11.1 Population, sample and informal inference — Practise with help

Work each question closed book first. Write down whether your answer describes the sample or the population.

Warm-up: remember

Find the mean of $4, 7, 9, 12, 13$ .
Write $18$ out of $60$ as a fraction in its simplest form.
Work out $15\%$ of $240$ .

Answers

$\dfrac{4 + 7 + 9 + 12 + 13}{5} = \dfrac{45}{5} = 9$ .
$\dfrac{18}{60} = \dfrac{3}{10}$ .
$0.15 \times 240 = 36$ .

If question 2 was hard, revise simplifying fractions before you start.

Practise with help

1. A council wants to know the mean recycling weight of households in a town. It weighs the recycling from 250 households. Name the population and the sample.

Hint

The population is the whole group the council wants to describe. The sample is the 250 households it actually weighed.

Next step

Write one sentence for each. Start with "The population is every household in…" and "The sample is the 250 households that…".

Full working

The population is every household in the town. The sample is the 250 households whose recycling was weighed. The mean weight of the sample is an estimate of the mean weight of the population.

2. A sample of 50 batteries has a mean life of 12 hours. Estimate the total life, in hours, of 800 batteries of the same type.

Hint

The estimate of a total is the sample mean multiplied by the population size.

Next step

Multiply $12$ by $800$ . The sample mean is in hours, so the total is in hours.

Full working

\text{estimated total} = 12 \times 800 = 9600 \text{ hours}.

This is an estimate. The 800 batteries were not all tested.

3. In a sample of 150 light bulbs, 18 are faulty. Estimate the number of faulty bulbs in a batch of 2000.

Hint

First find the sample proportion. It is the number faulty divided by the sample size.

Next step

The sample proportion is $\dfrac{18}{150} = 0.12$ . Multiply it by $2000$ .

Full working

\text{estimated number} = \frac{18}{150} \times 2000 = 0.12 \times 2000 = 240.

About 240 bulbs in the batch are faulty.

4. In a sample of 500 people, 85 support a new park. Estimate the percentage of the population who support the park.

Hint

A percentage is a proportion multiplied by 100.

Next step

Work out $\dfrac{85}{500}$ , then multiply by $100$ .

Full working

\frac{85}{500} = 0.17, \qquad 0.17 \times 100 = 17\%.

About $17\%$ of the population support the park. This is a conclusion about the population, drawn from one sample.

5. Two samples of size 40 give means of $3.1$ and $3.5$ . Estimate the population mean and state one reason why the two samples disagree.

Hint

Both samples are the same size, so their means carry the same weight.

Next step

Find the midpoint of $3.1$ and $3.5$ : add them and divide by 2.

Full working

\text{estimate} = \frac{3.1 + 3.5}{2} = 3.3.

The samples disagree because of sampling variability: each sample contains different members of the population.

6. A sample of 20 students gives a mean screen time of 4 hours. Would you expect a sample of 200 students to give exactly the same mean? Explain your answer.

Hint

Think about what happens when more students are included. The sample still does not contain everyone.

Next step

Compare the two sample sizes, then say whether a larger sample removes sampling variability completely.

Full working

No. A larger sample usually gives a mean closer to the population mean, and its sample-to-sample variation is usually smaller. It does not give the population mean exactly, because it is still only part of the population. The estimate from the sample of 200 is likely to be more reliable than the estimate from the sample of 20.