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3. 3. QUESTION 1Over the past 9 years a company has achieved the following market share Year Market Share % 1998 22.3 1999 21.6 2000 22.0 2001 21.1 2002 20.8 2003 19.9 2004 19.5 2005 18.4 2006 17.6(a) Plot the data on a graph and comment on the pattern shown. (4 marks)(b) Calculate and state the least squares regression line for Market Share dependent on Year. (10 marks)(c) Use the regression line found in (b) to estimate the year in which market share is likely to fall to 15% and comment upon the accuracy of your estimate. (6 marks) (Total 20 marks)3009/3/07/MA 2
4. 4. MODEL ANSWER TO QUESTION 1(a) Market Share % v Time 23 22 21 Market Share % 20 19 18 17 16 15 1998 1999 2000 2001 2002 2003 2004 2005 2006 YearComment: ‘The market share is decreasing over time.’(b) Year Market Share % x y x2 xy 1 22.3 1 22.3 2 21.6 4 43.2 3 22.0 9 66.0 4 21.1 16 84.4 5 20.8 25 104.0 6 19.9 36 119.4 7 19.5 49 136.5 8 18.4 64 147.2 9 17.6 81 158.4 45 183.2 285 881.4 Σx Σy Σx2 Σxy3009/3/07/MA 3 CONTINUED ON THE NEXT PAGE
5. 5. MODEL ANSWER TO QUESTION 1 CONTINUED n∑ xy − (∑ x )(∑ y )b= n∑ x 2 − (∑ x ) 2 (9 x881.4) − (45 x183.2)b= 9 x 285 − 45 2b = -311.4 = -0.577 (-0.58) 540a= ∑ y −b∑x n n If x goes from 1998 to 2006 the 183.2 45 equation is y = 1174.84-0.577xa= − −0.577 x 9 9a = 20.36 +2.89 = 23.25Y = 23.25 – 0.577x(c) Y = 23.25-0.557x 15 = 23.25 –0.577x x = 23.25 – 15 0.577 x = 14.29 Round up to 15 Therefore year is 2012The answer is an extrapolation and therefore subject to uncertainty. In real life the business mightchange strategy to protect its market share. A better estimate would have been obtained from usingthe regression line for year on market share.3009/3/07/MA 4
6. 6. QUESTION 2(a) Explain what is meant by a 90% confidence interval for a sample proportion. (4 marks)A company is concerned about the level of customer satisfaction with the service provided by itsdomestic and overseas call centre operations. A random sample of customers calling its domestic callcentre showed that 267 out of 400 were satisfied or very satisfied with the service. A random sampleof customers calling its overseas call centre showed that 350 out of 500 were satisfied or very satisfiedwith the service.(b) Test whether there is a difference in the proportion of customers satisfied or very satisfied with the service provided at the two call centres. (12 marks)(c) Explain what is meant by a type 2 error. Might such an error have been committed in part (b) above? (4 marks) (Total 20 marks)3009/3/07/MA 5
7. 7. MODEL ANSWER TO QUESTION 2(a) The 90% confidence interval means that if samples of the same size are taken the sample proportion will lie within the stated range 90 times out of 100.(b) Null hypothesis: there is no difference in the proportion satisfied or very satisfied with the customer service from its domestic and overseas call centres. Alternative hypothesis: there is a difference in the proportion satisfied or very satisfied with the customer service from its domestic and overseas call centres.Critical z value = 1.96/2.58p1 = 267/400 = 0.6675, p2 = 350/500 = 0.7 267 + 350 617 Pooled value of p = = = 0.6856 400 + 500 900 0.6675 − 0.7 − 0.0325 z= = = - 1.04 ⎛ 1 1 ⎞ 0.2156(0.0045) 0.6856(1 − 0.6856)⎜ + ⎟ ⎝ 400 500 ⎠Conclusions: There is insufficient evidence to reject the null hypothesis at the 5% significance level.There is no difference in the proportion satisfied or very satisfied with the customer service from itsdomestic and overseas call centres.(c) A type 2 error is to accept the null hypothesis when it is false. A type 2 error may have been committed.3009/3/07/MA 6
8. 8. QUESTION 3The consumer price index and the index of average earnings are as follows: Year Consumer Index of Average Price Index Earnings 1996 100.0 83.3 1997 101.8 86.8 1998 103.4 91.3 1999 104.8 95.7 2000 105.7 100.0 2001 106.9 104.4 2002 108.3 108.1 2003 109.8 111.7 2004 111.2 116.7 2005 113.5 121.4 2006 116.7 127.0 estimate(a) Rebase the index of average earnings to 1996 as the base year = 100 (4 marks)(b) Using the information in the table and your answer to part (a), calculate an index of average real earnings with 1996 as the base year. (4 marks)A worker’s wage has risen, between the years 1996 to 2006, from £350.64 to £420.78.(c) What further increase would be necessary to enable the worker’s wage to keep pace with the Real Earnings Index? (4 marks)(d) (i) Explain how you would carry out a Quota Sample and give one advantage and one disadvantage of the method. (ii) Explain how you would carry out a Systematic Sample and give one advantage and one disadvantage of the method. (8 marks) (Total 20 marks)3009/3/07/MA 7
