Sampling and estimation — Paper 7 · A Level Mathematics 9709

2 The weights in grams of oranges grown in a certain area are normally distributed with mean µ and standard deviation σ. A random sample of 50 of these oranges was taken, and a 97% confidence interval for µ based on this sample was (222.1, 232.1). (i) Calculate unbiased estimates of µ and σ2. [4] (ii) Estimate the sample size that would be required in order for a 97% confidence interval for µ to have width 8. [3]

7 marks

Mark scheme: 2 (i) µˆ = 227.(1) B1 Correct mean B1 2.17 seen σˆ 2 5 = 2.17 × M1 Solving an equation with 5 or 10 on the LHS 50 σˆ and some z value × on the RHS n ˆσ 2 = 265 or 266 A1 Correct answer [4] 163. (ii) 4 = 2.17 × B1ft Correct equation ft their wrong z if the same n as in part (i) and their σ M1 Solving an equation with their z and σ, and width 4 or 8 n = 78 A1 Correct answer (whole number) [3]

This question in 9709/71 May/June 2009

2 A random sample of n people were questioned about their internet use. 87 of them had a high-speed internet connection. A confidence interval for the population proportion having a high-speed internet connection is 0.1129 < p < 0.1771. (i) Write down the mid-point of this confidence interval and hence find the value of n. [3] (ii) This interval is an α% confidence interval. Find α. [4]

7 marks

Mark scheme: 2 (i) 0.145 B1 correct mid-point = 87 / n M1 equating their mid-point with 87 / n n = 600 A1 correct answer [3] .0 1451( − .0 145) (ii) 0.0321 = z × B1 0.0321 seen or implied 600 pq M1 Equating half-width with z × n z = 2.233 Φ(z) = 0.9872 M1 Correct method to find width of CI width of CI is 1 – 2 × (1 – 0.9872) A1 Correct answer α = 97.4% [4] 2 55 − 2 62

This question in 9709/72 May/June 2010

3 The weight, in grams, of a certain type of apple is modelled by the random variable X with mean 62 and standard deviation 8.2. A random sample of 50 apples is selected, and the mean weight in grams, X, is found. (i) Describe fully the distribution of X. [3] (ii) Find P(X > 64). [3]

6 marks

Mark scheme: 3 (i) (Approx) normal B1 mean 62 B1 2.8 2.8 2 sd = = 1.16 (3 sfs) B1 or var = = 1.34 (3 sfs) 50 50 [3] 64 − 62 (ii) (= 1.725 or 1.724) M1 For standardising ÷ 50 essential (no CC) .1" 16" 1 – Φ(“1.725”) M1 For correct area consistent with their mean = (1 – 0.9577) = 0.0423 (3 sfs) A1 [3] GCE AS/A LEVEL – May/June 2010 9709 73

This question in 9709/73 May/June 2010

1 In a survey of 1000 randomly chosen adults, 605 said that they used email. Calculate a 90% confidence interval for the proportion of adults in the whole population who use email. [3]

3 marks

Mark scheme: 1 0.605 ± z× .0605×10001(− .0 605 ) M1 z = 1.645 seen B1 Allow [0.58, 0.63]. [0.580, 0.630] A1 [3] Allow any brackets 10 ( 10 ) 4

This question in 9709/71 Oct/Nov 2010

1 In a survey of 1000 randomly chosen adults, 605 said that they used email. Calculate a 90% confidence interval for the proportion of adults in the whole population who use email. [3]

3 marks

Mark scheme: 1 0.605 ± z× .0605×10001(− .0 605 ) M1 z = 1.645 seen B1 Allow [0.58, 0.63]. [0.580, 0.630] A1 [3] Allow any brackets 10 ( 10 ) 4

This question in 9709/72 Oct/Nov 2010

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 9709/72 Oct/Nov 2010

2 (a) The time taken by a worker to complete a task was recorded for a random sample of 50 workers. The sample mean was 41.2 minutes and an unbiased estimate of the population variance was 32.6 minutes2. Find a 95% confidence interval for the mean time taken to complete the task. [3] (b) The probability that an α% confidence interval includes only values that are lower than the population mean is 16.1 Find the value of α. [2]

5 marks

Mark scheme: 32 6. 2 (a) 41.2 ± z × 50 M1 z = 1.96 B1 [39.6, 42.8] (3 sfs) A1 Allow any brackets or none, or < or “to” etc [3] (b) 2 × 1 or 1 or 0.125 or 12.5% M1 or 0.875 16 8 α = 87.5% A1 [2] 857. −85

