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1. 1. LCCI International Qualifications Business Statistics Level 3 Model Answers Series 3 2010 (3009)For further Tel. +44 (0) 8707 202909information Email. enquiries@ediplc.comcontact us: www.lcci.org.uk
3. 3. QUESTION 1A local authority conducts a random sample of the number of planning applications examined per dayby its planning officers over a period of five years. The data are shown below. Number of planning Number of applications examined days 40 and under 80 103 80 and under 120 68 120 and under 160 38 160 and under 200 22 200 and under 300 15(a) Calculate the mean and standard deviation for the number of planning applications examined per day. (9 marks)(b) Calculate a 95% confidence interval estimate for the mean daily number of planning applications examined. (5 marks)The government has set a standard of 120 planning applications to be examined per day.(c) Test whether the local authority reaches the government standard. (6 marks) (Total 20 marks)3009/3/10/MA Page 2 of 20
4. 4. MODEL ANSWER TO QUESTION 1(a)Number of planning Number ofapplications days 2examined x fx fx 2 f x x40 and under 80 103 59.5 6128.5 364645.8 215114.580 and under 120 68 99.5 6766.0 673217.0 2209.3120 and under 160 38 139.5 5301.0 739489.5 44706.6160 and under 200 22 179.5 3949.0 708845.5 121450.8200 and under 300 15 249.5 ..3742.5 ..933753.8 312337.4 246 25887.0 3419951.6 695818.6 2 2 f fx fx f x x fxArithmetic mean = x f = 25887 = 105.2 applications per day 246 2 fx 2 fxStandard deviation = s f f 2 3419951.6 25887s = 13902.2 11073.7 = 2828.5 246 246or 2 f x x 695818.6 f 246 == 53.18 per day 2 f x x f 2 f x x f 2 f x x f 2 f x x f == 1743409121 . 1653009/3/10/MA Page 3 of 20
5. 5. QUESTION 1 CONTINUED(b) 95% confidence interval z = 1.96 53.18ci x 1.96 105.2 1.96 n 246= 105.2 6.65= 98.58 to 111.88(c) Null Hypothesis: The local authority reaches the government standard. Alternative hypothesis: The local authority does not reach the government standard.One tail test z = 1.64 x 105.2 120z = 53.18 n 246= -14.27 = -4.21 3.39Conclusion: There is evidence to reject the null hypothesis; the local authority hasnot reached the government standard.3009/3/10/MA Page 4 of 20
6. 6. QUESTION 2A random sample of 12 households records the value of their house and the annual expenditure onrepairs to the house. Value of house Annual expenditure (£000) on repairs (£) 215 1850 323 2550 455 3150 329 1850 235 1750 425 1700 327 2600 385 3300 237 2750 155 1300 196 1600 176 1750(a) Calculate the least squares regression line of expenditure on repairs based on the value of a house. (10 marks)(b) Using the regression equation found in (a) estimate the annual expenditure on repairs to a house valued at £250,000. (2 marks)The value of the coefficient of determination is 0.403 or 40.3%(c) Explain what the coefficient of determination measures and comment on the accuracy of your answer in (b) above. (4 marks)(d) What factors, other than the value of the house, may affect the expenditure on repairs? (4 marks) (Total 20 marks)3009/3/10MA Page 5 of 20
7. 7. MODEL ANSWER TO QUESTION 2(a) Value of Annualhouse £000 expenditure 2 on repairs £ x xy 215 1850 46225 397750 323 2550 104329 823650 455 3150 207025 1433250 329 1850 108241 608650 235 1750 55225 411250 425 1700 180625 722500 327 2600 106929 850200 385 3300 148225 1270500 237 2750 56169 651750 155 1300 24025 201500 196 1600 38416 313600 176 1750 30976 308000 3458 26150 1106410 7992600 2 x y x xy n xy x yb 2 n x2 x 12 7992600 3458 26150 95911200 90426700b 12 1106410 34582 13276920 11957764b= 5484500 = 4.16 1319156 y x 26150 3458a b 4.16 n n 12 12a = 981.1y = 981.1 + 4.16x(b) Estimated costs of repair and maintenance = 981.1 + 4.16 x 250 = 981.1 + 1040. = 2021.1 (£)(c) The coefficient of determination measures the change in repairs expenditure due to the change in house value. Therefore, 40.3% of the change in repair expenditure is due to the change in house value. The estimated repairs expenditure is not very accurate.(d) Age of house, size of family, attitude to house repair, size and value may not be comparable.3009/3/10MA Page 6 of 20
