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Business Statistics Third LevelSeries 2 2002QUESTION 1In the year 2000 a random sample of 96 small companies revealed the following profits and lossesdistribution: Profits and Losses (£000) Number of companies -50 up to 0 10 0 and up to 10 12 10 and up to 20 21 20 and up to 40 26 40 and up to 80 19 80 and up to 150 8(a) Calculate the arithmetic mean and the standard deviation of these company profits and losses. (8 marks)In the previous year a random sample of 150 small companies showed a mean profit of £28,400 with astandard deviation of £28,352.(b) Test to see if there has been a significant increase in the average profit made by small companies. (7 marks)(c) Calculate a 99% confidence interval for the arithmetic mean in the year 2000. (5 marks) (Total 20 marks) 3
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Model Answer to Question 1(a) 2 f mid pt fx fx –50 up to 0 10 –25 –250 6,250 Over 0 and up to 10 12 5 60 300 Over 10 and up to 20 21 15 315 4,725 Over 20 and up to 40 26 30 780 23,400 Over 40 and up to 80 19 60 1,140 68,400 Over 80 and up to 150 8 115 920 105,800 96 2,965 208,875 åf å fx å fx 2 x= åx x = 2,965 = £30.89 (000) (£30,885) åf 96 2 æ ö 2 2 s = å fx – ç å fx ÷ çåf ÷ s= 208,875 æ 2,965 ö –ç ÷ åf è ø 96 è 96 ø = 2,175.78 – 953.91 = 1,221.87 = £34.96 (000) (£34,955)(b) Null hypothesis: There has not been an increase in the average level of profits Alternative hypothesis: There has been an increase in the average level of profits Critical z for 0.05 significance level I tail test 1.64 30.89 – 28.4 2.49 z = x1 – x2 = = 0.59 s12 + s22 34.962 28.3522 18.09 + n1 n2 96 150 Alternative answer: 0.58 Conclusions: Accept the null hypothesis, there is insufficient evidence to claim the average level of profits has increased. σ 34.955(c) ci = x ± 2.58 = 30.89 ± 2.58 n 96 = 30.89 ± 2.58 x 3.57 = 30.89 ± 9.21 = £21.68 to £40.1 (000) 4
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QUESTION 2The price of the standard family saloon car and the company market share was recorded for a randomsample of 12 car manufacturers. Selling 137 138 125 142 168 145 135 145 160 146 136 160 price £00 Market 14 15 10 8 9 7 11 5 3 5 7 2 share %(a) Plot the data on a scatter diagram and comment. (4 marks)(b) Calculate the product-moment correlation coefficient. (10 marks)(c) Test to find if the correlation coefficient differs significantly from zero. (6 marks) (Total 20 marks) 5
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Model Answer to Question 2(a) Price and Market Share 16 14 12Market Share % 10 Series 1 8 6 4 2 0 12,000 13,000 14,000 15,000 16,000 17,000 18,000 Price £ Comment : Some weak negative relationship. 137 14 18,769 196 1,918 138 15 19,044 225 2,070 125 10 15,625 100 1,250 142 8 20,164 64 1,136 168 9 28,224 81 1,512 145 7 21,025 49 1,015 135 11 18,225 121 1,485 145 5 21,025 25 725 160 3 25,600 9 480 146 5 21,316 25 730 136 7 18,496 49 952 160 2 25,600 4 320 1,737 96 253,113 948 13,593 åx åy å x2 å y2 å xy(b) r= n å xy – (å x ) (å y ) æ n å x 2 – (å x )2 ö æ n å y 2 – (å y )2 ö ç ÷ç ÷ è øè ø 12 x 13,593 – 1,737 x 96 r= (12 x 253,113 – 1,737 2 )(12 x 948 – 96 2 ) 163,116 – 166,752 r= (3,037,356 – 3,017,169 )(11,376 – 9,216 ) –3,636 r= = – 0.5506 (20,187 )(2,160 ) 6 CONTINUED ON NEXT PAGE
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Model Answer to Question 2 continued(c) Null hypothesis: The correlation coefficient does not differ from zero. Alternative hypothesis: The correlation coefficient does differ from zero. Degree of freedom = n – 2 = 12 – 2 = 10 Critical t0.025 = 2.23 r n–2 – 0.55 12 – 2 t= t= 1– r2 1– ( −0.552 ) –1.61 t= 1– 0.3025 –1.61 = 0.835 = –2.086 Conclusions: The calculated value of t is less than the critical value of t. There is insufficient evidence to reject the null hypothesis. The correlation coefficient does not differ from zero. 7
