Sampling and estimation — Paper 6 · A Level Mathematics 9709

2 Lengths of a certain species of lizard are known to be normally distributed with standard deviation 3.2 cm. A naturalist measures the lengths of a random sample of 100 lizards of this species and obtains an !% confidence interval for the population mean. He finds that the total width of this interval is 1.25 cm. Find !. [5] … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2 2 × z × 3.2 10 = 1.25 z = 1.953 A1 SOI ɸ(‘their 1.953’) (= 0.9746) M1 = 1 – 2(1 – ‘0.9746’) = 0.9492 M1 OE α = 94.9 or 95 A1 CWO 5

This question in 9709/62 Feb/March 2020

3 In the past, the mean time taken by Freda for a particular daily journey was 39.2 minutes. Following the introduction of a one-way system, Freda wishes to test whether the mean time for the journey has decreased. She notes the times, t minutes, for 40 randomly chosen journeys and summarises the results as follows. n = 40 Σt = 1504 Σt2 = 57 760 (a) Calculate unbiased estimates of the population mean and variance of the new journey time. [3] … … … … … (b) Test, at the 5% significance level, whether the population mean time has decreased. [5] … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 3(a) est (μ) = 37.6 or 1504 40 or 188 5 B1 est (σ2) = 2 40 57760 39 40 37.6   −  = 31.0154 = 2016 65 M1 Correct substitution in any correct formula 2 1504 1 39 40 57760   −   = 31.(0) (3 sf) A1 Accept 2016 65 or 1 65 31 3 3(b) H0: Pop mean (or μ) = 39.2 H1: Pop mean (or μ) < 39.2 B1 Both. Not just ‘mean’ 40 ' 0154 . 31 ' 2. 39 '6. 37 ' − M1 Allow use of biased variance (30.2), must have √40 = –1.817 A1 SC FT use of biased = –1.840 for A1 ‘1.817’ > 1.645 OE M1 Valid comparison ‘their 1.817’ with 1.645 or valid area comparison 0.0346 < 0.05 OE There is evidence that mean time has decreased A1FT FT their 1.817; in context, not definite, no contradictions SC For 2 tail test: H1: μ ≠ 39.2 and comp 1.96, max B0M1A1M1A0 (no FT for final mark) 5

This question in 9709/62 Feb/March 2020

2 In a survey, a random sample of 250 adults in Fromleigh were asked to fill in a questionnaire about their travel. (a) It was found that 102 adults in the sample travel by bus. Find an approximate 90% confidence interval for the proportion of all the adults in Fromleigh who travel by bus. [3] … … … … … … … … … … … … … … … … … … … … … … … (b) The survey included a question about the amount, x dollars, spent on travel per year. The results are summarised as follows. n = 250 Σx = 50 460 Σx2 = 19 854 200 Find unbiased estimates of the population mean and variance of the amount spent per year on travel. [3] … … … … … … … … … … A councillor wanted to select a random sample of houses in Fromleigh. He planned to select the first house on each of the 143 streets in Fromleigh. (c) Explain why this would not provide a random sample. [1] … … … … … … … … … …

7 marks

Mark scheme: 2(a) 102 250 102 250 250 250 − × (= 0.000966144) 102 '0.00096614' 250 z ± M1 Any z but must be a z value. One side of the interval scores M1. z = 1.645 B1 Confident Interval is 0.357 to 0.459 (3 sf) A1 Must be an interval. 3 2(b) Estimate of mean 50460 250       = $201.84 B1 Allow without units. Allow 3s.f. $202. 2 250 19854200 50460 249 250 250     −           or 2 1 50460 19854200 249 250   −     M1 Estimate of variance = 38 832.75 dollars2 or 38 800 (3 sf) A1 Allow with missing units. (Calculation of biased gives 38 700 scores M0A0) 3 2(c) e.g. Every house doesn’t have an equal chance of being selected or most houses have no chance of being selected. B1 Or other similar e.g. Houses in streets with few houses are more likely to be selected. Not just ‘biased’, OE, without explanation 1

