Diploma Part 2. Quantitative Methods. Examiner s Suggested Answers

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1 Diploma Part 2 Quantitative Methods Examiner s Suggested Answers Question 1 (a) The binomial distribution may be used in an experiment in which there are only two defined outcomes in any particular trial and the probability of each outcome is constant from trial to trial. (b) Using the binomial distribution with p = 0.6, q = 0.4 and n = 7, (i) P(5) = 7 C 5 (0.6) 5 (0.4) 2 = P(6) = 7 C 6 (0.6) 6 (0.4) = P(7) = (0.6) 7 = P(5 or more) = = (c) Using the Poisson distribution with µ = 2, P(0) = e 2 = P(1) = 2 e 2 = P(2) = (2 2 e 2 )/2 = P(3 or more) = 1 ( ) = (d) (i) z = ( )/2 = 1. Required probability = z = ( )/2 = 1.5 Required probability = Question 2 (a) The Laspeyres price index uses base-year quantities as weights. When a good s price rises, we expect quantity purchased to fall. So by using baseyear quantities as weights, the Laspeyres index tends to overweight those goods whose prices have risen. The Paasche price index uses current-year quantities as weights and so tends to give less weight (than the Laspeyres index) to those goods whose prices have risen, but to overweight those goods whose prices have fallen. p (b) (i) Aggregate price index = n = 100 = p

2 Price relatives (p n /p 0 ) 60/50 = /30 = /60 = /85 = So the arithmetic mean of price relatives is = (iii) pq = pq = pq = pq = Laspeyres price index = Paasche price index = = =. Fisher s Ideal Index = = Question 3 (a) SW: Mean = 170,200 Median = 172,000 Standard deviation (n) = 31,180 Standard deviation (n 1) = 31,330 NE: Mean = 153,000 Median = 149,140 Standard deviation (n) = 33,510 Standard deviation (n 1) = 33,680 The results suggest that houses are more expensive on average in the SW region. The degree of dispersion is similar in the two regions, though slightly higher in the NE. House prices are slightly positively skewed in the NE (where the median < the mean), but slightly negatively skewed in the SW (where the median > the mean).

3 (b) H 0 : µ sw = µ ne H 0 : µ sw µ ne At the 5% level the critical value of z is ± z = = Since 3.74 > 1.96, the null hypothesis can be rejected. There is evidence of a significant difference between house prices in the NE and SW. Question 4 (a) A one-tailed (or one-sided) test investigates whether a sample value is significantly greater than (or less than) an assumed population value at a given level of significance. The test uses an area in one tail of the normal curve. A two-tailed (or two-sided) test investigates whether a sample value is significantly different from an assumed population value and so uses areas in both tails of the normal curve. (b) (i) The standard error of the mean is the standard deviation of the sampling distribution of the mean. σ 1200 SE = = n % confidence interval = 7500 ± ( ) = 7,500 ± % confidence interval = 7500 ± ( ) (iii) H 0 : µ = 8000 H 1 : µ > 8000 = 7,500 ± At the 10% level, the critical value of z = 1.28 (one-tailed test) z = = The null hypothesis can be rejected. The students union s claim can be rejected at the 10% level.

4 Question 5 (a) (i) (iii) (b) (i) A very low degree of positive linear correlation. A very high degree of negative linear correlation. A low degree of negative linear correlation. Scatter Diagram 180 Selling Price ( 000) y Number of Rooms There appears to be a reasonable degree of positive linear correlation between number of rooms and selling price. Let y be Selling Price and x be Number of Rooms. Then, 2 2 x = 56, y = 1225, x = 352, y = , xy = 7245, n = 10 R = ( ) ( ) ( ) 56 ( ) 1225 [ ][ ] = This represents a very high degree of positive linear correlation. = (iii) R 2 = This suggests that 89.5% of the variation in selling prices in the sample is explained by variations in the number of rooms. (iv) The correlation coefficient measures the degree of linear correlation. It may be unreliable, therefore, if there is a non-linear relationship between two variables. Spurious correlation is possible with timeseries data where unrelated variables share a common trend over time.

5 Question 6 (a) Break-even analysis helps firms to identify how many units must be sold to avoid losses and start making profits. It can also help with other types of decisions, such as ensuring adequate capacity when embarking on a new product, or making a choice between buying and leasing equipment. (b) (i) Profit = R C = 12Q (8Q + 400) = 4Q 400 When Q = 150, Profit = = 200 So when 150 meals are sold per day, the expected daily profit is 200. At the break-even level, Total revenue = Total cost. 12Q = 8Q Q = 400 Q = 100 So the break-even level of output is 100 meals per day. (iii) When 4Q 400 = 200, Q = 50 Since the break-even level is 100, an additional 50 meals per day are required to break even. (c) Machine A: C = Q Machine B: C = Q The costs become equal when: Q = Q Q = 10,000 So the costs of operating the two machines are equal when both machines produce 10,000 units. Question 7 (a) Let the twelve quarters be x = 1, 2,..., 12. Then we have: 2 x = 78, y = 412, x = 650, xy = 2865, n = 12 ( ) ( ) 2280 b = 2 = = ( ) a = The linear regression trend is: y = x

6 (b) Time-series graph and Linear Trend Average Number of Calls per Hour Calls Trend Quarters (c) Years Qtrs Average (d) Forecasts: Q1: = Q2: = Q3: = Q4: = (e) The forecasts may not be accurate because they are based on a short time period (only three years) and do not take into account other possible influences. Question 8 (a) (i) The expected numbers in each age group are 66, 54 and 30. O E (O E) 2 /E At the 5% level of significance, the critical value of chi-squared with 2 degrees of freedom is Since > 5.99, we can reject the hypothesis that the age distributions are the same.

7 (b) (i) Row totals are 45, 75 and 30. Column totals are 95 and 55. Expected values are 28.5, 47.5, 19, 16.5, 27.5 and 11. O E (O E) 2 /E At the 5% level of significance, the critical value of chi-squared with 2 degrees of freedom is Since > 5.99, we can reject the hypothesis of independence. The results support the view that there is a link between age and the average size of current account balances.