Representation of data — Paper 7 · A Level Mathematics 9709

6 Photographers often need to take many photographs of families until they find a photograph which everyone in the family likes. The number of photographs taken until obtaining one which everybody likes has mean 15.2. A new photographer claims that she can obtain a photograph which everybody likes with fewer photographs taken. To test at the 10% level of significance whether this claim is justified, the numbers of photographs, x, taken by the new photographer with a random sample of 60 families are recorded. The results are summarised by Σ x 890 and Σ x2 13 780. = = (i) Calculate unbiased estimates of the population mean and variance of the number of photographs taken by the new photographer. [3] (ii) State null and alternative hypotheses for the test, and state also the probability that the test results in a Type I error. Say what a Type I error means in the context of the question. [3] (iii) Carry out the test. [4]

10 marks

Mark scheme: 6 (i) x = 148. (890/60 oe) B1 Correct answer 2 1  890 2  s = 59  13780 − 60  M1 Substituting in formula from book, o.e.   = 9.80 A1 [3] Correct answer (ii) H0: µ = 15.2 H1: µ < 15.2 B1 Correct H1 and H0 P(Type I error) = 0.1 (10%) B1 Correct answer Say the photographer has fewer discards when she doesn’t B1ft [3] o.e. must be related to question. No contradictions. ft their H1 14.83 − 152. (iii) Test statistic z = M1 Standardising must have 60 .9802 60 = –0.915 A1 Correct z (±0.91 to 0.92) or correct area 0.18 CV z = ± 1.282 M1 Valid comparison with correct CV must be + with + or – with – and consistent with their H1 oe comparison of areas Not enough evidence to support photographer’s claim. A1ft [4] Correct conclusion ft their z and their CV No contradictions GCE A/AS LEVEL – October/November 2009 9709 71 2

This question in 9709/71 Oct/Nov 2009

1 There are 18 people in Millie’s class. To choose a person at random she numbers the people in the class from 1 to 18 and presses the random number button on her calculator to obtain a 3-digit decimal. Millie then multiplies the first digit in this decimal by two and chooses the person corresponding to this new number. Decimals in which the first digit is zero are ignored. (i) Give a reason why this is not a satisfactory method of choosing a person. [1] Millie obtained a random sample of 5 people of her own age by a satisfactory sampling method and found that their heights in metres were 1.66, 1.68, 1.54, 1.65 and 1.57. Heights are known to be normally distributed with variance 0.0052 m2. (ii) Find a 98% confidence interval for the mean height of people of Millie’s age. [3]

4 marks

Mark scheme: 1 (i) doubling only gives even numbers so odd B1 Needs to be aware that odd numbers aren’t numbers not included [1] included .00052 (ii) 98% CI = 1.62 ± 2.326 × M1 Correct shape with 5 seen in denom and 5 their evaluated mean and sd (condone unbiased = 1.62 ± 0.0750 estimate of sample variance). B1 2.326 seen = (1.54, 1.70) A1 correct answer, cwo. Accept 1.55 and 1.70 . [3]

This question in 9709/72 Oct/Nov 2009

7 (a) Give a reason why sampling would be required in order to reach a conclusion about (i) the mean height of adult males in England, [1] (ii) the mean weight that can be supported by a single cable of a certain type without the cable breaking. [1] (b) The weights, in kg, of sacks of potatoes are represented by the random variable X with mean µ and standard deviation σ. The weights of a random sample of 500 sacks of potatoes are found and the results are summarised below. n = 500, Σx = 9850, Σx2 = 194 125. (i) Calculate unbiased estimates of µ and σ2. [3] (ii) A further random sample of 60 sacks of potatoes is taken. Using your values from part (b) (i), find the probability that the mean weight of this sample exceeds 19.73 kg. [4] (iii) Explain whether it was necessary to use the Central Limit Theorem in your calculation in part (b) (ii). [2]

11 marks

Mark scheme: 7 (a) (i) Pop too large Time consuming Not all pop accessible B1 [1] Or similar (ii) Testing involves destruction B1 [1] Or similar (b) (i) 9850/500 = (19.7) B1 500/499(194125/500 – (9850/500)2) M1 Allow with √. Method must be seen = 0.160(32) (3 sfs) or 80/499 A1 [3] or clearly implied. (ii) 19.73 −197. ".0 160" M1 For standardising 60 = 0.580 or 0.581 A1ft ft their mean and var in (b)(i) 1 – Φ(“0.580”) M1 Correct tail (= 1 – 0.7191) = 0.281 A1 [4] (iii) “Yes” must be seen or implied to gain mks X not nec’y normal B1 Sample large B1 [2] or X is approx N (SR Both reasons correct, but wrong or no conclusion scores SR B1)