This question in 9709/71 May/June 2011

2 35% of a random sample of n students walk to college. This result is used to construct an approximate 98% confidence interval for the population proportion of students who walk to college. Given that the width of this confidence interval is 0.157, correct to 3 significant figures, find n. [5]

5 marks

Mark scheme: .0 35 × .0 65 (pq ) q 2 2 × z × = 0.157 M1 For equation of the form n 2 × z × f(n) = 0.157 z = 2.326 B1 n = 4 × 2.3262 × 0.35 × 0.65 ÷ 0.1572 M1 Rearrange to form n = ... from a correct (=199.738…) equation in n, but allow any z and/or factor of “2” errors n = 200 A1 [5] cao

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

1 The weights, in grams, of packets of sugar are distributed with mean µ and standard deviation 23. A random sample of 150 packets is taken. The mean weight of this sample is found to be 494 g. Calculate a 98% confidence interval for µ. [3]

3 marks

Mark scheme: 1 z = 2.326 B1 seen 23 494 ± z × M1 Any z 150 = 490 to 498 (3 sfs) A1 [3]

This question in 9709/71 May/June 2012

7 Leila suspects that a particular six-sided die is biased so that the probability, p, that it will show a six is greater than 6.1 She tests the die by throwing it 5 times. If it shows a six on 3 or more throws she will conclude that it is biased. (i) State what is meant by a Type I error in this situation and calculate the probability of a Type I error. [3] (ii) Assuming that the value of p is actually 23, calculate the probability of a Type II error. [3] Leila now throws the die 80 times and it shows a six on 50 throws. (iii) Calculate an approximate 96% confidence interval for p. [4]

10 marks

Mark scheme: 7 (i) Conclude die is biased when it isn’t oe B1 In context 3 2 4 5 2 4 5   1   5   1   5   1    1   5    5   + 5 5C3    +        +    3 + 5  1  5    + 5 M1 or 1 –  5 C 2  6   6   6   6   6    6   6   6  6   6     23 A1 allow 1 end error = or 0.0355 (3 sf) 648 [3] 2 (ii) State or attempt P(0, 1, 2) with p = M1 Or 1– P(3,4,5) 3 2 3 4 5  2   1   2  1   1  M1 Attempt at correct expression 5C2     + 5   +    3   3   3  3   3  A1 Allow 0.21 17 = or 0.210 (3 sf) [3] 81 (iii) .0625 × 1( − .0625) Est Var(Ps) = M1 80 3 (=1024 ) z = 2.054 (or 2.055) B1 '3' 0.625 ± z× M1 Any z 1024 = 0.514 to 0.736 (3 sf) A1 [4]

This question in 9709/71 May/June 2013

1 A random sample of 80 values of a variable X is taken and these values are summarised below. n = 80 Σ x = 150.2 Σ x2 = 820.24 Calculate unbiased estimates of the population mean and variance of X and hence find a 95% confidence interval for the population mean of X. [6]

6 marks

Mark scheme: 1  4 6  10 100(=  25t −t  = ) 2500 6 0 3   16 100 2 M1 For E ( T2 ) – ( E ( T ) )2 )“ ”– ( 3 3

This question in 9709/73 Oct/Nov 2013

2 A die is biased. The mean and variance of a random sample of 70 scores on this die are found to be 3.61 and 2.70 respectively. Calculate a 95% confidence interval for the population mean score. [5]

5 marks

Mark scheme: 2 70 × 2.70 = 2.73913 M1A1 69 " 2.73913" .270 3.61 ± z M1 or 3.61 ± z M2A1(implied) 70 69 70 .270 without : 3.61 ± z M0A0M1 69 70 z = 1.96 B1 z = 1.96 B1 3.22 to 4.00 (3 sf) A1 [5] 3.23 to 3.99(4.00) (3 sf) A1 Answer must be an interval [Total: 5]

This question in 9709/71 May/June 2014

3 A die is thrown 100 times and shows an odd number on 56 throws. Calculate an approximate 97% confidence interval for the probability that the die shows an odd number on one throw. [4]

4 marks

Mark scheme: 3 p = 0.56 B1 Used .0 56×.0 44 ‘0.56’ ± z × M1 Equation of correct form condone just +ve or –ve 100 Must be z z = 2.17, or 2.169 or 2.171 B1 0.452 to 0.668 (3 s.f.) A1 [4] Seen Must be an interval

