Year 10 Extended · Unit 6 Statistics and the bell curve · Lesson 7 of 7

Judging a whole from a part

We seldom measure everything. This lesson asks how well a sample's mean and spread describe the population it came from, and what makes a sample worth trusting.

Statistics and ProbabilityGDC optional

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. From Lesson 3: a set of 5 values has Σ(x − μ)² = 45. What is its variance?
    Answer9. 45 ÷ 5 = 9.
  2. From Lesson 6: in a bell-shaped set, about what percentage lies within 2 standard deviations of the mean?
    Answer95%.
  3. From Year 9: a bag has 40 counters and 12 are red. What fraction are red, in its simplest form?
    Answer3/10. 12 ÷ 40 = 3 ÷ 10.
Words used in this lesson
population
all of the values or objects being studied. For example: every fish in a lake.
sample
a part of the population that is measured. For example: the 30 fish caught in a net.
estimate
a value found from a sample, used in place of an unknown. For example: x̄ estimates μ.
representative
like the population in the ways that matter. For example: a random sample from the whole register.
random sample
a sample in which every member has an equal chance of selection. For example: names drawn from a list.
capture-recapture
estimating a population by marking some and catching again. For example: tagged fish in a second catch.

The lake that cannot be counted

A warden at a lake in Kisumu wants to know how many fish live in it, and how much their lengths vary. She cannot catch every fish. She nets some, measures them, and uses that sample to say something about all of them: the population.

Predict first: if you list every possible small sample from a population and work out the variance of each one from its own mean, will those sample variances, on average, come out above, below or equal to the population's own variance? Test it in the figure.

2
11.67population variance
5.83average, dividing by n
11.67average, dividing by n − 1

Samples of size 2 from the population 2, 4, 6, 8, 10, 12 (36 samples): dividing by n averages 5.83, dividing by n − 1 averages 11.67, and the population variance is 11.67.

Show me the steps
  1. Keep n = 2. There are 36 samples. Dividing by n gives an average of 5.83, only half of the population's 11.67 (the factor is (n − 1) ÷ n = 1 ÷ 2). Dividing by n − 1 averages exactly 11.67.
  2. Set n = 3. There are 216 samples. Dividing by n averages 7.78, which is 2 ÷ 3 of 11.67. Dividing by n − 1 still averages 11.67.
  3. Set n = 4. Dividing by n averages 8.75, which is 3 ÷ 4 of 11.67. The shortfall is shrinking as the sample grows.
  4. Set n = 5. Dividing by n averages 9.33, which is 4 ÷ 5 of 11.67. In every case the n − 1 divisor averages 11.67, so on average it neither falls short nor overshoots.

A population is every value of interest; a sample is a part of it. The sample mean x̄ is an estimate of the population mean μ.

Measured from the sample's own mean, values sit closer together than they do around the true mean, so dividing by n falls short. The estimate of the population standard deviation divides by n − 1: sn−1 = √(Σ(x − x̄)² ÷ (n − 1)).

A sample is only as good as how it was chosen: it must be random enough to be representative, and a larger sample gives a smaller uncertainty. Every estimate carries uncertainty, and a good report says so.

n − 1 is for a sample used to estimate a population. If the data are the whole population, divide by n. Using n − 1 on a whole population overstates its spread, and using n on a small sample understates it.

Worked examples

1. A sample of 4 pencil lengths (cm) is 12, 15, 15, 18. Estimate the population standard deviation.

  1. x̄ = 60 ÷ 4 = 15, and the deviations are −3, 0, 0, 3, so Σ(x − x̄)² = 18.
  2. The estimate of the variance is 18 ÷ (4 − 1) = 6, so sn−1 = √6 = 2.4 cm to one decimal place.

2. Why n − 1? A population of just two values, 1 and 3, has variance 1.

  1. Take every sample of 2 with replacement: (1, 1), (1, 3), (3, 1), (3, 3). Their sums of squared deviations from their own means are 0, 2, 2, 0.
  2. Dividing by n = 2 gives 0, 1, 1, 0, which average 0.5, only half the true variance 1.
  3. Dividing by n − 1 = 1 gives 0, 2, 2, 0, which average 1, exactly the true variance.

3. Capture-recapture. 25 fish are caught, tagged and released. A later catch of 20 has 5 tagged fish. Estimate the population.

  1. The tagged share in the second catch, 5 ÷ 20 = 1 ÷ 4, should match the share in the whole lake, 25 ÷ N.
  2. 25 ÷ N = 1 ÷ 4, so N = about 100 fish. It rests on the fish mixing evenly, and on tags not being lost.

Your go. The first step is given. A sample of 5 values has x̄ = 8 and Σ(x − x̄)² = 28. Estimate the population standard deviation to one decimal place.

  1. The divisor is n − 1 = 4.
  2. Now divide the sum of squares by it and take the root.
Check the answer

28 ÷ 4 = 7 and √7 = 2.6. Dividing by n = 5 would have given √5.6 = 2.4, a little too small.

With a GDC (optional)

The one-variable statistics screen shows two standard deviations. One divides by n and is meant for a whole population; the other divides by n − 1 and is meant for a sample. When your data are a sample, read the one that divides by n − 1.

Check which is which by working one small data set by hand: the one that divides by n − 1 is the larger of the two whenever the values are not all equal. Everything here can be done by hand.

Check before you go on

1. A sample is used to estimate a population's standard deviation, and the sum of squared deviations is divided by n − 1. Why?

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. A sample is 4, 6, 7, 9, 9. Estimate the population standard deviation, to one decimal place.

3. A sample of 7 values has Σ(x − x̄)² = 90. What is the estimate of the population variance?

Level 2 · Building

4. A sample of 5 values has sn−1 = 4. What is Σ(x − x̄)²?

5. In a pond, 40 fish are tagged and released. Later a catch of 30 fish includes 6 tagged ones. Estimate the number of fish in the pond.

Level 3 · Stretch

6. A calculator reports a standard deviation of 2.4 when dividing by n, for a sample with n = 5. What is the standard deviation dividing by n − 1, to one decimal place?

7. A sample of 6 values has Σx = 90 and Σx² = 1420. Find the estimate of the population standard deviation, to one decimal place.

Something to think about

A sample of size 1 has no spread at all, and the formula with n − 1 gives 0 ÷ 0. What does that say about trying to estimate a spread from a single value? And why does the shortfall from dividing by n matter less when the sample is large?

One way to think about it

One value cannot show any spread, so no estimate of a spread is possible: the formula has nothing to work with, and 0 ÷ 0 is the formula saying so. The factor (n − 1) ÷ n is close to 1 when n is large, such as 99 ÷ 100, so dividing by n or by n − 1 hardly differs. The correction matters most for small samples.

Exit ticket

8. A pupil surveys 30 pupils from one class of a school of 900 about hours of sleep, and uses the mean to describe the school. What is the main threat to this estimate?