This question in 9709/71 Oct/Nov 2010

4 The volumes of juice in bottles of Apricola are normally distributed. In a random sample of 8 bottles, the volumes of juice, in millilitres, were found to be as follows. 332 334 330 328 331 332 329 333 (i) Find unbiased estimates of the population mean and variance. [3] A random sample of 50 bottles of Apricola gave unbiased estimates of 331 millilitres and 4.20 millilitres2 for the population mean and variance respectively. (ii) Use this sample of size 50 to calculate a 98% confidence interval for the population mean. [3] (iii) The manufacturer claims that the mean volume of juice in all bottles is 333 millilitres. State, with a reason, whether your answer to part (ii) supports this claim. [1]

7 marks

Mark scheme: 4 (i) Est(µ) = 331(.125) B1 8  "877179"  Est(σ2) =  −"331. 1252"  M1 Allow their Σx2 7  8  = 4.125 or 4.13 A1 [3] (ii) z = 2.326 B1 2.4 331 ± z × M1 Allow incorrect z (≠ 1, 0), not a prob 50 = 330 to 332 (3 sfs) A1 [3] Ignore brackets, if given. CWO (iii) No, because 333 is not within CI B1ft [1] GCE AS/A LEVEL – October/November 2011 9709 71

This question in 9709/71 Oct/Nov 2011

4 The volumes of juice in bottles of Apricola are normally distributed. In a random sample of 8 bottles, the volumes of juice, in millilitres, were found to be as follows. 332 334 330 328 331 332 329 333 (i) Find unbiased estimates of the population mean and variance. [3] A random sample of 50 bottles of Apricola gave unbiased estimates of 331 millilitres and 4.20 millilitres2 for the population mean and variance respectively. (ii) Use this sample of size 50 to calculate a 98% confidence interval for the population mean. [3] (iii) The manufacturer claims that the mean volume of juice in all bottles is 333 millilitres. State, with a reason, whether your answer to part (ii) supports this claim. [1]

7 marks

Mark scheme: 4 (i) Est(µ) = 331(.125) B1 8  "877179"  Est(σ2) =  −"331. 1252"  M1 Allow their Σx2 7  8  = 4.125 or 4.13 A1 [3] (ii) z = 2.326 B1 2.4 331 ± z × M1 Allow incorrect z (≠ 1, 0), not a prob 50 = 330 to 332 (3 sfs) A1 [3] Ignore brackets, if given. CWO (iii) No, because 333 is not within CI B1ft [1] GCE AS/A LEVEL – October/November 2011 9709 72

This question in 9709/72 Oct/Nov 2011

1 Test scores, X, have mean 54 and variance 144. The scores are scaled using the formula Y = a + bX, where a and b are constants and b > 0. The scaled scores, Y, have mean 50 and variance 100. Find the values of a and b. [4]

4 marks

Mark scheme: 1 50 = a + b × 54 B1 100 = b2 × 144 or 10 = b × 12 B1 b = 5 oe M1 Solving two simultaneous equations 6 a = 5 A1 [4] Both correct 0 35 × 0 65 M1 For √(pq/n) in equation

This question in 9709/73 Oct/Nov 2011

3 Jack has to choose a random sample of 8 people from the 750 members of a sports club. (i) Explain fully how he can use random numbers to choose the sample. [3] Jack asks each person in the sample how much they spent last week in the club caf´e. The results, in dollars, were as follows. 15 25 30 8 12 18 27 25 (ii) Find unbiased estimates of the population mean and variance. [3] (iii) Explain briefly what is meant by ‘population’ in this question. [1]

7 marks

Mark scheme: 3 (i) Number all members B1 Explain the selection of 3-digit random B1 numbers Omit repeats OR omit nos. over 750 (until B1 have 8 nos.) [3] (ii) Est (µ) = 20 B1 8  3636 2  M1 1/7 × (3636 – 1602/8) Est (σ2) =  − 20  7  8  436 A1 (7.89...)2 M1A1, but 7.89... only M1A0 = or 62.3 (3 sfs) 7 [3] (iii) Amounts spent last week in café by all club B1 members [1] 1 M1 Int = 1 ignore limits ∫

