Diploma Part 2. Quantitative Methods. Examiner s Suggested Answers

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1 Diploma Part Quantitative Methods Examiner s Suggested Answers Question 1 (a) The standard normal distribution has a symmetrical and bell-shaped graph with a mean of zero and a standard deviation equal to one. The area under the graph is equal to one. z 1 = (180 00)/5 = 0.8 z = (05 00)/5 = 0. Using the normal tables, the proportion between 180 and 05 = 1 ( ) = From the normal tables, 0.3 of the values lie below z = 0.54 approx. (X 00)/5 = 0.54 X = 187 approx (c) (i) z = (106 10)/4 = 1 Using the normal tables, the proportion above 106 = z = (95 10)/4 = 1.75 Using the normal tables, the proportion below 95 = Question (a) Dividing each index by 116 and multiplying by gives: 86.1, 94.83, 99.14,, A simple aggregate index (SAI) of prices (p) for 005, using 004 as the base year, is given by: P LI = P P SAI = P = 146 = A Laspeyres index (LI) for 005, using 004 as the base year, is given by: Q Q (30 40) + (80 80) + (36 96) = (0 40) + (75 80) + (40 96) = =

2 (iii) P PI = P A Paasche index (PI ) for 005, using 004 as the base year, is given by: Q Q (30 30) + (80 80) + (36 98) = (0 30) + (75 80) + (40 98) = = (iv) The simple aggregate index attaches equal weight to all three raw materials and so is pulled up by the relatively large percentage increase in the price of coal. The Laspeyres and Paasche indices give less weight to the price of coal, but since the Laspeyres index uses base year quantities as weights, while the Paasche index uses current year quantities as weights, the Laspeyres index tends to give more weight to those materials whose prices have risen. This explains why the Laspeyres index is slightly higher than the Paasche index. Question 3 (a) Using the appropriate mode on your calculator, and using 15, 5, 5, 15, 5 and 35 as the mid-point values, the mean of the distribution is 1.5 and the standard deviation (using n as the divisor) is The sample standard deviation (using n 1 as the divisor) is So we have: Mean = 1,500 Standard deviation (using n as the divisor) = 11,779 Standard deviation (using n 1 as the divisor) = 11,854 [Either answer for the standard deviation is acceptable here.] To estimate the median, we have: Median = L + So the median = 13,684. / F 40 6 i = 10 + f = n The average profit is 1,500, but there is a large degree of variability around this average, as indicated by the standard deviation. Since the mean is below the median, the distribution has a small negative skew. (b) ( ) = ( ) = 3 mean median SK = sd ( ) = Also accept Sk = As expected, the distribution has a slight negative skew (i.e. a longer tail to the left).

3 (c) H 0 : μ = 15 : μ z = = [Here, the sample standard deviation of should be used.] Since > 1.96, the null hypothesis cannot be rejected. There is not sufficient evidence to conclude that the average profit is different from 15,000. Question 4 (a) H 0 : π = 0.01 : π 0.01 p = 4/00 = 0.0 z = = Since 1.4 < 1.96, the null hypothesis cannot be rejected. The company s claim cannot be rejected. (b) H 0 : π 1 = π : π 1 π n 1 = 500, p 1 = 0. n = 500, p = 0.3 p ˆ = ( ) + ( ) = 05. z = = Since 3.65 < 1.96, the null hypothesis can be rejected. There is a significant difference between the two survey results.

4 (c) H 0 : μ 1 = μ : μ 1 μ n = 50, x = 180, s =, 500 n = 40, x = 175, s = 3, 600 z = = 04. Since 0.4 < 1.96, the null hypothesis cannot be rejected. There is no significant difference between the average expenditure on home insurance in the two regions. Question 5 (a) Correlation analysis allows business decision-makers to measure the degree of association between two variables. This is important as it will allow businesses to check, for example, that an increase in marketing expenditure is associated with an increase in sales, or that increases in staff training are associated with increases in productivity. This kind of analysis can help businesses to avoid unnecessary expenditure. ( ) = x = = s x = y = = 75. s = y So the mean of x is 4,860, with a standard deviation of 4,66. The mean of y is 7,50, with a standard deviation of,377. ( ) = (iii) ( ) b = = a = 7.5 ( ) = 4.84 So the equation is: y^ = x R = = This represents a high degree of positive linear correlation.

5 (iv) ŷ = ( ) = 0 So, we predict food and transport expenditure equal to 0,000. The sample size is large and the correlation coefficient is high. So, assuming that the sample was not biased, the prediction is likely to be accurate. The prediction might be improved by the inclusion of other factors, such as petrol prices, in the regression. Also, the prediction may be unreliable, as 50,000 is likely to be outside the range of the sample data. Question 6 (a) (i) Breaking even requires that total revenue equals total cost. I.e: 44Q = Q 0Q = Q = 000 components per month To make a monthly profit of 10,000 requires that: 0Q = 00 Q = 500 components per month (iii) Profit/loss = 0Q = (0 0) = 0000 So, there would be a loss of 0,000 per month. Equilibrium is reached when Q = Q S = Q D. I.e. when: Q = 4000 Q Q = 300 tonnes P = 3400 (i.e. 34 per tonne) After the tax, the supply function is: P = Q S Now equilibrium is reached when: Q = 4000 Q Q = 50 tonnes P = 3500 (i.e. 35 per tonne)

6 Question 7 (a) In the additive model, all the components of a time-series are summed to give the original data. So, we have: y = T + S + C + R where y represents the original data, T is the trend, S is the seasonal variation, C is cyclical variation and R represents random factors. In the multiplicative model, the components are multiplied together to give the original data. I.e. y = T S C R The centred moving averages are: 14.5, 15.5, 15.75, 16.5, 17.5, 18.5, 19.75, 0.5 The differences between the original values and the trend values for each quarter are averaged in the following table. The averages are then adjusted to ensure that they sum to zero. These are the required seasonal factors. Q1 Q Q3 Q Quarterly averages Quarterly averages adjusted to sum to zero So, the seasonal factors to three decimal places are: 1.81, 4.469, 1.969, The average quarterly increase in the trend is The forecasts for the four quarters of 006 are: Q1: ( ) = (35) Q: ( ) = (19) Q3: ( ) = 11.8 (1) Q4: ( ) = (31) The forecasts are not likely to be very accurate, as they are based on a short series (only three years) and do not take into account other factors that might affect the number of visitors to the ski resort (e.g. household disposable incomes and weather conditions).

7 Question 8 (a) A simple random sample gives every member of the population an equal chance of selection. Judgmental sampling is a form of non-random sampling in which the researcher selects participants in the belief that they are experts in a particular field or have appropriate experience to enable them to provide useful information. (b) The results obtained from a questionnaire survey may be biased (and therefore not representative of the relevant population), if those who fail to respond to the questionnaire differ in any relevant ways from those who do respond. This is known as non-response bias. If the sample of questionnaires fails to be representative of the population just by chance, then it is said to exhibit sampling error. (c) (i) A 95% confidence interval is given by: 5000 ± = ± The required sample size can be found by solving the following equation for n: = 400 n This gives n = 3,457.44, so a sample of 3,458 should be taken. (iii) The required sample size can be found by solving the following equation for n: = n This gives n =,435.4, so a sample of,436 should be taken.