This question in 9709/61 Oct/Nov 2020

2 In a survey, a random sample of 250 adults in Fromleigh were asked to fill in a questionnaire about their travel. (a) It was found that 102 adults in the sample travel by bus. Find an approximate 90% confidence interval for the proportion of all the adults in Fromleigh who travel by bus. [3] … … … … … … … … … … … … … … … … … … … … … … … (b) The survey included a question about the amount, x dollars, spent on travel per year. The results are summarised as follows. n = 250 Σx = 50 460 Σx2 = 19 854 200 Find unbiased estimates of the population mean and variance of the amount spent per year on travel. [3] … … … … … … … … … … A councillor wanted to select a random sample of houses in Fromleigh. He planned to select the first house on each of the 143 streets in Fromleigh. (c) Explain why this would not provide a random sample. [1] … … … … … … … … … …

7 marks

Mark scheme: 2(a) 102 250 102 250 250 250 − × (= 0.000966144) 102 '0.00096614' 250 z ± M1 Any z but must be a z value. One side of the interval scores M1. z = 1.645 B1 Confident Interval is 0.357 to 0.459 (3 sf) A1 Must be an interval. 3 2(b) Estimate of mean 50460 250       = $201.84 B1 Allow without units. Allow 3s.f. $202. 2 250 19854200 50460 249 250 250     −           or 2 1 50460 19854200 249 250   −     M1 Estimate of variance = 38 832.75 dollars2 or 38 800 (3 sf) A1 Allow with missing units. (Calculation of biased gives 38 700 scores M0A0) 3 2(c) e.g. Every house doesn’t have an equal chance of being selected or most houses have no chance of being selected. B1 Or other similar e.g. Houses in streets with few houses are more likely to be selected. Not just ‘biased’, OE, without explanation 1

This question in 9709/63 Oct/Nov 2020

1 A construction company notes the time, t days, that it takes to build each house of a certain design. The results for a random sample of 60 such houses are summarised as follows. Σ t = 4820 Σ t2 = 392 050 (a) Calculate a 98% confidence interval for the population mean time. [6] … … … … … … … … … … … … … … … … (b) Explain why it was necessary to use the Central Limit theorem in part (a). [1] … … … … …

7 marks

Mark scheme: 1(a) Est(μ) = 60 4820 or 241 3 or 80.3 (3 sf) B1 Est(σ2) = ) 2 ) 60 4820 ( 60 050 392 ( 59 60 − M1 Use of biased (80.72) score M0 A0. 82.0904 14530 177       to 82.635 or SD = 9.0604 to 9.0904 (3sf) A1 z = 2.326 B1 60 4820 ± z× 60 ' 0904 . 82 ' M1 Expression of the correct form – must be z value. 77.6 to 83.1 (3 sf) A1 CWO Use of biased 77.6 to 83.0(3) can score B1M1A1 (max 4/6). 6 1(b) Population distribution of times unknown B1 Accept ‘not normal’. 1

This question in 9709/62 Feb/March 2021

4 100 randomly chosen adults each throw a ball once. The length, l metres, of each throw is recorded. The results are summarised below. n = 100 Σl = 3820 Σl2 = 182 200 Calculate a 94% confidence interval for the population mean length of throws by adults. [6] … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 4 3820 100 [= 38.2] B1 2 100 182200 '38.2' 99 100   −     or 2 1 3820 182200 99 100   −     M1 Use of biased (362.76) scores M0 = 12092 33 or 366.424 or 366 (3 sf) A1 Accept SD=19.1422 or 19.1(3sf) ‘38.2’ ± z× '366.424' 100 M1 Expression of the correct form must be a z-value. z = 1.881 or 1.882 B1 Seen. 34.6 to 41.8 (3 sf) A1 Allow use of biased giving (34.6,41.8) Must be an interval. 6

This question in 9709/61 May/June 2021

3 A random sample of 75 students at a large college was selected for a survey. 15 of these students said that they owned a car. From this result an approximate !% confidence interval for the proportion of all students at the college who own a car was calculated. The width of this interval was found to be 0.162. Calculate the value of ! correct to 2 significant figures. [5] … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3 est(p) = 0.2 accept 15 75 2 × z × 0.2 0.8 75 × = 0.162 M1 Expression of the correct form. Condone missing 2x. z 75 0.081 0.2 0.8   = ×   ×   = 1.754 A1 Correct z. Condone 3sf accuracy. Φ(‘1.754’) = 0.96[03] '0.96' – (1 – '0.96') M1 OE. Using their z to find alpha. α = 92 A1 Following correct working. 5