8. 8. QUESTION 3(a) Explain the circumstances in which a paired t test would be used in preference to a two sample mean test for small independent samples. (4 marks)A consumer protection magazine wishes to compare the cost of repairs between two companies. Theyused a sample of 9 businesses to request estimates for the cost of a repair by both companies. Company X Company Y Business Estimated cost of Estimated cost of repairs £ repairs £ A 2400 2560 B 2600 2760 C 5696 5952 D 7936 7933 E 4000 4200 F 7040 7245 G 3930 3830 H 6000 5964 I 3576 3620(b) Test whether the estimated cost of repairs differs between the two companies. (12 marks)(c) What practical reasons may the two companies have for giving different estimates of repair costs? (4 marks) (Total 20 marks)3009/3/10MA Page 7 of 20
9. 9. MODEL ANSWER TO QUESTION 3(a) The paired t test is used when a single sample is subject to two “treatments” compared with two samples being compared.(b) Null hypothesis: The estimated cost of repair does not differ between the two companies. Alternative hypothesis: The estimated cost of repair does differ between the two companies.Degrees of Freedom = n - 1 = 9 - 1 = 8Critical t value 0.05 = 2.31: 0.01 = 3.36Company X Company YEstimated Estimatedcost of cost ofrepairs £ repairs £ Difference (d d) (d d )2 d 2 d 2400 2560 -160 -61.556 3789.1 25600 2600 2760 -160 -61.556 3789.1 25600 5696 5952 -256 -157.556 24823.8 65536 7936 7933 3 101.444 10291.0 9 4000 4200 -200 -101.556 10313.5 40000 7040 7245 -205 -106.56 11354.1 42025 3930 3830 100 198.444 39380.2 10000 6000 5964 36 134.444 18075.3 1296 3576 3620 ..44 54.444 2964.2 …1936 -886 124780 212002 2 2 Σd (d d) Σd d -886 = -98.44d = 9 n 2 2 d2 d 212002 886sd = = = 117.7 n n 9 9or 2 d d 124780sd = = 117.7 n 9 d 0 98.44t = = -2.36 sd n 1 117.7 9 13009/3/10MA Page 8 of 20
10. 10. QUESTION 3 CONTINUEDConclusion: The calculated value of t is greater than the critical value of t at the0.05 level; reject the null hypothesis accept the alternative hypothesis, theestimated cost of repair does differ between the two companies. The results is notsignificant at the 0.01 level accept the null hypothesis, the estimates do not differ betweenthe two companies.(c) The companies may have different cost structures due to eg labour costs, error in calculating costs, use of manufacturer’s or generic parts, quality of the work carried out.3009/3/10MA Page 9 of 20
11. 11. QUESTION 4A random sample of workers in the Marketing department of a large company were interviewed andthe degree of job satisfaction and their job status were recorded. Job Satisfaction High Medium Low Job Management 105 30 165 Status Skilled 69 32 89 Unskilled 56 38 96(a) Test whether there is an association between job satisfaction and job status. (12 marks)(b) National data show that marketing workers express the following levels of job satisfaction. Job Satisfaction High Medium Low 30% 16% 54% By combining the different job status of marketing workers’ test whether the views on job satisfaction of workers in the Marketing department differ from the national situation. (8 marks) (Total 20 marks)3009/3/10MA Page 10 of 20
12. 12. MODEL ANSWER TO QUESTION 4(a) Null hypothesis: there is no association between job satisfaction and job status. Alternative hypothesis: there is association between job satisfaction and job status.Degrees of freedom (r - 1)(c - 1) = (3 - 1)(3 - 1) = 2 x 2 = 4 2 0.05 = 9.49; 0.01 = 13.28Observed frequencies Job Satisfaction Job High Medium Low Status 105 30 165 69 32 89 56 38 96Expected frequencies 101.47 44.12 154.41 64.26 27.94 97.79 64.26 27.94 97.79 2Contributions to 0.1228 4.5176 0.7261 0.3489 0.5896 0.7908 1.0629 3.6212 0.0329 chi sq= 11.8127Conclusion: the calculated Chi-squared is greater than the critical value of Chi-squaredat the 0.05 but not at the 0.01 significance level, there is some association betweenjob satisfaction and job status.(b) Null hypothesis: marketing workers have the same levels of job satisfaction as marketing executives nationally. Alternative hypothesis: marketing workers do not have the same levels of job satisfaction as marketing executives nationally.Degrees of freedom (3 - 1) = 2 2Critical value of 0.05 = 5.99 0.01 = 9.21 High Medium Low Observed 230 100 350 Expected 204 108.8 367.2 3.314 0.712 0.806 Chi squared = 4.8313 2 2Conclusion: the calculated value of is less than the critical value of there is insufficientevidence to reject the null hypothesis that marketing executives have the same level ofjob satisfaction as marketing executives.3009/3/10MA Page 11 of 20