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QUESTION 3A company is planning the launch of a new product. It estimates the probability of good marketconditions to be 80%. If market conditions are good the probability of a successful launch is 75%, ifmarket conditions are poor the probability of a successful launch is 50%.(a) Find the probability that the launch is successful. (5 marks)(b) If the product launch is unsuccessful what is the probability that the market conditions were poor? (6 marks)The estimated returns from the new product launch are:Market conditions are good and the product launch is successful £55 millionMarket conditions are good and the product launch is unsuccessful – £13 millionMarket conditions are poor and the product launch is successful £37 millionMarket conditions are poor and the product launch is unsuccessful – £19 million(c) What is the expected profit from the new product launch? (4 marks)The company sells an established product that has variable levels of weekly sales with arithmeticmean of £5,000 and standard deviation of £600. You may assume that the sales are normallydistributed.(d) (i) Find the probability that in one week there are sales of over £6,500 (ii) The sales have to exceed £3,800 in each week for the product to break even, what is the probability of this happening? (5 marks) (Total 20 marks) 8
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Model Answer to Question 3(a) Good market conditions and successful = 0.8 x 0.75 = 0.6 Poor market conditions and successful = 0.2 x 0.5 = 0.1 Probability of successful launch = 0.7 Accept decision tree(b) Good and unsuccessful = 0.8 x 0.25 = 0.2 Poor and unsuccessful = 0.2 x 0.5 = 0.1 Probability unsuccessful = 0.3 Poor and unsuccessf ul 0.1 = = 0.33 Probability unsuccessf ul 0.3(c) Expected return Good market conditions and successful launch 0.8 x 0.75 x £55m = £33m Good market conditions and unsuccessful launch 0.8 x 0.25 x –£13m = -£2.6m Poor market conditions and successful launch 0.2 x 0.5 x £37m = £3.7m Poor market conditions and unsuccessful launch 0.2 x 0.5 – £19m = –£1.9m £32.2m(d) (i) z = x–µ z= 6,500 – 5,000 = 1,500 sd 600 600 = 2.5, prop = 1 – 0.994 = 0.006 (ii) z = x–µ z= 3,800 – 5,000 = 1,200 = 2 sd 600 600 Proportion = 0.977 9
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QUESTION 4(a) What are the benefits of an effective quality control system? (4 marks)A company in its quality control procedures sets the warning limit at the 0.025 probability point and theaction limit at the 0.001 probability point. This means for example that the upper action line is set sothat the probability of the mean exceeding the line is 0.001.The internal diameter of a bored hole is set at 35 mm with a known standard deviation of 0.1 mm.Random samples of 9 items at a time are taken from the production line to check the accuracy of themanufacturing process.(b) (i) Construct a quality control chart to monitor the manufacturing process. (8 marks) (ii) The results for 8 samples are given below. Plot these on your quality control chart and comment on the graph. (4 marks) Sample number 1 2 3 4 5 6 7 8 Sample mean (mm) 35.05 34.94 34.89 35.16 35.15 34.95 34.99 35.08(c) If the process mean changed to 35.02 mm and the standard deviation remained at 0.1 mm calculate the probability that the mean of a random sample of 9 components would be outside the warning limits. (4 marks) (Total 20 marks) 10
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Model Answer to Question 4(a) The benefits of a good quality control system are better quality products, fewer rejects and less waste, better customer relations, more sales Warning limits = x ± 1.96 σ = 35 ± 1.96 0 .1 = 35 ± 1.96 x 0.033 n 9 upper w lts 35.06 35.07(2) lower w lts 34.94 34.93(2) Action limits = x ± 3.09 σ = 35 ± 3.09 0 .1 = 35 ± 3.09 x 0.033 n 9 upper act lts 35.10 lower act lts 34.90(b) Quality Control Chart The process goes out of control twice. It seems unstable.(c) z = x–µ = 35.02 – 35.07 0.1 = 0.05 = 1.5 sd 0.033 9 Probability the mean lies outside the upper warning limit = 0.067 34.93 – 35.02 = 2 .7 0 .1 9 Probability the mean lies outside the lower warning limit = 0.004 Total probability = 0.071 or (0.072 if 0.0333 is used) 11
