Population/sample proportions | Year 12

01 / LEARN

Understand the idea.

Define a binary outcome

Define success precisely so every valid observation can be classified as success or failure. The population proportion $p$ is the fraction meeting that rule in the target population. For $x$ successes among $n$ valid observations, estimate it with $\hat{p} = \frac{x}{n}$ .

The parameter $p$ is fixed for the specified population and time, while $\hat{p}$ changes between samples. Write the estimate as a decimal or multiply by 100 to give a percentage. Keep the success definition and population attached to the number.

Use the correct denominator

Count valid answers in the denominator. Missing answers are unknown, not automatically failures. A proportion calculated among respondents describes those respondents directly; extending it to everyone selected, or to the whole population, requires attention to the non-response process.

A binomial model assumes fixed $n$ , two outcomes, independent observations and a common success probability. Without-replacement sampling from a finite population is not exactly independent. When the sampling fraction is substantial, the ordinary independent-trial model and standard errors may need adjustment.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Count to estimate

138 of 240 randomly selected students meet a travel criterion. Estimate the population proportion.

Worked solution

1
Identify successes and valid observations.
$x = 138, n = 240$
Success means meeting the stated travel criterion; the denominator is the total number observed.
2
Calculate the sample proportion.
$\hat{p} = \frac{138}{240} = 0.575$
The observed success count divided by sample size estimates the population proportion.
3
Convert to a percentage.
$0.575 \times 100 % = 57.5 %$
This is a sample estimate of 57.5%, rather than a known exact population percentage.

Answer

0.575, or 57.5%.

EXAMPLE 02

Missing answers

200 adults are selected; 150 respond and 90 say yes. Estimate the respondent proportion and give a limitation.

Worked solution

1
Identify the valid response denominator.
$n_{respondents} = 150, x = 90$
The task asks for the proportion among respondents, so the 50 unanswered selections are excluded from this denominator.
2
Calculate the respondent proportion.
${\hat{p}}_{respondents} = \frac{90}{150} = 0.60$
60% of the people who answered said yes.
3
Identify what remains unknown.
$200 - 150 = 50$
The answers of these 50 non-respondents are missing; treating them as no would invent information.
4
Limit the population conclusion.
The respondent estimate need not represent all adults if willingness or ability to respond relates to the answer, even when initial selection was random.

Answer

0.60 among respondents; response may relate to the answer.

03 / YOUR TURN

Try it yourself.

Check your understanding

Find the sample proportion for 45 successes among 75 observations.

Show answer & reasoning

0.6.

There are 45 successes among 75 valid observations, so $\hat{p} = \frac{45}{75} = 0.6$ . This is a sample estimate of 60%, not a claim that the population proportion is known exactly.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Find the sample proportion for 84 successes among 140.
2 marks
Find the success count for $n = 250$ and sample proportion 0.36.
2 marks
A report writes $p = 0.42$ because 84 of 200 sampled people said yes. Correct the notation.
2 marks
Of 180 selected people, 120 respond and 72 say yes. Give the respondent estimate.
2 marks
Why might 20 responses from one household not be independent observations of community preference?
2 marks
Why should a success rule not be changed after inspecting results?
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find the sample proportion for 84 successes among 140.
2 marks
$\frac{84}{140} = 0.6$ .
$0.6$ .
Find the success count for $n = 250$ and sample proportion 0.36.
2 marks
$250 \times 0.36 = 90$ .
$90$ .
A report writes $p = 0.42$ because 84 of 200 sampled people said yes. Correct the notation.
2 marks
Reserve $p$ for the population parameter.
$\hat{p} = 0.42$ estimates $p$ .
Of 180 selected people, 120 respond and 72 say yes. Give the respondent estimate.
2 marks
$\frac{72}{120} = 0.6$ .
$\frac{72}{180}$ instead measures the fraction selected who both responded and said yes.
$0.6$ .
Why might 20 responses from one household not be independent observations of community preference?
2 marks
The count alone does not establish binomial assumptions.
Shared circumstances and influence can correlate their answers.
Why should a success rule not be changed after inspecting results?
2 marks
Pre-specify measurement rules and transparently explain justified changes.
A selective redefinition can favour a desired conclusion.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods: sampling distributions and confidence intervals for means and proportions. Mean tasks distinguish known population spread from sufficiently large-sample estimated-spread normal intervals. Repeated-interval computational simulation, approximation conditions, design and interpretation are included.

Official curriculum source ↗

In this lesson

Understand the idea.

Define a binary outcome

Use the correct denominator

See the method in action.

Count to estimate

Identify successes and valid observations.

Calculate the sample proportion.

Convert to a percentage.

Missing answers

Identify the valid response denominator.

Calculate the respondent proportion.

Identify what remains unknown.

Limit the population conclusion.

Try it yourself.

Check your understanding

Keep practising