This question in 9709/73 Oct/Nov 2011

6 A survey taken last year showed that the mean number of computers per household in Branley was 1.66. This year a random sample of 50 households in Branley answered a questionnaire with the following results. Number of computers 0 1 2 3 4 > 4 Number of households 5 12 18 10 5 0 (i) Calculate unbiased estimates for the population mean and variance of the number of computers per household in Branley this year. [3] (ii) Test at the 5% significance level whether the mean number of computers per household has changed since last year. [5] (iii) Explain whether it is possible that a Type I error may have been made in the test in part (ii). [1] (iv) State what is meant by a Type II error in the context of the test in part (ii), and give the set of values of the test statistic that could lead to a Type II error being made. [2]

11 marks

Mark scheme: 6 (i) x = 1.96 B1 (Σx2f = 254) 50  254 2  S2 = x  − .1 96  M1 Correct sub in S2 formula 49  50  A1 = 1548 or 1.2637 [3] 1225 (ii) H0: Pop mean = 1.66 H0: Pop mean = 1.66 H1: Pop mean ≠ 1.66 B1 H1: Pop mean > 1.66 B0 .196 −.1 66 M1 .196 −.1 66 M1 .1 2637 .1 2637 50 50 A1 = 1.887 = 1.887 A1 z = 1.96 1.887<1.96 M1 z = 1.645 M1 No evidence that mean has changed A1ft In context Evidence mean has changed A1ft [5] (iii) No because H0 not rejected B1f If H0 rejected in (ii): Yes because H0 rejected [1] (iv) State mean not changed when it B1 In State mean not increased when it has context has B1 B1 test stat < 1.645 B1 –1.96 < test stat < 1.96 [2] GCE AS/A LEVEL – May/June 2012 9709 71

This question in 9709/71 May/June 2012

5 It is claimed that, on average, people following the Losefast diet will lose more than 2 kg per month. The weight losses, x kilograms per month, of a random sample of 200 people following the Losefast diet were recorded and summarised as follows. n = 200 Σ x = 460 Σ x2 = 1636 (i) Calculate unbiased estimates of the population mean and variance. [3] (ii) Test the claim at the 1% significance level. [5]

8 marks

Mark scheme: 5 (i) Est(µ) = 2.3 B1 2  2  200  1636  460  200  1636  460      − or 1.7043 for M1   − Est(ë2) = M1 Allow   199  199  200  200   200  200       Or 1/199 ( 1636 – 4602/200 ) = 2.90 (3 sf) or 2.91 or 578/199 A1 [3] (ii) H0: Pop mean wt loss = 2 kg H1: Pop mean wt loss > 2 kg B1 Allow ‘µ’ but not just ‘mean’ 3.2 − 2 3.2 − 2 Stand’ise with √200. Accept sd/var M1 .2'9045' .1'7043' 200 200 mixes Or xcrit = 2 + 2.326√( 2.9045/200 ) = 2.489 or ± 2.49 A1 or 0.0064 / 0.9936 for area comparison or xcrit = 2.28(03) comp z = 2.326 M1 For valid comparison ( z or area or xcrit ) Evidence that mean wt loss > 2 kg A1ft No contradictions Reject H0 / accept H1 only if H0 / H1 correctly defined 200 If not used in (i): var = 2.89, sd = 1.7, [5] 199 cr z = 2.496 can score all marks Total [8] GCE A LEVEL – October/November 2012 9709 73

This question in 9709/73 Oct/Nov 2012

4 The lengths, x m, of a random sample of 200 balls of string are found and the results are summarised by Σ x = 2005 and Σ x2 = 20 175. (i) Calculate unbiased estimates of the population mean and variance of the lengths. [3] (ii) Use the values from part (i) to estimate the probability that the mean length of a random sample of 50 balls of string is less than 10 m. [3] (iii) Explain whether or not it was necessary to use the Central Limit theorem in your calculation in part (ii). [2]

8 marks

Mark scheme: 4 (i) est(µ) = 2005/200 = (10.025) B1 1 20052 est(σ2) = 20175 – ) M1 Correct subst in correct formula 99 200 = 0.376 (3 sf) A1 [3] (ii) 10− '10. 025' (= –0.288) M1 Allow without √, but ÷√50 essential .0' 376256' 50 M1 1 – Φ(‘0.288’) A1 (Use of ‘biased’ variance can still score fully in (ii) ) = 0.387 (3 sf) [3] GCE AS/A LEVEL – May/June 2013 9709 71 (iii) Yes; (assumed distr of X normal) B1 although distr of X unknown B1 [2]

