Interpret simulated sampling distributions and interval variation lesson | Mathematical Methods (Year 12, VCE Units 3 & 4)

Q: Why is the simulation centre only near p?

A simulation contains a finite number of random samples, so its observed centre can differ from p even though repeated sampling is centred on p.

Q: How does sample size affect variation?

For a fixed population proportion, the variance p(1-p)/n becomes smaller as n increases, so sample proportions are less variable

Q: Why can confidence intervals from different samples have different endpoints?

Each sample can have a different sample proportion, and the interval is formed around that sample proportion with its own estimated margin

Q: Does this lesson assess execution in an external calculator or spreadsheet?

It checks algorithm tracing and interpretation. You can implement the stated procedure in your technology, but these checked entries do not prove that you independently ran it

Q: Can I enter a percentage for a proportion?

These scalar and generated proportion entries request a fraction or decimal. For example, enter 0.35 or 7/20 for 35%

Q: Can I use a calculator?

No. This topic is practised without a calculator

Q: Which year level and chapter is this?

Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 14: Statistical inference

Simulated sample proportions
Run	Success count	Sample proportion
Run 1	\(8\)	\(\frac{4}{25}\)
Run 2	\(9\)	\(\frac{9}{50}\)
Run 3	\(12\)	\(\frac{6}{25}\)
Run 4	\(12\)	\(\frac{6}{25}\)
Run 5	\(8\)	\(\frac{4}{25}\)
Run 6	\(12\)	\(\frac{6}{25}\)
Run 7	\(12\)	\(\frac{6}{25}\)
Run 8	\(9\)	\(\frac{9}{50}\)
Run 9	\(12\)	\(\frac{6}{25}\)
Run 10	\(19\)	\(\frac{19}{50}\)
Run 11	\(8\)	\(\frac{4}{25}\)
Run 12	\(9\)	\(\frac{9}{50}\)

Two simulated sample sizes
Run	Simulation A, n = 50	Simulation B, n = 10
Run 1	\(8: \frac{4}{25}\)	\(0: 0\)
Run 2	\(9: \frac{9}{50}\)	\(2: \frac{1}{5}\)
Run 3	\(12: \frac{6}{25}\)	\(2: \frac{1}{5}\)
Run 4	\(12: \frac{6}{25}\)	\(1: \frac{1}{10}\)
Run 5	\(8: \frac{4}{25}\)	\(2: \frac{1}{5}\)
Run 6	\(12: \frac{6}{25}\)	\(2: \frac{1}{5}\)
Run 7	\(12: \frac{6}{25}\)	\(3: \frac{3}{10}\)
Run 8	\(9: \frac{9}{50}\)	\(2: \frac{1}{5}\)
Run 9	\(12: \frac{6}{25}\)	\(2: \frac{1}{5}\)
Run 10	\(19: \frac{19}{50}\)	\(5: \frac{1}{2}\)
Run 11	\(8: \frac{4}{25}\)	\(1: \frac{1}{10}\)
Run 12	\(9: \frac{9}{50}\)	\(2: \frac{1}{5}\)

Simulated sample proportions
Run	Success count	Sample proportion
Run 1	\(3\)	\(\frac{3}{25}\)
Run 2	\(9\)	\(\frac{9}{25}\)
Run 3	\(7\)	\(\frac{7}{25}\)
Run 4	\(7\)	\(\frac{7}{25}\)
Run 5	\(7\)	\(\frac{7}{25}\)
Run 6	\(6\)	\(\frac{6}{25}\)
Run 7	\(7\)	\(\frac{7}{25}\)
Run 8	\(9\)	\(\frac{9}{25}\)
Run 9	\(7\)	\(\frac{7}{25}\)
Run 10	\(4\)	\(\frac{4}{25}\)
Run 11	\(6\)	\(\frac{6}{25}\)
Run 12	\(5\)	\(\frac{1}{5}\)
Run 13	\(5\)	\(\frac{1}{5}\)
Run 14	\(4\)	\(\frac{4}{25}\)