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QUESTION 5(a) In what circumstances is a significance test based on the ‘t’ distribution used in preference to a significance test based on the normal distribution? (4 marks)Data input clerks are sent on an intensive training course to increase their keyboarding speed. Theresults of a before and after test for a random sample of 10 clerks gave the following results. Speed ismeasured in key depressions per minute (kdp). Clerk a b c d e f g h i j Before training course (kdp) 625 598 685 754 658 690 559 840 758 685 After training course (kdp) 610 620 690 780 690 702 573 851 744 690(b) Test whether the training course has increased the clerks’ data input speed. (12 marks)(c) What is meant by a Type I and a Type II error? (4 marks) (Total 20 marks) 12
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Model Answer to Question 5(a) The ‘t’ distribution is used when n the sample size is small <30 and the standard deviation is estimated from the sample.(b) Null hypothesis: The course has not increased the speed of the data input clerks Alternative hypothesis: The course has increased the speed of the data input clerks Degree of freedom n – 1 10 – 1 = 9 Critical t0.05 value = 1.83 625 610 –15 225 615.04 598 620 22 484 148.84 685 690 5 25 23.04 754 780 26 676 262.44 658 690 32 1,024 492.84 690 702 12 144 4.84 559 573 14 196 17.64 840 851 11 121 1.44 758 744 –14 196 566.44 685 690 5 25 23.04 98 3,116 2,155.60 2 åd å d2 å æd – d ö ç ÷ è ø æ å çd – d ÷ ç ÷ ö2 å d2 – (å d )2 d = åd = 98 = 9 .8 sed = è ø n n 10 n –1 n –1 982 3,116 – 2,155.8 10 = = 239.53 = 15.48 or 10 – 1 10 – 1 If a two sample means t = d –0 = 9.8 – 0 15.48 = 9 .8 =2 test is used can score sed 4 .9 5: nh/ah, conclusions 10 Conclusions: The calculated value of ‘t’ is greater than critical value of ‘t’ reject the Null Hypothesis the speed of the data input clerks has increased(c) A Type I error is the error of rejecting a null hypothesis that is true and a Type II error is the error of accepting a false null hypothesis which should be rejected. 13
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QUESTION 6(a) (i) What is meant by the standard error of the mean? (4 marks) (ii) What is the difference between a one and two tail test? (4 marks)In September 2001, a Travel Agent used a random sample of 36 holiday makers to find out theaverage cost per person of a one week holiday in Ruritania. The following information was found: September 2001 Mean £372.40 Standard deviation £26.10 Sample size 36In the previous year the average cost of each holiday was £356.20.(b) Test whether the cost of a one week holiday to Ruritania has increased significantly since the previous year. (6 marks)The company based its views on customer satisfaction from the letters it receives. In the previousyear, 68% of the letters it received were of a positive nature.(c) The company wishes to adopt a more scientific approach to estimating customer satisfaction. What sample size would be needed to estimate the proportion of customers’ views to within 2% of the true figure at the 95% confidence level? (6 marks) (Total 20 marks) 14
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Model Answer to Question 6(a) (i) The standard error of the mean is the measure of dispersion of the sample means. It equals σ n (ii) A one tail test tests the direction of the difference between two statistics. A two tail test tests if there is a difference between the two statistics without regard to the direction.(b) Null hypothesis: There is no difference in the holiday cost between last year and September Alternative hypothesis: There is an increase in the holiday cost between last year and September Critical z value = 1.64 z = x –µ = 372.4 – 356.2 = 16.2 = 3.72 σ/ n 26.1 / 36 4.35 Conclusions: The calculated value of z is greater than the critical value of z. Reject the null hypothesis. There is evidence to suggest that the holiday cost has increased significantly. 0.02 0.02(c) ± 1.96 > ± 1.96 > p(1 – p) 0.68 x 0.32 n n 2 0.022 0.0004 3.8416 x 0.2176 1.96 > 3.8416 > n> > 2,089.8 0.68 x 0.32 0.2176 0.0004 n n n = 2,090 15