This question in 9709/71 May/June 2013

4 In a survey a random sample of 150 households in Nantville were asked to fill in a questionnaire about household budgeting. (i) The results showed that 33 households owned more than one car. Find an approximate 99% confidence interval for the proportion of all households in Nantville with more than one car. [4] (ii) The results also included the weekly expenditure on food, x dollars, of the households. These were summarised as follows. n = 150 Σx = 19 035 Σx2 = 4 054 716 Find unbiased estimates of the mean and variance of the weekly expenditure on food of all households in Nantville. [3] (iii) The government has a list of all the households in Nantville numbered from 1 to 9526. Describe briefly how to use random numbers to select a sample of 150 households from this list. [3]

10 marks

Mark scheme: 150 1504 (i) Var(Ps) = (= 0.001144) M1 150 Seen. Accept 2.574 to 2.579 z = 2.576 B1 33 ± z√‘0.001144’ M1 Expression of correct form. Any z 150 = 0.133 to 0.307 (3 sf) A1 4 Must be an interval 19035 (ii) (= 126.9 =127(3sf)) B1 150 150  4054716  19035  2   −    o.e. M1 For use of a correct formula 149  150  150     = 11001.17 or 11000(3 sf) A1 3 (iii) 4-digit nos. each digit 0-9 B1 Some valid way of generating 4 digit Ignore nos > 9526 B1 random nos Ignore repeats B1 3 from valid method from valid method SR If zero score, full explanation of method for drawing numbers out of a hat can score B1. NB Systematic sampling follows the scheme with first B1 for some way of generating a random starting point. Total: 10 8.4 4 8 2 √

This question in 9709/71 Oct/Nov 2014

4 In a survey a random sample of 150 households in Nantville were asked to fill in a questionnaire about household budgeting. (i) The results showed that 33 households owned more than one car. Find an approximate 99% confidence interval for the proportion of all households in Nantville with more than one car. [4] (ii) The results also included the weekly expenditure on food, x dollars, of the households. These were summarised as follows. n = 150 Σx = 19 035 Σx2 = 4 054 716 Find unbiased estimates of the mean and variance of the weekly expenditure on food of all households in Nantville. [3] (iii) The government has a list of all the households in Nantville numbered from 1 to 9526. Describe briefly how to use random numbers to select a sample of 150 households from this list. [3]

10 marks

Mark scheme: 150 1504 (i) Var(Ps) = (= 0.001144) M1 150 Seen. Accept 2.574 to 2.579 z = 2.576 B1 33 ± z√‘0.001144’ M1 Expression of correct form. Any z 150 = 0.133 to 0.307 (3 sf) A1 4 Must be an interval 19035 (ii) (= 126.9 =127(3sf)) B1 150 150  4054716  19035  2   −    o.e. M1 For use of a correct formula 149  150  150     = 11001.17 or 11000(3 sf) A1 3 (iii) 4-digit nos. each digit 0-9 B1 Some valid way of generating 4 digit Ignore nos > 9526 B1 random nos Ignore repeats B1 3 from valid method from valid method SR If zero score, full explanation of method for drawing numbers out of a hat can score B1. NB Systematic sampling follows the scheme with first B1 for some way of generating a random starting point. Total: 10 8.4 4 8 2 √

This question in 9709/72 Oct/Nov 2014

5 The volumes, v millilitres, of juice in a random sample of 50 bottles of Cooljoos are measured and summarised as follows. n = 50 Σv = 14 800 Σv2 = 4 390 000 (i) Find unbiased estimates of the population mean and variance. [3] (ii) An !% confidence interval for the population mean, based on this sample, is found to have a width of 5.45 millilitres. Find !. [4] Four random samples of size 10 are taken and a 96% confidence interval for the population mean is found from each sample. (iii) Find the probability that these 4 confidence intervals all include the true value of the population mean. [2]

