Applications: Sampling

A stratified sample splits the population into groups, called strata, and samples each group in proportion to its size, so that every year group has the same weight in the sample as it has in the school. Each share is the size of the group multiplied by the sampling fraction 50800. The shares are rarely whole numbers, and rounding each one on its own can leave the total one too high or one too low.

1. The sampling fraction is 50800 = 116: one pupil in every 16 is sampled, from every year group alike.
2. Divide each year group by 16: Year 7 gets 23516 = 14.6875, Year 8 gets 20516 = 12.8125, Year 9 gets 19016 = 11.875 and Year 10 gets 17016 = 10.625. Check: 14.6875 + 12.8125 + 11.875 + 10.625 = 50.
3. (a) Rounding each share to the nearest whole number gives 15 + 13 + 12 + 11 = 51 pupils, one more than the 50 the survey is for. Every share was rounded up, and the four small increases add up to more than a whole pupil.
4. Give each year group the whole-number part of its share: 14 + 12 + 11 + 10 = 47, so 3 more pupils are needed. The remainders are 0.6875, 0.8125, 0.875 and 0.625. The three largest belong to Years 9, 8 and 7, and each of those year groups gets one more pupil.
5. (b) The sample is 15 pupils from Year 7, 13 from Year 8, 12 from Year 9 and 10 from Year 10. Check: 15 + 13 + 12 + 10 = 50, and every year group is within one pupil of its exact share.

answer(a) the shares are 14.6875, 12.8125, 11.875 and 10.625, and rounding each one gives 51 pupils; (b) 15 from Year 7, 13 from Year 8, 12 from Year 9 and 10 from Year 10

techniqueSampling Methods · Populations and Samples

examsSAT · GCSE Higher · H2

A sample is biased when the way it is chosen makes some members of the population more likely to be picked than others, in a way that is linked to the answer. A systematic sample lists the whole population, splits the list into equal blocks, chooses a random start in the first block, and then takes every kth name, where k is the size of the population divided by the size of the sample.

1. Only people who are at a gym could be chosen for the first sample, and most people at a gym exercise. Adults who never exercise had no chance of being asked.
2. (a) The sample is biased. It gives 176200 = 88%, which makes the fraction who exercise look too large. Asking more people at the same gym would not help, because the bias comes from where the sample is taken, not from its size.
3. For the systematic sample, the interval is k = 24000200 = 120. The register is split into 200 blocks of 120 names, and one name is taken from each block.
4. Start at the 37th name and add 120 each time. The first three names chosen are the 37th, 157th and 277th, and the last is the 37 + 199 × 120 = 23917th. Check: the last block runs from the 23881st name to the 24000th, and 23917 is in it.
5. (b) The register sample gives 92200 = 0.46, so an estimated 0.46 × 24000 = 11040 adults exercise weekly, far fewer than the 88% from the gym suggests.

answer(a) it is biased, because only people at a gym can be asked, and the 88% it gives is too large; (b) the interval is 120; the 37th, 157th and 277th names, and the last is the 23917th; an estimated 11040 adults

techniqueSampling Methods · Populations and Samples

examsSAT · GCSE Higher · H2

In an observational study the researchers record what people already do; in an experiment the researchers decide who receives the treatment. When people choose for themselves, anything that affects both the choice and the result is a confounding variable, and it can make a treatment look better than it is. Random allocation spreads such variables evenly between the groups, including the ones nobody thought to measure.

1. The students chose for themselves whether to use the app, and the school only recorded what happened. This is an observational study.
2. (a) The pass rates are 84120 = 70% for the students who used the app and 99180 = 55% for the others, a difference of 15 percentage points. An observational study shows that using the app and passing go together; it cannot show that the app caused the passes.
3. A confounding variable affects both the choice and the result. Students who are more motivated are more likely to download a revision app, and they are also more likely to pass whether or not they use it. Motivation, or the time a student spends revising, is a confounding variable here.
4. In the new study a random draw decides who gets the app, so motivated students are as likely to be in one group as in the other. The pass rates are 66100 = 66% with the app and 60100 = 60% without it.
5. (b) The difference is now 66% − 60% = 6 percentage points. Most of last year's 15 points therefore came from which students chose the app, not from the app itself. What is left is small: on two groups of 100, a gap of 6 points could arise by chance alone, so this study shows at most a modest effect.

answer(a) an observational study; pass rates of 70% and 55%, a difference of 15 percentage points; (b) motivation, or the time spent revising, is a confounding variable; with random allocation the difference is 6 percentage points, so most of the first study's gap came from who chose the app, and a gap that size on 100 students each could be chance

techniqueObservational Studies and Experiments

examsSAT

The method assumes that the tagged fish mix freely with the others, and that the number of tagged fish does not change between the two catches. Then the share of tagged fish in the second catch estimates the share of tagged fish in the whole lake, and the one unknown in that equation is the number of fish.