9. 9. MODEL ANSWER TO QUESTION 3(a) Conversion of Index of Average Earnings to base year 1996 = 100 Divide by 83.3 multiply by 100 throughout Index of Average Index of Earnings Year Average Rebased Earnings 1996 =100 1996 83.3 100.0 1997 86.8 104.2 1998 91.3 109.6 1999 95.7 114.9 2000 100.0 120.0 2001 104.4 125.3 2002 108.1 129.8 2003 111.7 134.1 2004 116.7 140.1 2005 121.4 145.7 2006 127.0 152.5 estimate(b) Index of average real earnings Divide Rebased Index of Earnings by Consumer Price Index Index of Average Consumer Index of Earnings Year Price Real Rebased Index Earnings 1996 =100 1996 100 100.0 100.0 1997 101.8 104.2 102.4 1998 103.4 109.6 106.0 1999 104.8 114.9 109.6 2000 105.7 120.0 113.5 2001 106.9 125.3 117.2 2002 108.3 129.8 119.9 2003 109.8 134.1 122.1 2004 111.2 140.1 126.0 2005 113.5 145.7 128.4 2006 116.7 152.5 130.7 estimate(c) £350.64 x 130.7/100 = £458.29 -£420.78 = £37.513009/3/07/MA 8 CONTINUED ON THE NEXT PAGE
10. 10. MODEL ANSWER TO QUESTION 3 CONTINUED(d) (i) Quota sampling involves taking a sample with a given number of people. Often this is done by stopping people in the street. Interviewers are given quota controls i.e. the characteristics that the respondents should have. These often relate to age, gender and income. The advantages are the sample is relatively cheap; there is no need for a sampling frame and no need for call-backs. The disadvantages are there may be bias in the choice of respondents, theoretically the standard error cannot be calculated so significance tests are not valid. (ii) Systematic sampling involves taking a sample at a given interval. The interval is determined by the sampling proportion e.g. a 10% sampling proportion would mean items are chosen every 10 items. The first item to be chosen would be selected randomly between 1 and 10 e.g. 4 and the 4th, 14th, 24th etc item sampled. The advantages are the method is relatively cheap. The disadvantages are there may be a natural period in the subjects and the method excludes or includes particular subjects and theoretically the standard error cannot be calculated so significance tests are not valid.3009/3/07/MA 9
11. 11. QUESTION 4(a) Explain the difference between a paired t test and a two independent sample means test. (4 marks)Two random samples of twelve car hire companies were taken in countries X and Y. The car rentalcosts in £ per 10 day hire for a standard four door 2000 cc saloon are given in the table below. Car Rental Costs in Car Rental Costs in Country X Country Y 350 525 525 340 550 430 525 520 390 670 540 470 525 520 570 540 530 650 650 650 540 680 590 450(b) Test whether the car rentals paid in country X differ from those paid in Country Y. (12 marks)(c) Explain what is meant by the sampling distribution of the mean. (4 marks) (Total 20 marks)3009/3/07/MA 10
12. 12. MODEL ANSWER TO QUESTION 4(a) A paired t test is used when a single sample is subject to two treatments, for example, before and after: a two independent sample means test is used when the results from two separate samples are compared in respect of a stated parameter, the mean.(b) Null hypothesis: there is no difference in the car rentals paid in country X and country Y. Alternative hypothesis: There is a difference in the car rentals paid in country X and country YDegrees of freedom n + m –2 = 12+12 –2 = 22Critical t values 2.07/2.82 x y x2 y2 (x − x ) (y − y ) 2 2 350 525 122500 275625 30189.06 1.5625 525 340 275625 115600 1.5625 33764.06 550 430 302500 184900 689.0625 8789.063 525 520 275625 270400 1.5625 14.0625 390 670 152100 448900 17889.06 21389.06 540 470 291600 220900 264.0625 2889.063 525 520 275625 270400 1.5625 14.0625 570 540 324900 291600 2139.063 264.0625 530 650 280900 422500 39.0625 15939.06 650 650 422500 422500 15939.06 15939.06 540 680 291600 462400 264.0625 24414.06 590 450 348100 202500 4389.063 5439.063 6285 6445 3363575 3588225 71806.25 128856.3 Σx Σy Σx2 Σy2 ( Σ x−x ) Σ (y − y ) 2 2 6285 6445x= = 523.75 y = = 537.083 12 12 − − ( x − x ) 2 + ( y − y) 2 71806.25 + 128856.3 s = 95.5 (95.0)s= s= n+m−2 12 + 12 − 2 x− y 523.75 − 537.083 = -0.34t= t= 1 1 1 1 s + 95.5 + n m 12 12Conclusions: There is insufficient evidence to reject the null hypothesis. There is no difference in thecar rentals paid in country X and country Y.(c) When a large number of samples of a given size are taken from a population the mean values will vary from sample to sample. If the sample size is large (30 and greater) the sample means will be normally distributed. If the sample size is small (less than 30) the means will be distributed in accordance with the t distribution with n-1 degrees of freedom.3009/3/07/MA 11