This question in 9709/73 May/June 2014

5 The masses, m grams, of a random sample of 80 strawberries of a certain type were measured and summarised as follows. n = 80 Σm = 4200 Σm2 = 229 000 (i) Find unbiased estimates of the population mean and variance. [3] (ii) Calculate a 98% confidence interval for the population mean. [3] 50 random samples of size 80 were taken and a 98% confidence interval for the population mean, -, was found from each sample. (iii) Find the number of these 50 confidence intervals that would be expected to include the true value of -. [1]

7 marks

Mark scheme: 5 (i) 4200/80 (=52.5) B1 80  229 000 2  M1 = − '52.5' (= 107.595)   79  80  A1 [3] = 108 (3 sf) (ii) '52 '5.± z '10780. 595 ' M1 Correct form – must be z-value – allow one side only z = 2.326 B1 Seen 49.8 to 55.2 A1f [3] ft their 52.5 and 107.595. Must be an interval (iii) 49 B1 [1] [Total: 7] 2

This question in 9709/71 May/June 2015

4 In the past, the flight time, in hours, for a particular flight has had mean 6.20 and standard deviation 0.80. Some new regulations are introduced. In order to test whether these new regulations have had any effect upon flight times, the mean flight time for a random sample of 40 of these flights is found. (i) State what is meant by a Type I error in this context. [2] (ii) The mean time for the sample of 40 flights is found to be 5.98 hours. Assuming that the standard deviation of flight times is still 0.80 hours, test at the 5% significance level whether the population mean flight time has changed. [4] (iii) State, with a reason, which of the errors, Type I or Type II, might have been made in your answer to part (ii). [2]

8 marks

Mark scheme: 4 (i) Conclude flight times affected B1 Or accept pop mean changed from 6.2 when in fact they have not been. B1 2 although pop mean has not changed from 6.2 (ii) H0: Pop mean (or µ) = 6.2 H0: Pop mean (or µ) ≠ 6.2 B1 .5 98 − 2.6 M1 Allow with 40 instead of √40 Allow SD/Var mix 8.0 A1 (CV method 5.952 or 6.2279 M1 A1) 40 B1 For valid comparison = –1.739 (±) Accept (±)1.74 4 or P(z < –1.739) = 0.041 > 0.025 or 5.98 > 5.952 or comp z = 1.96 6.2 < 6.228 and correct conclusion No evidence that flight times affected (iii) H0 was not rejected oe B1* If in (ii) H0 was rejected, then: Type II B1*dep H0 rejected B1; Type I B1dep 2 Total 8

This question in 9709/72 May/June 2015

5 The mean breaking strength of cables made at a certain factory is supposed to be 5 tonnes. The quality control department wishes to test whether the mean breaking strength of cables made by a particular machine is actually less than it should be. They take a random sample of 60 cables. For each cable they find the breaking strength by gradually increasing the tension in the cable and noting the tension when the cable breaks. (i) Give a reason why it is necessary to take a sample rather then testing all the cables produced by the machine. [1] (ii) The mean breaking strength of the 60 cables in the sample is found to be 4.95 tonnes. Given that the population standard deviation of breaking strengths is 0.15 tonnes, test at the 1% significance level whether the population mean breaking strength is less than it should be. [4] (iii) Explain whether it was necessary to use the Central Limit theorem in the solution to part (ii). [2]

7 marks

Mark scheme: 5 (i) Cables broken or not all cables can be accessed oe or Too many cables oe e.g. previous days’ stocks may have gone or too time consuming oe B1 [1] (ii) H0: Pop mean brk str (or µ) = 5 B1 Not just “mean” H1: Pop mean brk str (or µ) < 5 .4 95 − 5 (± ) M1 Allow 60 instead of √60 .015 60 (= ±2.582) A1 comp ±2.326 There is evidence that mean breaking B1 ft Ft their –2.582 strength is less than it should be (No ft 2 tailed test) Or reject H0 (H0 correctly defined) [4] Correct comparison shown, no errors seen. Accept area comparison 0.0049 with 0.01 [CR method (x – 5)/(0.15/√60) = –2.326 M1 A1 leading to x = 4.955 compared to 4.95and correct conclusion B1ft OR ((x – 4.95)/0.15/√60) leading to 4.995 M1 A1 compared to 5and correct conclusion B1ft] (iii) Population not necessarily normal B1 SR B1 For “it” is not necc normal (no so yes B1dep [2] mention of population) AND Yes Total 7 3 5 5.3 3

This question in 9709/73 May/June 2015

5 The 150 oranges in a random sample from a certain supplier were weighed and the masses, X grams, were recorded. The results are summarised below. n = 150 Σx = 14 910 Σx2 = 1 525 000 (i) Calculate a 99% confidence interval for the population mean of X. [6] (ii) The supplier claims that the mean mass of his oranges is 100 grams. Use your answer to part (i) to explain whether this claim should be accepted. [1] (iii) State briefly why the sample should be random. [1]