9 marks

Mark scheme: 5 (i) 14800/50 or 296 B1 50  4390000 2  M1 Oe − '296' (= 187.755)   49  50  A1 3 = 188 (3 sf) (ii) '187.755' M1 '187.755' 2 × z × = 5.45 oe If ‘2 ×’ omitted: z × = 5.45 M1 50 A1 50 z = 1.406 or 1.405 z = 2.812 or 2.810 A0 Φ(‘1.406’) (= 0.92 or 0.9199) Φ(‘2.812’) (= 0.9975) M1 α = 99.5 or 99 or 100 M1 A0 α = 84 (2 sf) allow 83.98 A1 4 For complete method to find α SR use of biased var(184) scores M1A1(1.4205) Α=84.5 M1A1 (iii) 0.964 M1 = 0.849 (3 sf) A1 2 Total 9 15 ∫ 2

This question in 9709/72 May/June 2015

1 Jyothi wishes to choose a representative sample of 5 students from the 82 members of her school year. (i) She considers going into the canteen and choosing a table with five students from her year sitting at it, and using these five people as her sample. Give two reasons why this method is unsatisfactory. [2] (ii) Jyothi decides to use another method. She numbers all the students in her year from 1 to 82. Then she uses her calculator and generates the following random numbers. 231492 762305 346280 From these numbers, she obtains the student numbers 23, 14, 76, 5, 34 and 62. Explain how Jyothi obtained these student numbers from the list of random numbers. [3]

5 marks

Mark scheme: 1 (i) Eg: Only students who use canteen B1 or any reason that some are excluded The five will probably be friends B1 [2] B1 each sensible reason must be in context (ii) 2–digits B1 ignore > 82 (anything too big) B1 Ignore repeats B1 [3] [Total 5] 1 ( )

This question in 9709/73 May/June 2015

4 The marks, x, of a random sample of 50 students in a test were summarised as follows. n = 50 Σx = 1508 Σx2 = 51 825 (i) Calculate unbiased estimates of the population mean and variance. [3] (ii) Each student’s mark is scaled using the formula y = 1.5x + 10. Find estimates of the population mean and variance of the scaled marks, y. [3]

6 marks

Mark scheme:  1508 4 (i)   = 30 . 16 (30 2. ) B1 Allow any form  50  50  51 825   − ('30 . 16 2' ) M1 49  50  (129.46367) = 129 (3 sf) Or 130 A1 [3] (ii) (1.5 × ‘30.16’ + 10) = 55.24 B1ft ft their 30.16 (1.52 × ‘129….’) M1 1.52 × their(129) with nothing added at any stage = 291 ( 3 sf) A1ft [3] Allow 290 Total 6

This question in 9709/73 May/June 2015

7 The diameter, in cm, of pistons made in a certain factory is denoted by X, where X is normally distributed with mean - and variance 32. The diameters of a random sample of 100 pistons were measured, with the following results. n = 100 Σx = 208.7 Σx2 = 435.57 (i) Calculate unbiased estimates of - and 32. [3] The pistons are designed to fit into cylinders. The internal diameter, in cm, of the cylinders is denoted by Y, where Y has an independent normal distribution with mean 2.12 and variance 0.000 144. A piston will not fit into a cylinder if Y −X < 0.01. (ii) Using your answers to part (i), find the probability that a randomly chosen piston will not fit into a randomly chosen cylinder. [6]

9 marks

Mark scheme: 7 (i) est µ = 2.087 B1 allow 2.09 100  435.57 2  est σ2 =  − .2087  M1 1 / 99 (435.57 – 208.72 / 100 ) 99  100  100 without : 0.000131 M0A0 = 0.000132(3232) or 131 / 99 0000 A1 [3] 99 (ii) E(Y – X) = 2.12 – 2.087 ( = 0.033 ) B1 or 2.12 – 2.087 – 0.01 for Y – X – 0.01 < 0 allow 2.09 for 2.087 Var(Y – X) = 0.000144 + ‘0.00013232’ M1 or √(0.0122 + ‘0.00013232’) M1 = 0.000276(32) A1 = 0.016623 A1 .0 01− .0' 033' (= –1.384) M1 their E(Y – X) & Var(Y – X) .0' 00027632 ' var must be a combination of the two vars Φ(‘–1.384’) = 1 – Φ(‘1.384’) M1 correct area / prob consistent with their working = 0.0832 A1 [6] SR use of biased var ( 0.000131 ) in (i) and (ii) scores in (ii) B1M1 A1 for 0.000275 and M1M1 A1 for 0.0827 ( 6 / 6 available) Total [9] Total for paper [50]