This question in 9709/61 Oct/Nov 2021

1 The mass, in kilograms, of a block of cheese sold in a supermarket is denoted by the random variable M. The masses of a random sample of 40 blocks are summarised as follows. n = 40 Σm = 20.50 Σm2 = 10.7280 (a) Calculate unbiased estimates of the population mean and variance of M. [3] … … … … … … … … … … (b) The price, $P, of a block of cheese of mass M kg is found using the formula P = 11M + 0.50. Find estimates of the population mean and variance of P. [3] … … … … … … … … … …

6 marks

Mark scheme: 1(a) 20.5 40 = 0.5125 80 . Condone 20.5 40 . ( ) 2 40 10.728 '0.5125' 39 40   −     or 2 1 20.50 10.728 39 40   −     M1 Biased variance (0.005544 or 887 160 000 ) scores M0 A0. 0.0056859 or 0.00569 (3 sf) or 887 156 000 A1 CAO 3 1(b) [11 × ‘0.5125’ + 0.5]) = 6.1375 or 491 80 or 6.14 (3sf) B1 FT FT their 0.5125 112 × ‘0.0056859’ M1 With nothing added. Using their variance in (a) (no sd/var confusion) 0.688 (3sf) A1 CAO 3

This question in 9709/62 Oct/Nov 2021

3 The probability that a certain spinner lands on red on any spin is p. The spinner is spun 140 times and it lands on red 35 times. (a) Find an approximate 96% confidence interval for p. [3] … … … … … … … … … … … … From three further experiments, Jack finds a 90% confidence interval, a 95% confidence interval and a 99% confidence interval for p. (b) Find the probability that exactly two of these confidence intervals contain the true value of p. [3] … … … … … … … …

6 marks

Mark scheme: 3(a) 0.25 ± z 0.25 0.75 140 × M1 Expression of correct form (allow M1 for just one side stated). Must be a z-value. z = 2.054 or 2.055 B1 0.175 to 0.325 (3sf) A1 Must be an interval. 3 3(b) 0.90 × 0.95 × 0.01 + 0.90 × 0.05 × 0.99 + 0.10 × 0.95 × 0.99 M1 M1 M1 for one correct triple product. M1 for all correct and added. 0.147 A1 SC If zero scored award B1 for a 2 or 3 term expression of the form 0.90 × 0.95 [×c] OE. (0 < c ⩽ 1) 3

This question in 9709/62 Oct/Nov 2021

3 A random sample of 500 households in a certain town was chosen. Using this sample, a confidence interval for the proportion, p, of all households in that town that owned two or more cars was found to be 0.355 < p < 0.445. Find the confidence level of this confidence interval. Give your answer correct to the nearest integer. [5] … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3 est(p) = 0.4 B1 ( ) [ ] '0.4' 1 '0.4' '0.4' 0.445 500 z × − + = M1 OE Use of their 0.4 in a correct expression z ( ) '0.4' 1 '0.4' 0.045 500   × − = ÷       = 2.054 A1 Condone 2.053 and 2.05 0.98 – (1 – 0.98) M1 96% confidence A1 CWO, must be integer 5

This question in 9709/62 Feb/March 2022

1 (a) A javelin thrower noted the lengths of a random sample of 50 of her throws. The sample mean was 72.3m and an unbiased estimate of the population variance was 64.3m2. Find a 92% confidence interval for the population mean length of throws by this athlete. [3] … … … … … … … … … … … … … … … (b) A discus thrower wishes to calculate a 92% confidence interval for the population mean length of his throws. He bases his calculation on his first 50 throws in a week. Comment on this method. [1] … … … … … …

4 marks

Mark scheme: 1(a) 72.3 ± z 64.3 50 M1 Expression of correct form (allow only one side for M1). Must be a z value z = 1.751 B1 Accept 1.75 if nothing better seen CI is 70.3 to 74.3 metres (3 s.f.) A1 Allow without units Must be an interval 3 1(b) Not random sample B1 Need ‘random’ or ‘not representative/biased because…’ OE 1