8 marks

Mark scheme: 5 (i) Est(µ) = 14150910 (= 99.4) B1 Est(σ2) = 150149 ( 1525000150 – "99.4"2) M1 Allow M1 if 150149 omitted = 288.228 A1 z = 2.576 B1 Accept 2.574–2.579 "99.4" ± z × 288.228 ÷ 150 M1 Any z CI = 95.8 to 103 (3 sf) A1 [6] (NB Use of biased Var can score 5/6 max) (ii) 100 lies within this CI Hence yes B1 [1] Both needed, ft their CI (iii) To avoid bias or Necessary to enable statistical inference B1 [1] Or any equivalent

This question in 9709/72 Feb/March 2016

1 The time taken for a particular type of paint to dry was measured for a sample of 150 randomly chosen points on a wall. The sample mean was 192.4 minutes and an unbiased estimate of the population variance was 43.6 minutes2. Find a 98% confidence interval for the mean drying time. [3]

3 marks

Mark scheme: Qu Answer Marks Notes 1 192.4 ± z 43.6150 M1 Allow 43.6 Allow one side for M1 150 B1 z = 2.326 to 2.329 Condone √(43.6/149 ) oe 191 to 194 (3 sf) A1 [3] CWO

This question in 9709/73 May/June 2016

5 (a) The masses, in grams, of certain tomatoes are normally distributed with standard deviation 9 grams. A random sample of 100 tomatoes has a sample mean of 63 grams. Find a 90% confidence interval for the population mean mass of these tomatoes. [3] (b) The masses, in grams, of certain potatoes are normally distributed with known population standard deviation but unknown population mean. A random sample of potatoes is taken in order to find a confidence interval for the population mean. Using a sample of size 50, a 95% confidence interval is found to have width 8 grams. (i) Using another sample of size 50, an !% confidence interval has width 4 grams. Find !. [3] (ii) Find the sample size n, such that a 95% confidence interval has width 4 grams. [2]

8 marks

Mark scheme: 5 (a) 63 ± z × 9 M1 B1 Expression of correct form, any z 100 z = 1.645 B1 Seen 61.5 to 64.5 (3 sf) A1 [3] Must be an interval (b) (i) z = 1.962 (= 0.98) M1 Allow any2 z Φ(“0.98”) (= 0.8365) “0.8365” – (1 – “0.8365”) M1 (= 0.673) α = 67.3 (3 sf) A1 Allow 67 from correct working [3] (ii) 4=(2x’z’x’σ’)/√n M1 Attempt to solve equ of correct form n = 200 A1 [2] SR B1 for n = 100

This question in 9709/71 Oct/Nov 2016

6 A variable X takes values 1, 2, 3, 4, 5, and these values are generated at random by a machine. Each value is supposed to be equally likely, but it is suspected that the machine is not working properly. A random sample of 100 values of X, generated by the machine, gives the following results. n = 100 Σx = 340 Σx2 = 1356 (i) Find a 95% confidence interval for the population mean of the values generated by the machine. [6] (ii) Use your answer to part (i) to comment on whether the machine may be working properly. [2]

8 marks

Mark scheme: 6 (i) est(µ) = 3.4 B1 1 / 99 (1356 – 3402/100 ) est(σ2)= 10099 ( 1356100 − '3.4'2 ) M1 or 200/99 A1 = 2.02(0202) B1 z = 1.96 3.4 ± z × '2.020202'100 correct working only M1 = 3.12 to 3.68 ( 3 sf) allow from unbiased or biased variance A1 [6] (ii) Mean should be 3 B1* stated or implied CI does not include 3 Machine probably not working DB1 their CI properly or evidence that…. [2] 1

This question in 9709/73 Oct/Nov 2016

1 In a survey, 36 out of 120 randomly selected voters in Hungton said that if there were an election next week they would vote for the Alpha party. Calculate an approximate 90% confidence interval for the proportion of voters in Hungton who would vote for the Alpha party. [4] … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: Question Answer Marks Guidance 1 Var(Ps) = 0.3(1120− 0.3) (= 0.00175) M1 Attempt correct values in correct formula 0.3 ± z "0.00175" M1 must be a z-value, not a prob z = 1.645 B1 CI = 0.231 to 0.369 (3 sf) A1 Total: 4