This question in 9709/73 Oct/Nov 2015

1 The weights, in kilograms, of a random sample of eight 16-year old males are given below. 58.9 63.5 62.7 59.4 66.9 68.0 60.4 68.2 Find unbiased estimates of the population mean and variance of the weights of all 16-year old males. [3]

3 marks

Mark scheme: 5081 ( 8 ) = 63.5 B1 (Σx2 = 32360.12) 8 '32360.12' M1 oe 7 ( 8 − '63.5' 2 ) [3] From correct working = 14.6 (3 sf) or 2553/175 A1 1 1 1 1

This question in 9709/71 Oct/Nov 2016

1 The weights, in kilograms, of a random sample of eight 16-year old males are given below. 58.9 63.5 62.7 59.4 66.9 68.0 60.4 68.2 Find unbiased estimates of the population mean and variance of the weights of all 16-year old males. [3]

3 marks

Mark scheme: 5081 ( 8 ) = 63.5 B1 (Σx2 = 32360.12) 8 '32360.12' M1 oe 7 ( 8 − '63.5' 2 ) [3] From correct working = 14.6 (3 sf) or 2553/175 A1 1 1 1 1

This question in 9709/72 Oct/Nov 2016

3 Household incomes, in thousands of dollars, in a certain country are represented by the random variable X with mean - and standard deviation 3. The incomes of a random sample of 400 households are found and the results are summarised below. n = 400 Σ x = 923 Σx2 = 3170 (i) Calculate unbiased estimates of - and 32. [3] … … … … … … … (ii) A random sample of 50 households in one particular region of the country is taken and the sample mean income, in thousands of dollars, is found to be 2.6. Using your values from part (i), test at the 5% significance level whether household incomes in this region are greater, on average, than in the country as a whole. [5] … … … … … … … … … … … …

8 marks

Mark scheme: 3(i) B1 Est(σ2) = 2 400 3170 "2.3075" 399 400   −     OE M1 = 2.60696 or 2.61 (3 sf) A1 (Note: Biased Var= 2.600 scores M0) Total: 3 3(ii) H0: Pop mean (or µ) = "2.31" or "2310" H1: Pop mean (or µ) > "2.31" or "2310" B1 FT ± 2.6 "2.310" 2.60696 50 − ÷ = 1.27 M1 A1 Standardising using their values, Accept 1.28 Comp 1.645 (OE) M1 Valid comparison z values or areas No evidence that incomes in the region greater A1 FT OE FT their z. No contradictions (No FT for 2 tail test – max score B0 M1 A1 M1 for comp 1.96 A0) Note: Accept alternative CV method Total: 5

This question in 9709/72 May/June 2017

4 Last year the mean level of a certain pollutant in a river was found to be 0.034 grams per millilitre. This year the levels of pollutant, X grams per millilitre, were measured at a random sample of 200 locations in the river. The results are summarised below. n = 200 Σx = 6.7 Σx2 = 0.2312 (i) Calculate unbiased estimates of the population mean and variance. [3] … … … … … … … (ii) Test, at the 10% significance level, whether the mean level of pollutant has changed. [5] … … … … … … … … … … … … … …

8 marks

Mark scheme: 4(i) x = 6.7/200 (= 67/2000 = 0.0335) B1 s2 = 2 200 0.2312 "0.0335" 199 200   × −     M1 s2 = 2 0.2312 0.0335 200 − M0 = 0.0000339(2) = 27/796000 A1 = 0.00003375 A0 Total: 3 4(ii) H0: Pop mean level = 0.034 H1: Pop mean level ≠ 0.034 B1 not just "mean", but allow just “µ” "030335" 0.034 "0.00003392" 200 − M1 must have 200 "0.00003375" 200 0.0335 0.034 − M1 = –1.21(4) (3 sfs) (–1.22 ↔–1.21) A1 = –1.217 (3 sfs) A1 Comp with z = −1.645 (or 0.1124>0.05) M1 0.112 > 0.05 valid comparison z or areas No evidence that (mean) pollutant level has changed, accept H0 (if correctly defined) A1FT correct conclusion no contradictions SR: One tail test: B0, M1A1 as normal, M1 (comparison with 1.282 consistent signs) A0 Total: 5