This question in 9709/62 May/June 2022

3 Batteries of type A are known to have a mean life of 150 hours. It is required to test whether a new type of battery, type B, has a shorter mean life than type A batteries. (a) Give a reason for using a sample rather than the whole population in carrying out this test. [1] … … … … … A random sample of 120 type B batteries are tested and it is found that their mean life is 147 hours, and an unbiased estimate of the population variance is 225 hours2. (b) Test, at the 2% significance level, whether type B batteries have a shorter mean life than type A batteries. [5] … … … … … … … … … … … … … … … (c) Calculate a 94% confidence interval for the population mean life of type B batteries. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 3(a) Batteries unusable after testing or Population too big or too costly or too time consuming to use the whole population oe B1 1 3(b) H0: μ = 150 H1: μ < 150 B1 Or population mean = 150; not just ‘mean’ = 150 147 150 225 120   M1 Allow with continuity correction Need 120 –2.191 A1 Condone – 2.19 –2.191 < –2.054 [or –2.055] M1 OE. For valid comparison with 2.054 or 2.055 Or 0.0143 (or 0.0142) < 0.02 For two tail test allow comp –2.326 OE if H1: μ ≠ 150 (can score B0M1A1M1A0 max 3/5 ) [Reject Ho] There is evidence that the (mean) life of type B is less than type A (or less than 150) A1 FT In context, not definite with no contradictions Accept critical value method 147.19 M1A1 147 < 147.19 M1 conclusion A1 Or 150 > 149.81 5 Question Answer Marks Guidance 3(c) 147  z × 15 120 M1 Expression of correct form must be a z value z = 1.881 [or 1.882] B1 144 to 150 (3 s.f.) A1 Must be an interval Incorrect z value can only score M1B0A0 3

This question in 9709/63 May/June 2022

5 A builders’ merchant sells stones of different sizes. (a) The masses of size A stones have standard deviation 6 grams. The mean mass of a random sample of 200 size A stones is 45 grams. Find a 95% confidence interval for the population mean mass of size A stones. [3] … … … … … … … … (b) The masses of size B stones have standard deviation 11 grams. Using a random sample of size 200, an !% confidence interval for the population mean mass is found to have width 4 grams. Find !. [4] … … … … … … … … … … … …

7 marks

Mark scheme: 5(a) 6 M1 For expression of correct form, any z. 45 z Accept one side of interval for M1. 200 z = 1.96 B1 Must be seen. 44.2 to 45.8 (3 sf) A1 Must be an interval. 3 5(b) 11 M1 Or … = 4 for M1 z = 2 200 z = 2.571 A1 Accept 3sf if nothing better seen. ɸ(their '2.571') = 0.9949 M1 OE For area consistent with their values. and their '0.9949' – (1 – their '0.9949') [= 0.9898] Must be seen. α = 99.0 (3 sf) A1 Allow 99. cwo Final answer of 0.99 scores A0. 4

This question in 9709/61 Oct/Nov 2022

1 Each of a random sample of 80 adults gave an estimate, h metres, of the height of a particular building. The results were summarised as follows. n = 80 Σh = 2048 Σh2 = 52 760 (a) Calculate unbiased estimates of the population mean and variance. [3] … … … … … … … … (b) Using this sample, the upper boundary of an !% confidence interval for the population mean is 26.0. Find the value of !. [4] … … … … … … … … … … … …

7 marks

Mark scheme: Question Answer Marks Guidance 1(a) 2048 128 B1 Est μ = 25.6 or or 80 5 80  52760  2048  2  1  20482  M1 Substitution into a correct formula. Est σ2 =  −    or  52760 −  Biased 4.14 scores M0.  79  80  80  79  80    1656 A1 = 4.19 (3 sf) or 395 3 1(b) '4.19' M1 Use of correct equation with their values. ‘25.6’ + z = 26.0 80 z = 1.748 or 1.747 A1 Accept 3sf. FT Biased z = 1.758. (ɸ(‘1.748’) = 0.960) ‘0.960’ – (1 – ‘0.960’) M1 Correct area using their values. α = 92.0 or 91.9 A1 Allow 92 . FT Biased 92.1. A final answer of 0.92 or 0.919 scores A0. 4

This question in 9709/62 Oct/Nov 2022

5 X is a random variable with distribution B 10, 0.2 . A random sample of 160 values of X is taken. (a) Find the approximate distribution of the sample mean, including the values of the parameters. [3] … … … … … … (b) Hence find the probability that the sample mean is less than 1.8. [3] … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 5(a) For X, μ = 2 σ2 = 1.6 Mean = 2 B1 1.6 1 B1 Accept Var = 0.12 (accept sd=0.1 if clearly identified). Variance = or or 0.01 160 100 Normal B1 3 5(b) 1.8 − 3201 − '2' 1.8 − '2' M1 Allow with wrong continuity correction.  or  [= –2.03 or –2] M1 can be implied by correct final answer or for '0.01' '0.01' –2.03 / –2.0 or 0.9788 / 0.9772 seen. or ± (287.5 – ’320’) / '256' or ± (288 – ‘320’) / '256' [= –2.03 or –2] ɸ(‘–2.03’) = 1 – ɸ(‘2.03’) M1 Correct area consistent with their values. M1 can be implied by correct final answer. = 0.0212 or 0.0228 (3 sf) A1 3