This question in 9709/72 Feb/March 2017

3 (a) The waiting time at a certain bus stop has variance 2.6 minutes2. For a random sample of 75 people, the mean waiting time was 7.1 minutes. Calculate a 92% confidence interval for the population mean waiting time. [3] … … … … … … … … … … … … … … … … … … … … … … … … (b) A researcher used 3 random samples to calculate 3 independent 92% confidence intervals. Find the probability that all 3 of these confidence intervals contain only values that are greater than the actual population mean. [2] … … … … … … … … … … … … … … (c) Another researcher surveyed the first 75 people who waited at a bus stop on a Monday morning. Give a reason why this sample is unsuitable for use in finding a confidence interval for the mean waiting time. [1] … … … … … … … …

6 marks

Mark scheme: 3(a) 7.1 ± z × 2.6 75 M1 score M1) seen z = 1.751 B1 6.77 to 7.43 (3 sfs) A1 Must be an interval Total: 3 3(b) 0.043 M1 Allow 0.083 for M1 = 0.000064 A1 Total: 2 3(c) e.g. Particular day or time of day B1 Allow "Not random" Total: 1

This question in 9709/71 May/June 2017

1 In a survey of 2000 randomly chosen adults, 1602 said that they owned a smartphone. Calculate an approximate 95% confidence interval for the proportion of adults in the whole population who own a smartphone. [4] … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 1 0.801 1 0.801 2000 × − (= 0.0000797) M1 0.801± z × "0.0000797" M1 Allow any z-value z = 1.96 B1 0.784 to 0.818 (3 sf) A1 As final answer. Must be an interval Allow 0.783 to 0.819 Total: 4

This question in 9709/72 May/June 2017

3 After an election 153 adults, from a random sample of 200 adults, said that they had voted. Using this information, an !% confidence interval for the proportion of all adults who voted in the election was found to be 0.695 to 0.835, both correct to 3 significant figures. Find the value of !, correct to the nearest integer. [4] … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 1533 200 × 200200−153 M1 153 200 + z × = 0 . 8 35 2 00 (Var(Ps) = 0.000898875) (s.d. 0.02998) z = 2.335 A1 allow 2.33 or 2.34 2Φ ( z ) − 1 M1 or equivalent method indep α = 98 A1 allow 98.0 but not e.g. 98.04 4

This question in 9709/71 Oct/Nov 2017

4 The lengths, in millimetres, of rods produced by a machine are normally distributed with mean - and standard deviation 0.9. A random sample of 75 rods produced by the machine has mean length 300.1 mm. (i) Find a 99% confidence interval for -, giving your answer correct to 2 decimal places. [3] … … … … … … … … … … … … … … … The manufacturer claims that the machine produces rods with mean length 300 mm. (ii) Use the confidence interval found in part (i) to comment on this claim. [1] … … … … … …

4 marks

Mark scheme: 4(i) 300.1 ± z × 0.9 M1 allow any value of z 75 z = 2.576 B1 allow 2.574 to 2.579 299.83 to 300.37 (2 dps) A1 answer must be seen to 2 dps need an interval 3 4(ii) CI includes 300 so claim supported or B1 FT or equivalent justified or probably true FT from CI in (i) 1

This question in 9709/71 Oct/Nov 2017

2 The number of words in History essays by students at a certain college has mean - and standard deviation 1420. (i) The mean number of words in a random sample of 125 History essays was found to be 4820. Calculate a 98% confidence interval for -. [3] … … … … … … … … … … (ii) Another random sample of n History essays was taken. Using this sample, a 95% confidence interval for - was found to be 4700 to 4980, both correct to the nearest integer. Find the value of n. [3] … … … … … … … … … … …

6 marks

Mark scheme: 2(i) 1420 4820 125 z ± × z = 2.326 B1 Accept 2.326 - 2.329 4524/4525 to 5115/5116 or 4520 to 5120 (3 sf) A1 Must be an interval 3 Question Answer Marks Guidance 2(ii) 4840 x = B1 or width = 280 or half width = 140 4840 + 1.96 × n 1420 = 4980 OE M1 or 140 = 1.96 × n 1420 OE n = 395 A1 CAO must be an integer 3

This question in 9709/72 Oct/Nov 2017

3 The masses, m kg, of packets of flour are normally distributed. The mean mass is supposed to be 1.01 kg. A quality control officer measures the masses of a random sample of 100 packets. The results are summarised below. n = 100 Σm = 98.2 Σ m2 = 104.52 (i) Test at the 5% significance level whether the population mean mass is less than 1.01 kg. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (ii) Explain whether it was necessary to use the Central Limit theorem in your answer to part (i). [1] … … … … … …

