidence interval = 97.33 ± (2.575 2.22) 95% confidence interval = 97.33 ± 5.72 (e) H 0 : µ 95 H 1 : µ > 95 At the 5% level, the critical value of z = 1.645 (one-tailed test). 97. 33 95 z = = 105. 222. Cannot reject H 0.
Q3 (a) Let y be Aptitude and x be Age. Then, x = 289, y = 1085, xy = 30495, x = 9159, y = 118875, n = 10 2 2 R = ( 10 30495) ( 289 1085) {( 10 9159) 289 } ( 10 118875) 1085 { } = 2 2 = 0.89 This represents relatively strong, negative linear correlation. (b) 8615 b = = 107. 8069 8615 8069 11525 = 8615 9643. 4 a = y bx = 108. 5 (( 1. 07) 28. 9) = 139. 36 So, yˆ = 139. 36 1. 07x (c) The regression equation suggests that younger applicants gain higher scores in the aptitude test, and the equation could be used to predict aptitude score from an applicant s age. This is not likely to be very useful because the sample is small and other influences on aptitude are not considered. (d) Aptitude: 105 110 90 120 125 95 100 120 110 110 Ranking: 4 6 1 8.5 10 2 3 8.5 6 6 (e) 6 13 Spearman s R = 1 = 092. 10( 100 1) This suggests a high degree of positive correlation.
Q4 (a) Seasonal variation refers to any consistent pattern in a variable over some time period within a year. Data is said to be seasonally adjusted when the seasonal variation has been removed so that only the trend and random variations remain. (b) (i) Let y be the number of new stereo systems sold: y Trend (centred 4-pt MA) 130 105 85 112.5 120 116.25 150 118.75 115 122.5 95 126.875 140 129.375 165 130.625 120 131.25 100 133.125 140 137.5 180 141.25 140 145 110 160 (ii) Qtrs 1 2 3 4 Years 1 27.5 3.75 2 31.25 7.5 31.875 10.625 3 34.375 11.25 33.125 2.5 4 38.75 5 Average : 34.792 7.917 30.833 5.625 Adjusted : 34.37 8.33 31.2 5.21 (iii) The seasonally adjusted series is: 95.625, 113.33, 116.25, 114.79, 115.625, 123.33, 126.25, 134.79, 130.625, 128.33, 131.25, 134.79, 145.625, 148.33, 141.25, 154.79
(iv)
Q5 (a) Break-even analysis enables business people to identify the level of output at which the total cost of production is equal to total revenue, so that no profit is being made. It may be useful as a means of indicating the point at which profits will start to be made and by concentrating managers minds on the determinants of costs and revenues. The simplifying assumptions upon which the model is based may limit its usefulness in practice. (b) (i) At the break-even level, Total revenue = Total cost. 60Q = 200000 + 10Q 50Q = 200000 00Q = 4000 So the break-even level of output is 4,000 units per month. (ii) Profit = R C = 50Q 200000 So profit is 5,000 when: 50Q = 200000 + 5000 = 205000 Q = 4100 So the required quantity is 4,100 units per month. (iii) When 50Q 200000 = 10000, Q = 3800 Since the break-even level is 4,000, an additional 200 units are required to break even. (iv) Now 70Q = 10Q + 264000 Now 70Q = 4400 So the new break-even level is 4,400 units per month.
Q6 (a) (i) H 0 : π 0.5 H 1 : π > 0.5 where π is the proportion of male compact car buyers. (ii) At the 5% level of significance, the critical z-value is 1.645. So the null hypothesis will be rejected if z > 1.645 (one-tailed test). (iii) z = 052. 05. 05. 05. 325 = 0. 721 (iv) Since 0.721 < 1.645, the null hypothesis cannot be rejected. The evidence does not support the claim that more than half the buyers of compact cars are men. (b) H 0 : µ N = µ S H 1 : µ N µ S where µ N and µ S are the mean house prices in the north and south respectively. In a two-tailed test, at the 5% level of significance, the critical z-value is ±1.96. z = 150000 160000 20000 50 2 2 + 25000 80 = 2. 515 Reject the null hypothesis. The evidence suggests that house prices are not the same in the two regions.
Q7 (a) A weighted index takes into account the relative importance of each item included in the index. Unweighted price index numbers suffer from the serious disadvantage that they do not make any allowance for the relative importance of each item in the average household s budget. To calculate a more useful index, therefore, the unit prices should be weighted in such a way that the weights reflect the relative importance of each item. Quantities purchased are often used as weights for price indices. (b) (i) Aggregate quantity index = qn q = 282 100 100 = 88 125 0 320. (ii) Price relatives (q n /q 0 ) 120/150 = 0.800 100/110 = 0.909 62/60 = 1.033 2.742 So the arithmetic mean of quantity relatives is 2. 742 100 = 91. 4 3 (iii) (q n /q 0 ) w 0.8 0.42 = 0.336 0.909 0.08 = 0.0727 1.033 0.5 = 0.5167 Weighted Index = (0.336 + 0.0727 + 0.5167) 100 = 92.54
Q8 (a) Relevant points might include: Postal questionnaires are useful and convenient ways of obtaining primary data. They can cover a wide geographical area at low cost and can include a wide range of questions. However, it is difficult to design easy-to-understand and non-ambiguous questions. There may be a low response rate and non-response bias. Interviewing is a useful and very direct way of obtaining primary data. But it can be expensive and time-consuming to obtain a sufficiently large and random sample. Interviewers need to be well trained. Leading questions should be avoided. Respondents may not answer truthfully and non-response bias may be a problem. (b) (i) Need to obtain a sampling frame from which a sample can be selected (this might come from the manufacturer s list of past customers). Then describe a sampling method (such as cluster sampling) that will be less costly than simple random or stratified sampling. (ii) One possibility would be to use personal interviews in which consumers are shown the designs and asked to rank them in order of preference, based on different criteria (e.g. colour, texture, etc.) (iii) An average ranking or score for each design for each criteria could be calculated and presented in a table, together with some overall conclusions and recommendations for the manufacturer.