13. 13. QUESTION 5A, B and C are candidates for a role as design manager. The Managing Director recommends acandidate and the company board must ratify the decision. The probability the Managing Directorrecommends A is 55%, B 25% and C 20%. The probability that the board ratifies A is 60%, B 30% and C 10% and is independent of the recommendation made by the Managing Director.(a) Find the probability that (i) B is a successful candidate (ii) No candidate is successful (iii) Given that one candidate is successful, it is candidate B. (10 marks)A container ship terminal can unload standard containers at the rate of 2000 per 8 hour day with astandard deviation of 200 containers. Assume the rate of unloading is normally distributed.(b) Find the probability that more than 2360 containers are unloaded in a day. (3 marks)(c) A ship carries 6500 containers what is the probability it will be unloaded in less than 3 days. (7 marks) (Total 20 marks)3009/3/10MA Page 12 of 20
14. 14. MODEL ANSWER TO QUESTION 5(a) (i) B successful = 0.25 x 0.3 = 0.075 (ii) No candidate is successful A = 0.55 x 0.4 = 0.22 B = 0.25 x 0.7 = 0.175 C = 0.20 x 0.9 = 0.18.. 0.575 (iii) Probability a candidate successful = 1 - 0.575 = 0.425Probability B successful = 0.075 = 0.176 Probability success 0.425(b) More than 2360 in a day x 2000 2360z z = 360/200 = 1.8 sd 200Table probability = 0.964 required probability = 1 - 0.964 = 0.036(c) A joint normal distribution approach is needed.Joint mean x1 x2 x3 = 2000 + 2000 + 2000 = 6000Joint standard deviation = sd12 2 sd 2 sd32 = 2002 2002 2002 =346 x 6500 6000z z = 500/346 = 1.4 (1.445) sd 346Table probability = 0.919 required probability = 1- 0.919 = 0.081 (0.075)3009/3/10/MA Page 13 of 20
15. 15. QUESTION 6(a) Explain why, when the sample size increases, the sample proportions cluster more closely about the population proportion. (4 marks)A company is concerned about the difference in sick leave between the US factories and the UKfactories. A random sample of 70 US staff showed 11 took more than 10 days sick leave absent,whilst a sample of 90 UK staff showed 18 took more than 10 days sick leave.(b) (i) Test whether the proportion of staff taking more than 10 days sick leave is higher in the UK than the US (12 marks) (ii) Explain what is meant by a Type 1 error and whether a Type 1 error may have been committed in your conclusions in part (b) (i). (4 marks) (Total 20 marks)3009/3/10/MA Page 14 of 20
16. 16. MODEL ANSWER TO QUESTION 6(a) When samples of a given size are taken, the distribution of the sample proportions is referred to as the sampling distribution of the proportion . p (1 pThe formula is se = as n increases for any given proportion the value of se nwill decrease.(b) (i) Null hypothesis: the proportion of staff with more than 10 days sick leave in the US does not differ from the proportion of staff with more than 10 days sick leave in the UK. Alternative hypothesis: the proportion of staff with more than 10 days sick leave in the US is less than the proportion of staff with more than 10 days sick leave in the UK.Critical z value for 0.05 significance level = -1.64p1 = 11 = 0.1571; p2 = 18 = 0.20 70 90 n1 p1 n2 p 2p n1 n2= 0.1571 x 70 + 0.20 x 90 = 11 + 18 = 0.181 70 + 90 160 p1 p2 0.1571 0.20 z 1 1 1 1 p1 p 0.181 0.819 n1 n2 70 90= -0.0429 = - 0.698 0.0614Conclusions: the calculated z value is less than the critical value of z. There is insufficientevidence to reject the null hypothesis. The proportion of staff with more than 10 days sickleave in the US does not differ from the proportion of staff with more than 10 days sick leavein the UK. (ii) A type 1 error is when a true null hypothesis is rejected when it should be accepted. As the null hypothesis was accepted a type 1 error cannot have been made.3009/3/10/MA Page 15 of 20