8 marks

Mark scheme: 3(i) m = 98.2 100 = 0.982 s = 2 982 .0 100 52 . 104 99 100 − × (= 0.28582) or var = 0.08169 M1 H0: Pop mean mass = 1.01 H1: Pop mean mass < 1.01 B1 not just ‘mean’, but allow just ‘µ’ 0.28582 100 0.982 1.01 − ± M1 0.284387 100 0.982 1.01 − ± M1 = −0.980 (3 sf) accept ± A1 = –0.985 (3 sfs) accept ± A1 Comp with z = − 1.645 (or areas 0.1635 > 0.05) M1 Valid comparison of z’s or area’s No evidence that (mean) mass is less than 1.01 A1 FT Correct conclusion FT their z 7 Question Answer Marks Guidance 3(ii) Distr of X normal (so distr of X normal) Must state or imply No B1 X/parent population 1

This question in 9709/72 Oct/Nov 2017

3 After an election 153 adults, from a random sample of 200 adults, said that they had voted. Using this information, an !% confidence interval for the proportion of all adults who voted in the election was found to be 0.695 to 0.835, both correct to 3 significant figures. Find the value of !, correct to the nearest integer. [4] … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 1533 200 × 200200−153 M1 153 200 + z × = 0 . 8 35 2 00 (Var(Ps) = 0.000898875) (s.d. 0.02998) z = 2.335 A1 allow 2.33 or 2.34 2Φ ( z ) − 1 M1 or equivalent method indep α = 98 A1 allow 98.0 but not e.g. 98.04 4

This question in 9709/73 Oct/Nov 2017

4 The lengths, in millimetres, of rods produced by a machine are normally distributed with mean - and standard deviation 0.9. A random sample of 75 rods produced by the machine has mean length 300.1 mm. (i) Find a 99% confidence interval for -, giving your answer correct to 2 decimal places. [3] … … … … … … … … … … … … … … … The manufacturer claims that the machine produces rods with mean length 300 mm. (ii) Use the confidence interval found in part (i) to comment on this claim. [1] … … … … … …

4 marks

Mark scheme: 4(i) 300.1 ± z × 0.9 M1 allow any value of z 75 z = 2.576 B1 allow 2.574 to 2.579 299.83 to 300.37 (2 dps) A1 answer must be seen to 2 dps need an interval 3 4(ii) CI includes 300 so claim supported or B1 FT or equivalent justified or probably true FT from CI in (i) 1

This question in 9709/73 Oct/Nov 2017

3 A researcher wishes to estimate the proportion, p, of houses in London Road that have only one occupant. He takes a random sample of 64 houses in London Road and finds that 8 houses in the sample have only one occupant. Using this sample, he calculates that an approximate !% confidence interval for p has width 0.130. Find ! correct to the nearest integer. [5] … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3 8 8 (1 ) 64 64 64 × − (= 7 4096 or 0.00171) M1 OE, e.g. 7 1 8 8 64 × 2 × z 7 " " 4096 = 0.130 M1 Correct equation using their variance z = 1.572 A1 ɸ("1.572") (= 0.942) (0.942 – (1 – 0.942) = 0.884) M1 2ɸ(their z) -1 α = 88 A1 CAO 5

This question in 9709/73 May/June 2018

1 The standard deviation of the heights of adult males is 7.2 cm. The mean height of a sample of 200 adult males is found to be 176 cm. (i) Calculate a 97.5% confidence interval for the mean height of adult males. [3] … … … … … … … … … … … … (ii) State a necessary condition for the calculation in part (i) to be valid. [1] … … …

4 marks

Mark scheme: Question Answer Marks Guidance 1(i) 176 ± z × 7.2 M1 need correct form must be z 200 z = 2.24 B1 allow 2.241 and 2.242 175 to 177 A1 cwo 3 1(ii) Sample random B1 oe. both words essential 1

This question in 9709/71 Oct/Nov 2018

2 The standard deviation of the volume of drink in cans of Koola is 4.8 centilitres. A random sample of 180 cans is taken and the mean volume of drink in these 180 cans is found to be 330.1 centilitres. (i) Calculate a 95% confidence interval for the mean volume of drink in all cans of Koola. Give the end-points of your interval correct to 1 decimal place. [3] … … … … … … … … … (ii) Explain whether it was necessary to use the Central Limit theorem in your answer to part (i). [1] … … …

4 marks

Mark scheme: 2(i) z = 1.96 B1 seen 330.1 ± z × 4.8 180 M1 Must be of correct form. Any z = 329.4 to 330.8 (1 dp) A1 Must be to 1 dp. Must be an interval. 3 2(ii) Yes, because vol of all cans not stated to be normal B1 Or Yes, population not stated to be normal 1