This question in 9709/62 Oct/Nov 2022

5 A builders’ merchant sells stones of different sizes. (a) The masses of size A stones have standard deviation 6 grams. The mean mass of a random sample of 200 size A stones is 45 grams. Find a 95% confidence interval for the population mean mass of size A stones. [3] … … … … … … … … (b) The masses of size B stones have standard deviation 11 grams. Using a random sample of size 200, an !% confidence interval for the population mean mass is found to have width 4 grams. Find !. [4] … … … … … … … … … … … …

7 marks

Mark scheme: 5(a) 6 M1 For expression of correct form, any z. 45 z Accept one side of interval for M1. 200 z = 1.96 B1 Must be seen. 44.2 to 45.8 (3 sf) A1 Must be an interval. 3 5(b) 11 M1 Or … = 4 for M1 z = 2 200 z = 2.571 A1 Accept 3sf if nothing better seen. ɸ(their '2.571') = 0.9949 M1 OE For area consistent with their values. and their '0.9949' – (1 – their '0.9949') [= 0.9898] Must be seen. α = 99.0 (3 sf) A1 Allow 99. cwo Final answer of 0.99 scores A0. 4

This question in 9709/63 Oct/Nov 2022

1 In a survey of 200 randomly chosen students from a certain college, 23% of the students said that they owned a car. Calculate an approximate 93% confidence interval for the proportion of students from the college who own a car. [3] … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: 1 0.23 ± z × 0.23 (1 0.23) 200  M1 Expression of correct form. Any z, but z = 0.8328 scores B0M0. z = 1.811 or 1.812 B1 0.176 to 0.284 (3 sf) A1 Must be an interval. 3

This question in 9709/62 May/June 2023

3 The masses, in kilograms, of newborn babies in country A are represented by the random variable X, with mean - and variance 32. The masses of a random sample of 500 newborn babies in this country were found and the results are summarised below. n = 500 Σx = 1625 Σx2 = 5663.5 (a) Calculate unbiased estimates of - and 32. [3] … … … … … … … … … … … … … … … … … … … … … … A researcher wishes to test whether the mean mass of newborn babies in a neighbouring country, B, is different from that in country A. He chooses a random sample of 60 newborn babies in country B and finds that their sample mean mass is 2.95kg. Assume that your unbiased estimates in part (a) are the correct values for - and 32. Assume also that the variance of the masses of newborn babies in country B is the same as in country A. (b) Carry out the test at the 1% significance level. [5] … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 3(a) Est (μ) = 3.25 = 13/4 or 1625/500 B1 Est(σ2) = 2 500 5663.5 ( "3.25" ) 499 500  or 2 1 1625 5663.5 499 500        M1 Expression of correct form. = 0.766 (3 sf) or 1529/1996 A1 Biased variance of 0.7645 scores M0A0. 3 Question Answer Marks Guidance 3(b) H0: Pop mean (or μ) = ‘3.25’ H1: Pop mean (or μ) ≠ ‘3.25’ B1FT Not just ‘mean’. FT their 3.25 . 2.95 "3.25" "0.766" 60   M1 Standardising with their values. Must have √60. = –2.655 A1 Or P(𝑋ത < 2.95) = 0.0039 or 0.00396 or 0.00397 . SC FT their biased est(σ2), i.e. 0.7645 to give z = 2.658 A1. ‘2.655’ > 2.576 or ‘–2.655’ < –2.576 M1 For valid comparison, e.g. 0.0039 or 0.00396 or 0.00397 < 0.005, or 0.0078 < 0.01, or 0.00792 < 0.01 . [Reject H0] There is evidence that (mean) mass in (country B) is different (from country A). A1FT OE. Must be in context and not definite, e.g., not ‘Mean mass is not different’, No contradictions. Context needs either ‘mass’ or ‘countries’ OE. SC, Use of one-tail test. ‘2.655’ > 2.326 or 0.0039 < 0.01 M1A0 (Max B0M1A1M1A0 3/5). Accept critical value method. Either: Xcrit=2.959 M1A1 2.95<2.959 M1A1FT with correct conclusion, or Xcrit=3.241 M1A1 3.25>3,241 M1A1FT with correct conclusion. 5