This question in 9709/73 May/June 2017

1 A random sample of 75 values of a variable X gave the following results. n = 75 Σ x = 153.2 Σx2 = 340.24 Find unbiased estimates for the population mean and variance of X. [3] … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: Question Answer Marks Guidance 1 est(µ) (= 153.2 ÷ 75) = 2.04 (3 sf) B1 est(σ2) = 7475 ( 340.2475 − "2.04267"2 ) oe M1 = 0.369 (3 sf) A1 Accept 0.368 3

This question in 9709/71 May/June 2018

4 The mean mass of packets of sugar is supposed to be 505 g. A random sample of 10 packets filled by a certain machine was taken and the masses, in grams, were found to be as follows. 500 499 496 495 498 490 492 501 494 494 (i) Find unbiased estimates of the population mean and variance. [3] … … … … … … … … … … The mean mass of packets produced by this machine was found to be less than 505 g, so the machine was adjusted. Following the adjustment, the masses of a random sample of 150 packets from the machine were measured and the total mass was found to be 75 660 g. (ii) Given that the population standard deviation is 3.6 g, test at the 2% significance level whether the machine is still producing packets with mean mass less than 505 g. [5] … … … … … … … … … … … … … … … … … … … … … … … … (iii) Explain why the use of the normal distribution is justified in carrying out the test in part (ii). [1] … … … … … … … … …

9 marks

Mark scheme: 4(i) Est(µ) = 495.9 B1 Accept 496 Est(σ2) = 2 10 2459283 9 10 ( "495.9" ) − M1 Attempt Σx2 and subst in correct formula (1/9(“2459283” – “4959”2/10)). May be implied by correct answer = 12.8 (3 sf) or 383/30 A1 (Note: Biased var “11.49” scores M0 A0) 3 4(ii) H0: µ = 505 H1: µ < 505 75660 505 150 3.6 150 − ÷ B1 Allow ‘Pop mean’ but not just ‘mean’ = –2.04 M1 Correct stand'n; must have √150. No sd/var mixes. Condone sample SD (3.58/3.39) Accept standardisation of totals ((75660-75750)/44.091) Accept CV method A1 Accept +2.04 (Note: if valid area comparison done 0.0207/0.0206 or 0.979 needed for A1) comp z = –2.054 M1 Valid comparison of z’s or area (0.0207/6>0.02; 0.979(3)<0.98) No evidence (at 2%) that machine pkts mean mass < 505 A1ft oe No contradictions. SC Two tail test can score B0 M1 A1 M1 for comparison with 2.326 A0 (max 3/5) 5 Question Answer Marks Guidance 4(iii) Large sample, so sample mean approx normally distr'd B1 Allow just ‘Sample is large’ or ‘n is large’ n>30 1

This question in 9709/72 May/June 2018

2 Amy has to choose a random sample from the 265 students in her year at college. She numbers the students from 1 to 265 and then uses random numbers generated by her calculator. The first two random numbers produced by her calculator are 0.213 165 448 and 0.073 165 196. (i) Use these figures to find the numbers of the first four students in her sample. [2] … … … There were 25 students in Amy’s sample. She asked each of them how much money, $x, they earned in a week, on average. Her results are summarised below. n = 25 Σ x = 510 Σ x2 = 13 225 (ii) Find unbiased estimates of the population mean and variance. [3] … … … … … … … … … … … (iii) Explain briefly what is meant by ‘population’ in this question. [1] … … … …

6 marks

Mark scheme: 2(i) 213, 165, 73, 196 Allow 073 B1 For 3-digit no, < 265, consisting of three consecutive integers from given digits, backwards or forward. (73 or 073 counts as a 3-digit no.) B1 For another three such. Other answers may be valid. If other method used, method must be clear 2 Question Answer Marks Guidance 2(ii) 510 25 = 102 5 or 20.4 B1 2 25 13225 102 24 25 5       −        M1 2 1 510 13225 24 25     −     118 (3 sf) or 2821 24 A1 3 2(iii) (Average) weekly earnings of all students in Amy’s year B1 Not ‘All students in Amy’s year’ 1

This question in 9709/73 May/June 2018

5 The numbers of basketball courts in a random sample of 70 schools in South Mowland are summarised in the table. Number of basketball courts 0 1 2 3 4 >4 Number of schools 2 28 26 10 4 0 (i) Calculate unbiased estimates for the population mean and variance of the number of basketball courts per school in South Mowland. [4] … … … … … … … … … … … The mean number of basketball courts per school in North Mowland is 1.9. (ii) Test at the 5% significance level whether the mean number of basketball courts per school in South Mowland is less than the mean for North Mowland. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … (iii) State, with a reason, which of the errors, Type I or Type II, might have been made in the test in part (ii). [1] … … … … … …