Simulated sample proportions
Run	Success count	Sample proportion
Run 1	\(2\)	\(\frac{1}{10}\)
Run 2	\(6\)	\(\frac{3}{10}\)
Run 3	\(1\)	\(\frac{1}{20}\)
Run 4	\(6\)	\(\frac{3}{10}\)
Run 5	\(4\)	\(\frac{1}{5}\)
Run 6	\(2\)	\(\frac{1}{10}\)
Run 7	\(4\)	\(\frac{1}{5}\)
Run 8	\(2\)	\(\frac{1}{10}\)
Run 9	\(7\)	\(\frac{7}{20}\)
Run 10	\(1\)	\(\frac{1}{20}\)
Run 11	\(6\)	\(\frac{3}{10}\)
Run 12	\(3\)	\(\frac{3}{20}\)

Simulated sample proportions
Run	Success count	Sample proportion
Run 1	\(41\)	\(\frac{41}{50}\)
Run 2	\(38\)	\(\frac{19}{25}\)
Run 3	\(42\)	\(\frac{21}{25}\)
Run 4	\(39\)	\(\frac{39}{50}\)
Run 5	\(38\)	\(\frac{19}{25}\)
Run 6	\(43\)	\(\frac{43}{50}\)
Run 7	\(37\)	\(\frac{37}{50}\)
Run 8	\(41\)	\(\frac{41}{50}\)
Run 9	\(43\)	\(\frac{43}{50}\)
Run 10	\(40\)	\(\frac{4}{5}\)
Run 11	\(41\)	\(\frac{41}{50}\)
Run 12	\(39\)	\(\frac{39}{50}\)

Simulated sample proportions and intervals
Run	Success count	Sample proportion	Approximate interval for p
Run 1	\(10\)	\(\frac{2}{5}\)	\(\left[0.21,0.59\right]\)
Run 2	\(4\)	\(\frac{4}{25}\)	\(\left[0.02,0.30\right]\)
Run 3	\(12\)	\(\frac{12}{25}\)	\(\left[0.28,0.68\right]\)
Run 4	\(7\)	\(\frac{7}{25}\)	\(\left[0.10,0.46\right]\)
Run 5	\(6\)	\(\frac{6}{25}\)	\(\left[0.07,0.41\right]\)
Run 6	\(7\)	\(\frac{7}{25}\)	\(\left[0.10,0.46\right]\)
Run 7	\(8\)	\(\frac{8}{25}\)	\(\left[0.14,0.50\right]\)
Run 8	\(10\)	\(\frac{2}{5}\)	\(\left[0.21,0.59\right]\)
Run 9	\(9\)	\(\frac{9}{25}\)	\(\left[0.17,0.55\right]\)
Run 10	\(8\)	\(\frac{8}{25}\)	\(\left[0.14,0.50\right]\)

Simulated sample proportions
Run	Success count	Sample proportion
Run 1	\(3\)	\(\frac{3}{25}\)
Run 2	\(7\)	\(\frac{7}{25}\)
Run 3	\(7\)	\(\frac{7}{25}\)
Run 4	\(7\)	\(\frac{7}{25}\)
Run 5	\(13\)	\(\frac{13}{25}\)
Run 6	\(5\)	\(\frac{1}{5}\)
Run 7	\(4\)	\(\frac{4}{25}\)
Run 8	\(7\)	\(\frac{7}{25}\)

Simulated sample proportions
Run	Success count	Sample proportion
Run 1	\(5\)	\(\frac{1}{2}\)
Run 2	\(5\)	\(\frac{1}{2}\)
Run 3	\(5\)	\(\frac{1}{2}\)
Run 4	\(7\)	\(\frac{7}{10}\)
Run 5	\(8\)	\(\frac{4}{5}\)
Run 6	\(4\)	\(\frac{2}{5}\)
Run 7	\(5\)	\(\frac{1}{2}\)
Run 8	\(4\)	\(\frac{2}{5}\)