This question in 9709/62 May/June 2023

1 A random variable X has the distribution N 410, 400 . Find the probability that the mean of a random sample of 36 values of X is less than 405. [3] … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: Question Answer Marks Guidance 1 405 − 410 M1 [= –1.5] For standardising, must have 36. 20 14580 − 6 Allow totals method 14760. 14400 No mixed methods. ɸ('–1.5') = 1 – ɸ('1.5') M1 For area consistent with their working. = 0.0668 A1 3

This question in 9709/61 Oct/Nov 2023

2 In a survey of 300 randomly chosen adults in Rickton, 134 said that they exercised regularly. This information was used to calculate an % confidence interval for the proportion of adults in Rickton who exercise regularly. The upper bound of the confidence interval was found to be 0.487, correct to 3 significant figures. Find the value of correct to the nearest integer. [4] … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 2 M1 For expression of the correct form. 134  166 134 300 300 + z = 0.487 300 300 z = 1.405 A1 Accept 1.404, or anything that rounds to 1.39 to 1.41. ɸ–1('1.405') = 0.9199 or 0.92; 1 − 2(1 – 0.92) M1 Attempt area above or below their 1.405 and convert to a confidence level. α = 84 A1 Allow α = 84%. cwo Note: final answer 0.84 scores A0. 4

This question in 9709/61 Oct/Nov 2023

2 The length, in minutes, of mathematics lectures at a certain college has mean - and standard deviation 8.3. (a) The total length of a random sample of 85 lectures was 4590 minutes. Calculate a 95% confidence interval for -. [3] … … … … … … … … … … The length, in minutes, of history lectures at the college has mean m and standard deviation s. (b) Using a random sample of 100 history lectures, a 95% confidence interval for m was found to have width 2.8 minutes. Find the value of s. [2] … … … … … … … … …

5 marks

Mark scheme: 2(a) 4590 8.3 M1 For expression of correct form. Any z (but not ϕ(z)). z 85 85 z = 1.96 B1 52.2 to 55.8 (3 sf) A1 Must be an interval. 3 Question Answer Marks Guidance 2(b) s s M1 Equation of correct form (any z). 1.96  = 1.4 or 2  1.96  = 2.8 Allow factor of 2 error (i.e. first equation = 2.8). 100 100 50 A1 s = 7.14 (3 sf) or 7 2 Question Answer Marks Guidance

This question in 9709/62 Oct/Nov 2023

2 In a survey of 300 randomly chosen adults in Rickton, 134 said that they exercised regularly. This information was used to calculate an !% confidence interval for the proportion of adults in Rickton who exercise regularly. The upper bound of the confidence interval was found to be 0.487, correct to 3 significant figures. Find the value of ! correct to the nearest integer. [4] … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 2 M1 For expression of the correct form. 134  166 134 300 300 + z = 0.487 300 300 z = 1.405 A1 Accept 1.404, or anything that rounds to 1.39 to 1.41. ɸ–1('1.405') = 0.9199 or 0.92; 1 − 2(1 – 0.92) M1 Attempt area above or below their 1.405 and convert to a confidence level. α = 84 A1 Allow α = 84%. cwo Note: final answer 0.84 scores A0. 4

This question in 9709/63 Oct/Nov 2023

2 A random sample of 250 people living in Barapet was chosen. It was found that 78 of these people owned a BETEC phone. (a) Calculate an approximate 98% confidence interval for the proportion of people living in Barapet who own a BETEC phone. [3] … … … … … … … … … … … … … … … … (b) Manjit claims that more than 40% of the people living in Barapet own a BETEC phone. Use your answer to part (a) to comment on this claim. [1] … … … … … … …

4 marks

Mark scheme: 2(a) 78 78 M1 Use of a correct formula (any z). −(1 ) 78 250 250 ± z × 250 250 z = 2.326 B1 = 0.244 to 0.38[0] (3 sf) A1 Must be an interval. 3 2(b) Unlikely to be true because confidence interval does not contain 0.4. B1 ft FT their confidence interval. Must include this reason and ‘unlikely’, oe. Allow “not true because 0.4 is not in the confidence interval.” But “Confidence interval only goes up to 0.38 so not true” and “‘it’ lies outside the confidence interval” both score B0. 1