This question in 9709/72 Oct/Nov 2018

6 In the past, Angus found that his train was late on 15% of his daily journeys to work. Following a timetable change, Angus found that out of 60 randomly chosen days, his train was late on 6 days. (i) Test at the 10% significance level whether Angus’ train is late less often than it was before the timetable change. [5] … … … … … … … … … … … … … … … … … … … … … … … Angus used his random sample to find an !% confidence interval for the proportion of days on which his train is late. The upper limit of his interval was 0.150, correct to 3 significant figures. (ii) Calculate the value of ! correct to the nearest integer. [5] … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 6(i) Ho: p = 0.15 H1: p < 0.15 (N(60 × 0.15, 60 × 0.15 × 0.85) ) = N(9, 7.65) B1 H1: µ < 9 Use of Normal approximation: (N(0.15, 0.15 0.85 60 × )) = N(0.15, 0.002125) 6.5 '9' '7.65' − M1 For standardising (or 6 0.5 60 60 '0.15' '0.002125' + − = –0.904) Allow wrong or no cc = –0.904 A1 Accept ± ‘0.904’ < 1.282 M1 Valid comparison of z values or ɸ('–0.904')= 0.183 > 0.1 ft their 0.904 No evidence train late less often A1ft Use of Bin (60,0.15) to give Pr (< = 6) = 0.1848 M1A1 Valid comparison with 0.1 M1 Conclusion A1ft 5 6(ii) 0.1 + z × 0.1 0.9 60 × = 0.150 M1 For √ (0.1 × 0.9 / 60) seen M1 for 0.1 + z × ... = 0.150 or 2z… = 0.1 z = 1.291 A1 φ(‘1.291’) (= 0.90(16)) M1 for correct method to find α α = 80 A1ft ft their z. Must be a +ve non-zero integer < 100 5

This question in 9709/72 Oct/Nov 2018

1 The standard deviation of the heights of adult males is 7.2 cm. The mean height of a sample of 200 adult males is found to be 176 cm. (i) Calculate a 97.5% confidence interval for the mean height of adult males. [3] … … … … … … … … … … … … (ii) State a necessary condition for the calculation in part (i) to be valid. [1] … … …

4 marks

Mark scheme: Question Answer Marks Guidance 1(i) 176 ± z × 7.2 M1 need correct form must be z 200 z = 2.24 B1 allow 2.241 and 2.242 175 to 177 A1 cwo 3 1(ii) Sample random B1 oe. both words essential 1

This question in 9709/73 Oct/Nov 2018

1 The masses of a certain variety of plums are known to have standard deviation 13.2 g. A random sample of 200 of these plums is taken and the mean mass of the plums in the sample is found to be 62.3 g. (i) Calculate a 98% confidence interval for the population mean mass. [3] … … … … … … … … … … … … … … … (ii) State with a reason whether it was necessary to use the Central Limit theorem in the calculation in part (i). [1] … … … … … …

4 marks

Mark scheme: 1(i) z = 2.326 B1 62.3 ± z 13.2 200 M1 Any z. Expression of correct form. Must be a ‘z’ 60.1 to 64.5 (3 sfs) A1 Must be an interval 3 1(ii) Yes, because pop not (given to be) normal, or pop distribution unknown B1 No contradictions 1

This question in 9709/72 Feb/March 2019

3 It is claimed that, on average, a particular train journey takes less than 1.9 hours. The times, t hours, taken for this journey on a random sample of 50 days were recorded. The results are summarised below. n = 50 Σt = 92.5 Σt2 = 175.25 (i) Calculate unbiased estimates of the population mean and variance. [3] … … … … … … … … … … … … … … … … … … … … … … (ii) Test the claim at the 5% significance level. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 3(i) Est(µ) = 1.85 B1 Est(σ2) = 2 50 175.25 '1.85' 49 50   −     M1 Allow 2 50 175.25 '1.85' 49 150   −     or 0.0290 for M1 = 0.0842 (3 sf) or 33 392 A1 Cao If 50 49 omitted (giving var = 0.0825 or sd = 0.287) M0A0 3 3(ii) H0: Pop mean time = 1.9 (h) H1: Pop mean time < 1.9 (h) B1 Allow ‘µ’ but not just ‘mean’ 1.85 1.9 '0.0842' 50 − ± M1 ±1.85 1.9 '0.290' 50 − Accept totals method (92.5–95) / 4.21 = –1.22 A1 = –1.22 comp z = –1.645 M1 Or other valid comparison 0.888 or 0.889 < 0.95 OR 0.111 or 0.112>0.05 No evidence that mean time < 1.9 h A1 FT their z. Correct conclusion. No contradictions If 50 49 not used in (1): var = 0.8225, sd = 0.907, cr = 1.17 can score all marks in (ii) Note- 2 tail test can score B0 M1 A1 M1 (comparison with 1.96) A0 (no ft) max3/5 5