13. 13. QUESTION 5(a) Explain what the difference is between a one tail and a two tail test. (4 marks)The records of a company for two manufacturing plants X and Y, contain the following data on ordervalue for the past year: Plant X Plant Y Mean order value £ 760 698 Median order value £ 720 665 Standard deviation £ 145 135 Sample size 35 40(b) Calculate the coefficient of skewness for both plants and comment on your answers. (4 marks)(c) Test whether the mean order value is significantly higher in plant X than in plant Y. (8 marks)When the order values of the two plants are combined the mean order value for the company as awhole is found to be £726.93 and the standard deviation is £139.76.(d) Calculate the 95% confidence interval for the overall mean order value. (4 marks) (Total 20 marks)3009/3/07/MA 12
14. 14. MODEL ANSWER TO QUESTION 5(a) A one tail test tests if the difference between two samples varies in a single direction either greater or smaller; a two tail test tests if the difference between two samples varies in either direction (greater or smaller).(b) Coefficient of Skew = 3(Mean – Median) Standard Deviation Plant X = 3(760 -720) = 120/145 = 0.828 145 Plant Y = 3(698 – 665) = 99/135 = 0.733 135Both distribution are positively skewed, Plant X is more skewed than plant Y.(c) Null hypothesis: There is no difference in the mean order value between plant X and plant Y. Alternative hypothesis: The mean order value in plant X is greater than in plant Y.Critical z value = 1.64/2.33 x1 − x 2 760 − 698 62z= = = = 1.908 s12 s 2 2 1452 1352 600.71 + 455.625 + + n1 n2 35 40Conclusions: There is evidence to support the alternative hypothesis at the 0.05 level; the mean ordervalue at plant X is greater than the mean order value at plant Y. At the 0.01 level there is insufficientevidence to support the alternative hypothesis; the mean order value does not differ between the twoplants. σ 139.76(e) 95% confidence interval = x ± 1.96 = 726.93 ± 1.96 n 75 = 726.93 ± 31.63 = £695.30 to £758.563009/3/07/MA 13
15. 15. QUESTION 6A random sample is taken of workers in different sectors of the economy and the number of days theyare absent on sick leave in 2006 are recorded in the table below. Sector of the Economy Public Sector Domestic Owned Foreign Owned Sick Leave Taken Private Private Less than 5 days sick 35 45 20 leave 5 days and less than 10 95 102 35 days sick leave 10 or more days sick 70 53 45 leave(a) Test whether there is any association between the sector of the economy and the amount of sick leave taken. (12 marks)In a similar survey carried out 5 years earlier the proportions of people taking sick leave were asfollows. Less than 5 and less than 10 10 or more days days sick leave days sick leave sick leave 17% 54% 29%(b) Test whether the pattern of sick leave taken has changed over the period. (8 marks) (Total 20 marks)3009/3/07/MA 14
16. 16. MODEL ANSWER TO QUESTION 6(a) Null hypothesis: there is no association between the sector of the economy and the number of days sick leave taken. Alternative hypothesis: there is association between the sector of the economy and the number of days sick leave taken.Degrees of freedom (R-1)(C-1) = (3-1)(3-1) = 4Critical Χ2 = 9.49/13.28Observed 35 45 20 95 102 35 70 53 45Expected 40 40 20 92.8 92.8 46.4 67.2 67.2 33.6Contributions to X2 0.625 0.625 0 0.052155 0.912069 2.800862 0.116667 3.000595 3.867857 X2 = 12.00021Conclusions: The calculated value of X2, 12.00 is more than the critical value of X2 at the0.05 level: reject the null hypothesis there is evidence to support the alternative hypothesis. There isassociation between the work sector and the number of days sick leave taken.The calculated value of X2, 12.00 is less than the critical value of X2 at the 0.01 level, there isinsufficient evidence to reject the null hypothesis: there is no association between the work sector andthe number of day’s sick leave taken.(b) Null hypothesis: the pattern of sick leave has not changed. Alternative hypothesis: the pattern of sick leave has changed.Degrees of freedom = n-1 = 3-1 = 2Critical X2 = 5.99/9.21Observed 100 232 168Expected 85 270 145Contributions to X2 2.65 5.35 3.65 X2 = 11.65Conclusions: The calculated value of X2 is greater than the critical value of X2 at both the 0.05 and0.01 levels: reject the null hypothesis accept the alternative hypothesis. There is strong evidence tosupport the alternative hypothesis, the pattern of sick leave has changed.3009/3/07/MA 15