1. Let N be the number of fish in the lake. The share of the lake that is tagged is 120N, and the share of the second catch that is tagged is 24150 = 0.16.
2. Set the two shares equal: 120N = 24150. Multiply both sides by 150N: 24N = 120 × 150 = 18000, so N = 1800024 = 750.
3. (a) There are about 750 fish in the lake. Check: 120 tagged fish out of 750 is 120750 = 0.16, the same share as in the catch.
4. If one tag in five falls off, only 45 × 120 = 96 fish still carry a tag at the second catch. Fewer tagged fish are caught than the lake size would give, and dividing by 120 rather than 96 makes the lake look larger than it is.
5. (b) The estimate in (a) is too high. With 96 tagged fish, 96N = 24150 gives N = 96 × 15024 = 600 fish.

answer(a) about 750 fish; (b) too high: with 96 tags still in place, the estimate is 600 fish

techniqueCapture and Recapture · Populations and Samples

examsSAT · GCSE Higher

The five households are the population, and its mean μ is a fixed number. A sample mean X changes from one sample to the next. Its sampling distribution lists every value that X can take, with its probability. Here it is found by listing every sample, since each of the 52 = 10 samples is equally likely.

1. Call the households A, B, C, D and E, with 0, 1, 1, 2 and 3 cars. The ten samples and their means are AB 0.5, AC 0.5, AD 1, AE 1.5, BC 1, BD 1.5, BE 2, CD 1.5, CE 2 and DE 2.5.
2. Count how often each mean occurs. X takes the values 0.5, 1, 1.5, 2 and 2.5 with probabilities 210, 210, 310, 210 and 110. Check: 2 + 2 + 3 + 2 + 1 = 10.
3. (a) P(X ≥ 2) = 210 + 110 = 310, from the samples BE, CE and DE.
4. Weight each value by its probability: E(X) = 0.5 × 2 + 1 × 2 + 1.5 × 3 + 2 × 2 + 2.5 × 110 = 1410 = 1.4.
5. (b) The mean for the street is μ = 0 + 1 + 1 + 2 + 35 = 75 = 1.4 cars, so E(X) = μ. The sample mean is an unbiased estimator of the population mean, even though no single sample has a mean of exactly 1.4.

answer(a) X is 0.5, 1, 1.5, 2 or 2.5 with probabilities 210, 210, 310, 210 and 110, and P(X ≥ 2) = 310; (b) E(X) = 1.4, the same as the mean for the street, 1.4 cars per household

techniqueThe Sampling Distribution · Mean and Variance of X-bar · Populations and Samples

examsH2

For a random sample of n from a population with mean μ and variance σ², E(X) = μ and Var(X) = σ²n. When the population is normal, X is normal too, for any n. The mean of many eggs varies much less than a single egg does, because the light and heavy eggs in a sample balance each other.

1. The mean of the sample has the same expected value as a single egg: E(X) = μ = 62 g.
2. (a) Var(X) = σ²n = 5²25 = 2525 = 1. So E(X) = 62 and Var(X) = 1, and the standard deviation of the mean is √1 = 1 g.
3. The masses are normal, so X ∼ N(62, 1). Standardize with the standard deviation of the mean: z = 60 − 621 = −2, so P(X < 60) = 1 − Φ(2) = 1 − 0.9772 = 0.0228.
4. For a single egg the standard deviation is 5 g: z = 60 − 625 = −0.4, so P(X < 60) = 1 − Φ(0.4) = 1 − 0.6554 = 0.3446.
5. (b) The mean of the sample is below 60 g with probability 0.0228, while a single egg is below 60 g with probability 0.3446, about 15 times as likely. Check: the curve for the mean is 5 times as narrow, so 60 g is 2 of its standard deviations below 62 g, rather than 0.4 of them.

answer(a) E(X) = 62 g and Var(X) = 1; (b) P(X < 60) = 0.0228, against 0.3446 for a single egg

techniqueMean and Variance of X-bar · Standard Error

examsH2

The standard error of an estimate is its standard deviation from one sample to the next. For a mean it is σ√n, and for a proportion it is √p(1 − p)n. Both get smaller as n grows, but only in step with √n: to halve a standard error takes four times the sample.