This question in 9709/62 Feb/March 2024

4 (a) A random sample of 8 boxes of cereal from a certain supplier was taken. Each box was weighed and the masses in grams were as follows. 261 249 259 252 255 256 258 254 Find unbiased estimates of the population mean and variance. [3] … … … … … … … … … … … … … … … … … … … … … … … … … (b) The supplier claims that the mean mass of boxes of cereal is 253 g. A quality control officer suspects that the mean mass is actually more than 253 g. In order to test this claim, he weighs a random sample of 100 boxes of cereal and finds that the total mass is 25 360 g. (i) Given that the population standard deviation of the masses is 3.5 g, test at the 5% significance level whether the population mean mass is more than 253 g. [5] … … … … … … … … … … … … … … … … … … … An employee says, ‘This test is invalid because it uses the normal distribution, but we do not know whether the masses of the boxes are normally distributed.’ (ii) Explain briefly whether this statement is true or not. [1] … … …

9 marks

Mark scheme: 4(a) Est(μ) = 2044 8 [=255.5] B1 Accept 3sf if nothing better seen. Est(σ2) = 2 8 522348 "255.5" 7 8        or 1 7 (‘522348’ – 2 ‘2044’ 8 ) M1 Attempt to find Σx2 and substitute in correct formula. May be implied by correct answer. Biased 13.25 scores M0. = 15.1 (3 sf) or 106 7 A1 OE 3 4(b)(i) H0: μ = 253 H1: μ > 253 B1 Allow ‘Population mean’ but not just ‘mean’. 25360 100 253 3.5 100   M1 Standardising must have 100. = 1.714 A1 1.714 > 1.645 or 0.0432 < 0.05 M1 OE [Reject H0 ] There is sufficient evidence (at 5% level) to suggest [mean] mass is greater than 253 A1FT OE FT their ‘1.714’ in context, not definite, no contradictions. Accept critical value method of 253.57 < 253.60 or 253.02 > 253. Use of a two-tailed test scores B0 M1 A1 M1 A0 (comp with 0.025 1.96 ). 5 Question Answer Marks Guidance 4(b)(ii) Not true. Large sample, [so sample mean is approx normally distributed]. B1 OE Allow ‘Not true. Large sample’ or ‘Not true. n is large’ or ‘Not true. CLT used’. 1

This question in 9709/61 May/June 2024

2 The widths, w cm, of a random sample of 150 leaves of a certain kind were measured. The sample mean of w was found to be 3.12 cm. Using this sample, an approximate 95% confidence interval for the population mean of the widths in centimetres was found to be [3.01, 3.23]. (a) Calculate an estimate of the population standard deviation. [3] … … … … … … … … … … … … … … … … … … … … (b) Explain whether it was necessary to use the Central Limit theorem in your answer to part (a). [1] … … …

4 marks

Mark scheme: 2(a) 3.12 + z × 150  = 3.23 Any z, but must be a z value. z = 1.96 B1 σ = 0.687 (3sf) [cm] A1 3 2(b) Yes, because population [of widths] not given to be normally distributed B1 Or ‘underlying distribution’ instead of population. Allow ‘yes, because population distribution not known’. Need both statements. 1

This question in 9709/63 May/June 2024

6 A manufacturer of cell phones claims that 25% of students own a Pumpkin phone. Jeyeraj thinks that the proportion of students at his large college who own a Pumpkin phone is less than 25%. He plans to test the manufacturer’s claim. He chooses a random sample of 30 students at his college. If the number of students who own a Pumpkin phone is less than 5, Jeyeraj will reject the manufacturer’s claim. (a) State suitable hypotheses for the test. [1] … … … … … (b) Given that the true proportion of students at the college who own a Pumpkin phone is 10%, use a binomial distribution to find the probability of a Type II error. [3] … … … … … … … … … … … … … … … … … … At Florence’s college, in a random sample of 40 students, it was found that 5 own a Pumpkin phone. (c) Calculate an approximate 95% confidence interval for the proportion of students at Florence’s college who own a Pumpkin phone. [3] … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 6(a) H0: Proportion (at college) owning Pumpkin phone = 0.25 B1 Allow p = 0.25, p < 0.25. Allow 25%. H1: Proportion (at college) owning Pumpkin phone < 0.25 1 6(b) B(30, 0.1) and P(X ⩾ 5) attempted M1 May be implied. 1− (0.930 + 30×0.929×0.1 + 30C2×0.928×0.12 + 30C3×0.927×0.13 + 30C4×0.926×0.14) M1 For expression or terms. No end errors. = 1 – (0.042391 + 0.141304 + 0.22766 + 0.236088 + 0.177066) = 0.175 (3 sf) accept 0.176 A1 SC Unsupported working and correct answer scores M1B1. 3 6(c) M1 Any z must be a z. 1 7 × Only one side calculated can score M1. 1 8 8 ± z 8 40