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QUESTION 7(a) In what circumstances is the multiplicative model preferable to the additive model in time series calculations? (4 marks)The table below shows the quarterly sales figures for a company: Quarter 1 Quarter 2 Quarter 3 Quarter 4 1998 48 56 70 96 1999 64 76 86 132 2000 72 100 108 188(b) By the method of centred moving averages, calculate the trend values for the time series and plot the trend on a graph. (10 marks)The average seasonal components calculated by the multiplicative method are shown below: Quarter 1 Quarter 2 Quarter 3 Quarter 4 0.768 0.903 0.980 1.349(c) Using the average seasonal components provided above and the original sales figures, calculate the seasonally adjusted sales and plot the results on the graph constructed for part (b). Comment on your results. (6 marks) (Total 20 marks) 16
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Model Answer to Question 7(a) The multiplicative model is preferable when the data has a strong trend with the differences between the trend and the original data values varying proportionally to the trend values rather than absolutely.(b)(c) Qtr Sales Total m avg 1 m avg 2 Trend 1 48 2 56 3 70 270 67.5 69.5 4 96 286 71.5 74 5 64 306 76.5 78.5 6 76 322 80.5 85 7 86 358 89.5 90.5 8 132 366 91.5 94.5 9 72 390 97.5 100.25 10 100 412 103 110 11 108 468 117 12 188 17 CONTINUED ON NEXT PAGE
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Model Answer to Question 7 continued Sales Seasonal adjustment Seasonal Adjusted Sales 48 0.768 62.50 56 0.902 62.08 70 0.980 71.43 96 1.349 71.16 64 0.768 83.33 76 0.902 84.26 86 0.980 87.76 132 1.349 97.85 72 0.768 93.75 100 0.902 110.86 108 0.980 110.20 188 1.349 139.36 Comment: The trend and seasonally adjusted data values show little difference. This implies there is no change in the underlying trend. 18
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QUESTION 8(a) When might a χ2 test be used? (4 marks)From its records, a company analyses its sales by size and region. The table below shows the results: Size of order Region £0 and up to £1,000 and up £3,000 and up £10,000 and £1,000 to £3,000 to £10,000 over Northern 40 33 20 15 Midlands 50 45 40 25 Southern 40 32 40 20(b) Test whether there is a relationship between size of order and region. (12 marks)(c) Combining the three regions estimate a 99% confidence interval for the proportion of orders that are valued at less than £1,000. (4 marks) (Total 20 marks) 19
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Model Answer to Question 8(a) The Chi-squared test is used to test for association between characteristics rather than variables, for randomness or changes in proportions.(b) Null hypothesis: There is no association between the size of order and region Alternative hypothesis: There is association between the size of order and region Degrees of freedom = (4 – 1)(3 – 1) = 6 Critical χ2 = 12.59 <£1,000 £1,000<3,000 £3,000<10,000 £10,000+ Region 40 33 20 15 50 45 40 25 40 32 40 20 Total 130 110 100 60 expected freq 35.1 29.7 27.0 16.2 52.0 44.0 40.0 24.0 42.9 36.3 33.0 19.8 contribution 0.684 0.367 1.815 0.089 to chi 0.077 0.023 0.000 0.042 0.196 0.509 1.485 0.002 5.288 2 χ = å (O – E ) = 5.288 E Conclusions: The calculated χ2 is less than the critical χ2 . There is insufficient evidence to reject the null hypothesis. There is no association between the size of order and region.(c) 99% confidence z = ± 2.58 p = 130/400 = 0.325, (1 – p) = 0.675 ci = p ± 2.58 p(1 – p ) / n = 0.325 ± 2.58 0.325 x 0.675 / 400 = 0.325 ± 2.58 x 0.0234 = 0.325 ± 0.0604 0.265 to 0.385 20
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