1. For the mean, the standard error is 48√n minutes, and the target is 48√n ≤ 4.
2. Multiply both sides by √n and divide both sides by 4: √n ≥ 12, so n ≥ 144.
3. (a) She must survey at least 144 teenagers. Check: 48√144 = 4812 = 4 minutes exactly.
4. For the proportion, the standard error is √0.4 × 0.6n = √0.24n. Square both sides of √0.24n ≤ 0.02: 0.24n ≤ 0.0004, so n ≥ 0.240.0004 = 600.
5. (b) The proportion needs at least 600 teenagers, and a survey of 600 meets both targets. Check: √0.24600 = √0.0004 = 0.02, and with n = 600 the standard error of the mean is 48√600 ≈ 1.96 minutes, well under 4.

answer(a) at least 144 teenagers; (b) at least 600 for the proportion, so the survey must include 600 teenagers

techniqueStandard Error

examsSAT · H2

The central limit theorem says that for a large sample the mean X is approximately normal, X ∼ N(μ, σ²n) approximately, whatever the shape of the population. How large is large depends on the population: n ≥ 30 is a rule of thumb, and a strongly skewed population needs a larger sample than a symmetric one.

1. The service times are not normal, but n = 64 is large, so by the central limit theorem X is approximately normal. Its mean is μ = 4 minutes and its standard deviation is σ√n = 4√64 = 48 = 0.5 minutes.
2. Standardize: z = 4.8 − 40.5 = 1.6.
3. (a) P(X > 4.8) = 1 − Φ(1.6) = 1 − 0.9452 = 0.0548.
4. For 4 customers the standard deviation of the mean would be 4√4 = 2 minutes. A normal curve with mean 4 and standard deviation 2 puts 0 minutes at z = 0 − 42 = −2, so it gives P(X < 0) = 1 − Φ(2) = 1 − 0.9772 = 0.0228.
5. (b) The method gives a probability of 0.0228 to a negative mean time, which is impossible. For a population this skewed, 4 customers are far too few for the mean to be close to normal. With 64 customers, 0 minutes is 8 standard deviations below the mean, and the normal curve puts almost no area there.

answer(a) P(X > 4.8) = 0.0548; (b) 0.0228 for a negative mean time, which is impossible, so 4 customers are too few for the central limit theorem to apply

techniqueThe Central Limit Theorem · Applying the Theorem · Standard Error

examsH2

The total T of n independent values has E(T) = nμ and Var(T) = nσ². The total is n times the sample mean, so the central limit theorem applies to it as well: for large n, T is approximately normal, whatever the distribution of a single parcel's mass.

1. Let T be the total mass of the 50 parcels. Its mean is E(T) = 50 × 11 = 550 kg.
2. The masses are independent, so their variances add: Var(T) = 50 × 8 = 400, and the standard deviation of T is √400 = 20 kg. With n = 50, the central limit theorem gives T ∼ N(550, 20²) approximately.
3. Standardize the safe load: z = 600 − 55020 = 5020 = 2.5.
4. (a) P(T > 600) = 1 − Φ(2.5) = 1 − 0.9938 = 0.0062, about 6 loads in every 1000.
5. (b) The top 2.5% of totals lies above z = 1.96, since Φ(1.96) = 0.975. That total is 550 + 1.96 × 20 = 550 + 39.2 = 589.2 kg. Check: 589.2 − 55020 = 1.96.

answer(a) P(T > 600) = 0.0062; (b) 589.2 kg

techniqueApplying the Theorem · The Central Limit Theorem

examsH2

When n is large and p is not close to 0 or 1, X ∼ B(n, p) is close to a normal distribution with the same mean np and variance np(1 − p). The binomial takes only whole numbers, each drawn as a bar of width 1, while the normal curve is continuous. So each whole number k is replaced by the interval from k − 0.5 to k + 0.5: this is the continuity correction.

1. X ∼ B(100, 0.5) has mean np = 50 and variance np(1 − p) = 100 × 0.5 × 0.5 = 25. Both np and n(1 − p) are 50, well above 5, so X is approximately N(50, 5²).
2. The event X ≥ 58 includes the whole bar for 58, which starts at 57.5. With the continuity correction, P(X ≥ 58) ≈ P(Y > 57.5), where Y ∼ N(50, 25), and z = 57.5 − 505 = 1.5.
3. (a) P(X ≥ 58) ≈ 1 − Φ(1.5) = 1 − 0.9332 = 0.0668. The exact binomial sum is 0.0666, so the approximation is out by only 0.0002.
4. Exactly 50 is the single bar from 49.5 to 50.5. Standardize both edges: z = 49.5 − 505 = −0.1 and z = 50.5 − 505 = 0.1.
5. (b) P(X = 50) ≈ Φ(0.1) − Φ(−0.1) = 2 × 0.5398 − 1 = 0.0796, which agrees with the exact value 10050 × 0.5¹⁰⁰ = 0.0796. Without the correction the interval would have no width, and a continuous curve gives a probability of 0 to any single value.

answer(a) P(X ≥ 58) ≈ 0.0668, against the exact 0.0666; (b) P(X = 50) ≈ 0.0796, the same as the exact value to 4 decimal places; without the correction a single value would have probability 0

techniqueApproximating a Binomial · Continuity Correction