This question in 9709/61 May/June 2025

3 A machine dispenses coffee into cups. The volume, V cm 3, of coffee in a cup was measured for a random sample of 150 cups. The results were summarised as follows. / v = 46 350 / v2 = 14 410 800 (a) (i) Calculate unbiased estimates of the population mean, n, and population variance, v 2. [3] … … … … … … … (ii) Calculate a 95% confidence interval for n. [3] … … … … … … … Another random sample of n cups of coffee is taken, where 100 1 n 1 120 . A 95% confidence interval for n is calculated using this sample. You may assume that, for large samples, unbiased estimates of v 2 are very similar. (b) Without calculation, state whether this confidence interval would be wider or narrower than the confidence interval found in part (a)(ii). Give a reason for your answer. [2] … … … … … …

8 marks

Mark scheme: 3(a)(i) 46350 B1 Est(μ) = or 309 150  14410800  46350  2   150  M1 Use of correct formula. 2 150 2  Est( ) =  −     =  591 1 46350 149  150  150    149  ( 14410800 – ). 149 150 OE. = 595 (3 sf) or 88650/149 A1 3 3(a)(ii) z = 1.96 B1 Seen. '595' M1 Any z. 309  z  Use of correct formula. 150 = 305 to 313 (3 sf) A1 Use of biased (591) gives same interval but scores B1M1A0. Unsupported correct answer scores max B1B1. 3 3(b) Either Smaller sample means more uncertainty about the value of µ. OE. 1 1 B1* OE. or n < 120 means that  n 150 Therefore Wider B1dep Dep on previous B1. 2

This question in 9709/63 May/June 2025

4 The masses of a certain species of animal are known to be normally distributed with standard deviation v kg. A researcher obtains the masses of a random sample of n animals of this species and uses these masses to find two confidence intervals (a% and 90%) for the population mean. The width of the a% confidence interval is .1414 # the width of the 90% confidence interval. (a) Find a. [3] … … … … … … … … … … … … (b) Find the probability that the 90% confidence interval contains the population mean given that the a% confidence interval contains the population mean. [1] … … … … … … … … … … …

4 marks

Mark scheme: 4(a)   M1 Form an equation in z (accept z = 1.414 × 1.645 [2 ×] z × = 1.414 × [2 ×] 1.645 × [ z = 2.326 ] n n OE). Factor of 1.414 on the wrong side of the equation can still score M1. Condone 1.282 instead of 1.645 for M1. 2Φ(‘2.326’) –1 M1 OE. α = 98 (3 sf) A1 Allow 98%. 3 4(b) 90 45 B1FT FT their α. = or 0.918 (3 sf) '98' 49 no ft if alpha < 90 in 4(a). 1

This question in 9709/61 Oct/Nov 2025

4 The masses of a certain species of animal are known to be normally distributed with standard deviation v kg. A researcher obtains the masses of a random sample of n animals of this species and uses these masses to find two confidence intervals (a% and 90%) for the population mean. The width of the a% confidence interval is .1414 # the width of the 90% confidence interval. (a) Find a. [3] … … … … … … … … … … … … (b) Find the probability that the 90% confidence interval contains the population mean given that the a% confidence interval contains the population mean. [1] … … … … … … … … … … …

4 marks

Mark scheme: 4(a)   M1 Form an equation in z (accept z = 1.414 × 1.645 [2 ×] z × = 1.414 × [2 ×] 1.645 × [ z = 2.326 ] n n OE). Factor of 1.414 on the wrong side of the equation can still score M1. Condone 1.282 instead of 1.645 for M1. 2Φ(‘2.326’) –1 M1 OE. α = 98 (3 sf) A1 Allow 98%. 3 4(b) 90 45 B1FT FT their α. = or 0.918 (3 sf) '98' 49 no ft if alpha < 90 in 4(a). 1

This question in 9709/63 Oct/Nov 2025