10 marks

Mark scheme: 5(i) ˆµ = 126 70 or 9 5 or 1.8 oe B1 Σx2f = 286 B1 Seen or implied 2 Est( ) σ = 2 2 70 69 70 ( '1.8' ) Σ − x f M1 oe attempted = 0.858 or 296 / 345 A1 Note: Final answer for var 0.846 (biased) and no working implies B1 for 286 4 Question Answer Marks Guidance 5(ii) H0: µ = 1.9 H1: µ < 1.9 B1 Or ‘pop mean’; not just ‘mean’ '0.858' 70 1.8 1.9 − M1 Standardise with their values from (i). Must have sqr 70. No SD / Var mix = –0.903 A1 Accept ± 0.903 < 1.645 M1 comp 1.645 allow comp 1.96 if H1: µ ≠ 1.9 or comp 1 – φ(‘0.903’)=0.182 or 0.183 with 0.05 (or 0.025 if H1: µ ≠ 1.9) No evidence that mean no courts in S is less than in N A1ft No contradictions. ft their 0.903, but not comp 1.96 i.e. no ft for a 2 tail test Accept cv method: cv = 1.718 M1A1 1.718 < 1.8 M1 conclusion A1 (cv centred on 1.8 gives 1.982 M1A1 and M1 for 1.982 > 1.9 A1 conclusion) 5 5(iii) Type II because H0 was not rejected B1ft ft their conclusion, i.e. if H0 rejected, ‘Type I because H0 rejected’ B1 Answer must be consistent with their conclusion. No conclusion in (ii) will score B0 1

This question in 9709/72 Oct/Nov 2018

6 Ramesh plans to carry out a survey in order to find out what adults in his town think about local sports facilities. He chooses a random sample from the adult members of a tennis club and gives each of them a questionnaire. (i) Give a reason why this will not result in Ramesh having a random sample of adults who live in the town. [1] … … … … (ii) Describe briefly a valid method that Ramesh could use to choose a random sample of adults in the town. [2] … … … … … … … … Ramesh now uses a valid method to choose a random sample of 350 adults from the town. He finds that 47 adults think that the local sports facilities are good. (iii) Calculate an approximate 90% confidence interval for the proportion of all adults in the town who think that the local sports facilities are good. [4] … … … … … … … … … … … … … (iv) Ramesh calculates a confidence interval whose width is 1.25 times the width of this 90% confidence interval. Ramesh’s new interval is an x% confidence interval. Find the value of x. [3] … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 6(i) Biased towards people who like tennis Excludes people who don't like tennis B1 1 6(ii) Obtain a list of all people in the town B1 Use random numbers B1 or, e.g. pick numbers from a hat or other sensible 2 6(iii) Var(p) = 47 47 350 350 (1 ) 350 − (= 0.000332152) M1 z = 1.645 B1 47 47 350 350 (1 ) 47 350 350 z − ± M1 Must be a z value 0.104 to 0.164 (3 sf) A1 Must be an interval 4 6(iv) 1.25 × 1.645 (= 2.056) M1 or 1.25 × their width ÷ 2 ÷ their 47 47 350 350 (1 ) 350 − (Complete method) Φ(‘2.056’) (= 0.980) M1 Attempt Φ(their z) x = 96 (2 sf) A1 Allow 0.96 (2 sf) CWO 3

This question in 9709/71 May/June 2019

2 The length of worms is denoted by X cm. The lengths of a random sample of 50 worms were measured. Some of the results were lost, but the following results are available. ³ Σx2 = 4361 ³ An unbiased estimate of the population variance of X is 9.62. Calculate the mean length of the 50 worms. [3] … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: 2 2 49 50 ( ) x − = 9.62 M1 or 2 ( ) 4361 49 50 49 ( ) Σ × − x = 9.62 BOD regarding symbols used 2 x = 4361 50 - 9.62× 49 50 = 77.7924 A1 (Σx)2 = 4361 × 50 – 9.62 × 50 × 49 = 194481or Σx =441 ( ) Σx or ( ) x must be correctly identified x = 8.82 (3 sf) A1 SC use of ‘biased’ leading to 8.81 B1 3

This question in 9709/73 May/June 2019