2003 2011 p0 q0 pn qn p0q0 pnq0 p0qn pnqn 7.50 10.00 8.00 18.00 120 41 25 21 9.00 12.50 10.00 22.40 158 52 30 25 900 410 200 378 1080 512.5 250 470.4 1185 520 240 450 1422 650 300 560 sum 1888 2312.9 2395 2932 Laspeyres’ 122.51 Paasche’s 122.42 11. 2006 2011 p0 q0 pn qn p0q0 pnq0 p0qn pnqn 3.63 2.11 10.03 4.01 3 4 1 7 4.49 3.26 12.05 5.21 2 6 1 5 10.89 8.44 10.03 28.07 13.47 13.04 12.05 36.47 7.26 12.66 10.03 20.05 8.98 19.56 12.05 26.05 57.43 75.03 50.00 66.64 Laspeyres’ 130.65 Paasche’s 133.28 12(a) Month Aug Sept Oct Nov Dec Jan 155 143 120 139 165 162 100.00 92.26 77.42 89.68 106.45 104.52 (b) (i) August to January change = 104:52 - 100 = 4:52% (ii) September to December change = 13. Year Average wage Average RPI deflated to 2008 values 2008 2009 2010 255.1 271.3 290.7 214.8 213.7 223.6 255.1 272.7 279.3
π = 40 ± 1.96 x 1.732 = 36.6% to 43.4% 9. P = 80% Sep = 2.8284 95% confidence interval is 80 ±1.96×2.8284 = 74.5% to 85.5% 10. n = 400 = £230 σ = 42 SEP = = 2.1 (a) 95% confidence interval is by 230 ± 1.96 x 2.1 = 230 ± 4.12 or 225.1 to 234.1 (b) 99% confidence interval is given by 230 ± 2.58 x 2.1 = 230 ± 5.42 or 224.6 to 235.4
18. H0: No association between age and preference H1 : is an association between age and preference O E O-E 20 35 -15 6.428 40 42 2 0.095 80 63 17 4.587 30 15 15 15.000 20 18 2 0.222 10 27 -17 10.703 37.037 Test statistic χ2 = 37:037 5% critical value, df = (2 – 1)(3 – 1) = 2, from tables χ2= 5:991 Reject H0 Evidence to suggest that age and preference are associated (would also reject at 1% and 0.1% levels) More under-21s and less over-35s than expected appear to dislike the design. 19.
H0:Pattern of demand hasn’t changed H1 Pattern of demand has changed The critical value on 2 degrees of freedom at 5% SL= 5.991 Reject H0 and conclude that there has been a significant change in the pattern of demand. 20. H0 : µD = 0 H1 : µD > 0 (one tailed since it is hoped that the campaign would not decrease sales) Critical value of t = 1:895 at 5% (one tailed on 7 degrees of freedom) Differences are: A B C D E F G H 28 -25 40 100 60 -20 0 50 SD = 42.633 Standard = = 15.073 = 1.932 Since 1.932 is greater than 1.895 H0 can be rejected at 5%. Therefore it seems likely that the campaign has worked.
1000×x + 10 = 50 x = .04 which is a 2 in 50 chance. 8. (a) Best decision is to wait (b) Probability should decrease to 0,5 9. £19m 2 £20.8m 3 Rival store 0.7 0.3 No rival store branch store 0.7 0.3 No rival store branch £10m £40m £16m £32m To maximize expected profit, move to the new site
10 (a) Tut Fles 12.9 11 Don’t Dictator .2 9.97 12.9 –5 Earthquake .1 .9 Earthquake .1 9.7 10 –6 15 0.01 0.99 0.7 0.3 –20 10 –15 1
1. 2. Let P = number of units of product P and Q = number of units of product Q Max 600P + 400Q Subject to 2P + Q ≤ 10 P + Q ≤ 7 P, Q ≥ 0 Solution is P = 3 and Q = 4 giving a profit of £3400. 3. Let D = number of diesel engines and P = number of petrol engines to be produced each day Max 60D + 45P Subject to 4D + 2P ≤ 16 2D + 2P ≤ 12 P ≥ 3
(a) 50 radio and 30 TV ads giving a total rating of 28000 (b) Radio and budget constraints binding. Shadow price of radio ads constraint is 50 so If radio ads increased by 50 then extra rating would be 50 x 50 = 2500 (c) Can’t reduce TV ads rating. 8. (a) Let x1 = Amount invested in government Let x2 = Amount invested in corporate bonds Let x3 = Amount invested in FTSE 100 stocks Let x4 = Amount invested in Aim stocks Max .04x1 + .05x2 + .06x3 + .08x4 + .03(550000 - x1 - x2 - x3 - x4) i.e. Max 16500 + .01x1 + .02x2 + .03x3 + .05x4 s.t x1 + x2 + x3 + x4 ≤ 550 000 x2 ≥ 50 000 x2 + x3 + x4 ≤ 300 000 x1 + x2 - x3 - x4 ≤ 0 x1 + x2 + x3 - 3x4 ≥ 0 x1 , x2 , x3 , x4 ≥ 0
G 3 Activity to tp tm jJ (J2 C D J 2 2 3 7 6 6 3 3 5 3.5 3.5 4.8 0.6944 0.4444 0.25 (b) Details of the EST and LST and floats are given below. Activity EST EFT LFT LST Critical? A B C D E F G H I J 0 2 2 2 5.5 6.5 0 16.5 22.5 24.5 2 5 5.5 5.3 6.5 16.5 3 22.5 24.5 29.3 2 16.5 5.5 5.5 6.5 16.5 16.5 22.5 24.5 29.3 0 13.5 2 2.2 5.5 6.5 13.5 16.5 22.5 24.5 0 11.5 0 0.2 0 0 13.5 0 0 0 Y N Y N Y Y N Y Y Y Activities A, C, E, F, H, I are critical. Expected time before the national campaign can be launched is 29.3 days. (c) The variance of the critical path is 0.6944 + 0.25 = 0.9444 (Note: Activity D is not on the critical path), so the standard deviation is 0.9718 To find the probability that expected time is greater than 32 days we need to calculate the Z value which is = 2.77 From tables this give a probability of 0.0028 So there is a 0.28% chance that the expected time will exceed 32 days, which means that there is a probability of 100 – 0.28 = 99.72% that the launch can take place within 32 days.