This question in 9709/72 May/June 2019

1 A coin is thrown 100 times and it shows heads 60 times. Calculate an approximate 98% confidence interval for the probability, p, that the coin shows heads on any throw. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: 1 0.6 ± z 0.4 0.6 100 × M1 z = 2.326 B1 2.326 to 2.329 0.486 to 0.714 (3 sf) A1 Must be an interval 3

This question in 9709/73 May/June 2019

2 The heights of a certain species of animal have been found to have mean 65.2 cm and standard deviation 7.1 cm. A researcher suspects that animals of this species in a certain region are shorter on average than elsewhere. She takes a large random sample of n animals of this species from this region and finds that their mean height is 63.2 cm. She then carries out an appropriate hypothesis test. (i) She finds that the value of the test statistic z is −2.182, correct to 3 decimal places. (a) Stating a necessary assumption, calculate the value of n. [4] … … … … … … … … … … … … … … … … … … … … … … (b) Carry out the hypothesis test at the 4% significance level. [3] … … … … … … … … … … … … … … … … … … … (ii) Explain why it was necessary to use the Central Limit theorem in carrying out the test. [1] … … … … …

8 marks

Mark scheme: 2(i)(a) Assume standard deviation for the region is 7.1 B1 Or standard deviation is same as for whole population OE 7.1 63.2 65.2 2.182 n − = − M1 Attempt to find correct equation (accept +2.182) n = {–2.182 × 7.1 ÷ (–2)}2 A1 Any correct expression for n or n . SOI n = 60 A1 CWO. Must be an integer 4 2(i)(b) H0: population mean (or µ) = 65.2 H1: population mean (or µ) < 65.2 B1 Not just ‘mean’ 2.182 > 1.751 M1 Or valid area comparison. There is evidence that animals are shorter in this region A1 CWO. No contradictions 3 2(ii) Population unknown or population not given as normal B1 Allow population not normal. Accept distribution of X unknown. 1

This question in 9709/72 Oct/Nov 2019

3 The masses, in grams, of bags of flour are normally distributed with mean -. The masses, m grams, of a random sample of 50 bags are summarised by Σm = 25 110 and Σm2 = 12 610 300. (i) Calculate a 96% confidence interval for -, giving the end-points correct to 1 decimal place. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … Another random sample of 50 bags of flour is taken and a 99% confidence interval for - is calculated. (ii) Without calculation, state whether this confidence interval will be wider or narrower than the confidence interval found in part (i). Give a reason for your answer. [1] … … … … … … … … … … … …

7 marks

Mark scheme: 3(i) est(µ) = 25110 50 (= 502.2) ( ) 2 2 50 12610300 25110 50 58 est 1.1836 49 50 50 49 50 σ    = − = × =       M1 OE 1.18 (3 sf) or 58 49 A1 Accept SD = 1.0879 z = 2.054 or 2.055 B1 '1.1836' 502.2 50 z ± × M1 Must be of correct form. 501.9 to 502.5 (1dp) A1 CWO. Must be in interval. SC accept use of biased variance (1.16) for M1 A1 6 3(ii) More confident or z would be greater, Hence wider. B1 OE Reason needed 1

This question in 9709/72 Oct/Nov 2019

3 The times, in minutes, taken by competitors to complete a puzzle have mean - and standard deviation 3. The times taken by a random sample of 10 competitors are noted and the results are given below. 25.2 26.8 18.5 25.5 30.1 28.9 27.0 26.1 26.0 24.9 (i) Stating a necessary assumption, calculate a 97% confidence interval for -. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (ii) Two more random samples, each of 10 competitors, are taken. Their times are used to calculate two more 97% confidence intervals for -. Find the probability that neither of these intervals contains the true value of -. [1] … … … … … … … … …

6 marks

Mark scheme: 3(i) Assume population is normally distributed B1 25.9 = x B1 Allow 259 10 z =2.17 B1 3 '25.9' 10 ± × z M1 Must have correct form and z. 23.8 to 28.0 (3 sf) A1 CWO 5 3(ii) 0.032 (=0.0009) B1 1

This question in 9709/73 Oct/Nov 2019