17. 17. QUESTION 7A company’s largest selling product has average weekly sales of 2500 units with standard deviation250 units. Assume that weekly sales are normally distributed.(a) Find the probability that weekly sales: (i) exceed 3050 units (ii) lie between 3100 units and 2200 units (8 marks)(b) Find the level of sales which will be exceeded on 90% of occasions. (3 marks)The same company is planning the launch of a new product. It estimates that the probability of goodeconomic conditions is 70%. If economic conditions are good the probability of the workforceaccepting a new wage offer is 30%; if economic conditions are poor the probability of the workforceaccepting a new wage offer is 50%.(c) Find the probability that the wage offer is accepted. (5 marks)(d) If the wage offer is accepted what is the probability that the economic conditions were poor. (4 marks) (Total 20 marks)3009/3/07/MA 16
18. 18. MODEL ANSWER TO QUESTION 7(a) (i) probability that sales exceed 3050 units x−x 3050 − 2500 550 z= = = = 2.2 σ 250 250 Probability = 1 – 0.986 = 0.014 (ii) probability that sales lie between 3100 units and 2200 units.Probability of sales between 2500 and 3100 x−x 3100 − 2500 600z= = = = 2.4 σ 250 250Probability = 0.992 –0.5 = 0.492Probability of sales between 2500 and 2200 x−x 2200 − 2500 300z= = = = 1.2 σ 250 250Probability = 0.885 –0.5 = 0.385Probability of sales 2200 to 3100 = 0.492 +0.385 = 0.877(b) Nearest table value for 90%/0.90 is z = 1.3 x−x z= , σ x − 2500− 1.3 = 250x = (-1.3 x 250) + 2500 - 325 +2500 = 2175 units(c)Probability that the wage offer is accepted.Good economic conditions and wage offer accepted = 0.7 x 0.3 = 0.21+Poor economic conditions and wage offer accepted = 0.3 x 0.5 = 0.15 = 0.36(d)Probability poor economic conditions and wage offer accepted = 0.15 = 0.4167 (0.42) Probability wage offer accepted 0.363009/3/07/MA 17
19. 19. QUESTION 8(a) Explain how the mean and range chart are used in quality control. (4 marks)Quality control procedures are used which set the warning limits at the 0.025 probability point andaction limits at the 0.001 probability point. This means, for example, that the upper action limit is set sothat the probability of the means exceeding the limit is 0.001. The weight of a packet of biscuits is setat 150 grams with a standard deviation of 8 grams. Samples of 8 items at a time are taken from theproduction line to check the accuracy of the manufacturing process.(b) (i) Calculate the values of the action and warning limits and construct a quality control chart to monitor the manufacturing process. (8 marks) (ii) The results for 7 samples are given below. Plot these on your quality control chart and comment on the results. (4 marks)sample number 1 2 3 4 5 6 7sample mean (grams) 150.5 156.4 143.9 152.9 148.13 159.5 157.6(c) Find the probability that a single item taken from the production line lies outside the action limits. You may assume that the control chart was correctly set up with mean 150 grams and standard deviation 8 grams. (4 marks) (Total 20 marks)3009/3/07/MA 18
20. 20. MODEL ANSWER TO QUESTION 8(a) When a product is made to a standard the aim is that average dimension of a sample is close to the specified value. It is also intended that the variability of the item should remain constant. The mean chart checks that the average stays within acceptable limits and the range chart checks that the variability of the product stays within acceptable limits. σ 8(b) Warning Limits = x ± 1.96 = 150 ± 1.96 = 144.5 to 155.5 n 8 σ Action Limits = x ± 3.09 = 141.3 to 158.7 n Quality Control Chart 165.0 160.0 UAL 155.0 UWL Weight grams Mean 150.0 145.0 LWL LAL 140.0 135.0 1 2 3 4 5 6 7 Sample NumberThe process is outside the action limits at sample 6. It needs to be stopped and adjusted.(c) Probability an item lies above the upper action limit x−μ 158.7 − 150z= = z= = 8.7/8 = 1.09 σ 8Probability for table value for z = 1.09 (1.1) = 0.864Required probability = 1- 0.864 = 0.136Probability an item lies below the lower action limit = 0.136Total Probability = 0.2723009/3/07/MA 19 © Education Development International plc 2007