17. 17. QUESTION 7 An investigation was carried out into the number of errors per shift made in dispatching internet orders from warehouses in London and in Hong Kong. Random samples were taken in both London and Hong Kong. A summary of the results is given below. London Hong Kong Arithmetic mean 225 211 Standard Deviation 31 23 Median 218 217 Sample size 55 47(a) Calculate a measure of skewness for each of the cities. (4 marks) (b) Write a brief memo to the Sales Manager of the company highlighting the main features in the error statistics between the two cities. (8 marks) (c) Test whether the average number of errors in Hong Kong is less than the average number of errors in London. (8 marks) (Total 20 marks) 3009/3/10/MA Page 16 of 20
18. 18. MODEL ANSWER TO QUESTION 7(a) The measure of skew = 3(mean – median) Standard deviationLondon = 3(225 – 218) Hong Kong= 3(211 – 217) 31 23 = 0.677 = -0.783(b) Memo layout: Title, Date, From, To and SubjectA range of content is possible. The following is suggestedThe arithmetic mean and standard deviation are both larger for London than Hong Kong.Coefficient of variations 0.138 and -0.109 London is more variable than Hong Kong.The London Median is greater than the Hong Kong Median. The London Mean is greater than theHong Kong Mean.The London mean must have more extreme high values whilst Hong Kong must havemore extreme low values.The Hong Kong data is negatively skewed and London data is positively skewed but toapproximately the same degree.(c) Null hypothesis: there is no difference in the number of errors made in dispatching internet orders from warehouses in London and in Hong Kong. Alternative hypothesis: there is a greater number of errors made in dispatching internet orders from warehouses in London than in Hong Kong.Critical value of z for 0.05 significance level = 1.64/2.33 x1 x2 225 211z = s2 1 s 2 2 312 232 n1 n1 55 47 14 14z = = 2.62 17.47 11.26 5.23Conclusions: The calculated value of z is more than the critical value of z. There is evidenceto reject the null hypothesis. There is a greater number of errors made in dispatching internetorders from warehouses in London than in Hong Kong.3009/3/10/MA Page 17 of 20
19. 19. QUESTION 8(a) Explain the impact of a business setting its quality control limits too wide. (4 marks)Quality control procedures are used which set the warning limits at the 0.025 probability point and actionlimits at the 0.001 probability point. This means, for example, that the upper action limit is set so that theprobability of the means exceeding the limit is 0.001. The average weight of pre-cut tuna is set at 175grams with a standard deviation of 5 grams. Samples of 9 items at a time are taken from the productionline to check the accuracy of the manufacturing process.(b) (i) Draw a control chart to monitor the process (7 marks) (ii) 7 samples taken from the production line had the following mean weight (grams): Weight per piece 178.6 180.8 170.5 168.4 172.5 175.8 175.2 (grams) Plot these data on your control chart and comment appropriately (5 marks)(iii) If the sample mean had been wrongly set at 177 cm, assuming the standard deviation remains the same and the sample size is 9, what is the probability that a sample mean lies outside the upper action limit? (4 marks) (Total 20 marks)3009/3/10/MA Page 18 of 20
20. 20. MODEL ANSWER TO QUESTION 8 (a) By setting the quality control limits too wide the number of samples rejected will decrease and costs of resetting the process will be reduced and form a lower level of rejected items. The business may gain a reputation for poor quality which may damage its potential sales.(b) 5 (i) Warning limits x 1.96 175 1.96 = 175 ± 3.26 = 171.7 to 178.3 n 9 (171.67 to 178.33 for z =2) 5 Action limits x 3.09 175 3.09 = 175 ± 5.15 = 169.85 to 180.15 n 9 (170 to 180 for z = 3) (ii) Quality Control Chart 180 Weight per piece grams 178 176 174 172 170 168 1 2 3 4 5 6 7 Sample number Comment: Samples 2 and 4 lie outside the action limits and the process is unstable. It needs to be stopped and adjusted. x 180.15 177 (iii) z z 5 n 9 = 3.15 = 1.9 table proportion = 0.971 therefore answer = 0.029 1.67 3009/3/10/MA Page 19 of 20 © Education Development International plc 2010
21. 21. LEVEL 33009/3/10/MA Page 20 of 20 © Education Development International plc 2010
22. 22. EDI International House Siskin Parkway East Middlemarch Business Park Coventry CV3 4PE UK Tel. +44 (0) 8707 202909 Fax. +44 (0) 2476 516505 Email. enquiries@ediplc.com www.ediplc.com1517/2/10/MA Page 21 of 12 © Education Development International plc 2010