Simulated sample proportions
Run	Success count	Sample proportion
Run 1	\(19\)	\(\frac{19}{50}\)
Run 2	\(29\)	\(\frac{29}{50}\)
Run 3	\(21\)	\(\frac{21}{50}\)
Run 4	\(34\)	\(\frac{17}{25}\)
Run 5	\(15\)	\(\frac{3}{10}\)
Run 6	\(22\)	\(\frac{11}{25}\)
Run 7	\(22\)	\(\frac{11}{25}\)
Run 8	\(22\)	\(\frac{11}{25}\)

Simulated sample proportions and intervals
Run	Success count	Sample proportion	Approximate interval for p
Run 1	\(7\)	\(\frac{7}{25}\)	\(\left[0.10,0.46\right]\)
Run 2	\(14\)	\(\frac{14}{25}\)	\(\left[0.37,0.75\right]\)
Run 3	\(14\)	\(\frac{14}{25}\)	\(\left[0.37,0.75\right]\)
Run 4	\(16\)	\(\frac{16}{25}\)	\(\left[0.45,0.83\right]\)
Run 5	\(15\)	\(\frac{3}{5}\)	\(\left[0.41,0.79\right]\)
Run 6	\(14\)	\(\frac{14}{25}\)	\(\left[0.37,0.75\right]\)
Run 7	\(15\)	\(\frac{3}{5}\)	\(\left[0.41,0.79\right]\)
Run 8	\(17\)	\(\frac{17}{25}\)	\(\left[0.50,0.86\right]\)
Run 9	\(12\)	\(\frac{12}{25}\)	\(\left[0.28,0.68\right]\)
Run 10	\(15\)	\(\frac{3}{5}\)	\(\left[0.41,0.79\right]\)
Run 11	\(11\)	\(\frac{11}{25}\)	\(\left[0.25,0.63\right]\)
Run 12	\(14\)	\(\frac{14}{25}\)	\(\left[0.37,0.75\right]\)
Run 13	\(13\)	\(\frac{13}{25}\)	\(\left[0.32,0.72\right]\)
Run 14	\(13\)	\(\frac{13}{25}\)	\(\left[0.32,0.72\right]\)

Repeated sampling and interval variation

Specify independent trials, separate n from R, and distinguish fixed population truth from varying samples and intervals

\widehat p=X/n,\quad E(\widehat p)=p,\quad\operatorname{Var}(\widehat p)=p(1-p)/n

Choose the experiment Specify a fixed population success probability p, n independent trials per sample and R repeated samples. Decide which outcomes count as success before drawing
Start a fresh sample count For each repetition r = 1,…,R, set X = 0. Keep the same p and n while comparing repeated samples
Generate each trial Draw U uniformly from [0,1). If U < p, add 1 to X. Repeat this step n times using fresh independent draws
Record and repeat Store X/n as one sample proportion, then return to X = 0 for the next sample. Do not carry successes into the next sample
Summarise the repetitions Inspect the distribution of the R proportions. When comparing sample sizes, keep p fixed; when comparing probabilities, keep n fixed
Form each interval from its own sample Use that sample’s observed proportion and n. The usual normal interval below is approximate and requires an appropriate independent random sample and sufficiently large success/failure counts \(\displaystyle \widehat p\pm1.96\sqrt{\frac{\widehat p(1-\widehat p)}{n}}\)
Check coverage only when p is known In a simulation the chosen p is known, so count an interval if lower ≤ p ≤ upper. In a real inference problem p is unknown, so the realised interval’s coverage cannot usually be observed
Keep confidence and observed coverage separate A nominal 95% procedure is intended to cover the fixed p in about 95% of repetitions under its approximation conditions. A finite run need not give exactly 95% and the procedure need not have exact nominal coverage

Worked example

Trace two independent five-trial samples with p = 0.6 using the ten supplied uniform draws

U=(0.12,0.81,0.43,0.02,0.65;\ 0.58,0.31,0.97,0.24,0.49)

\(\displaystyle X_1=3,\quad\widehat p_1=3/5\)Count the first group below 0.6
\(\displaystyle X_2=4,\quad\widehat p_2=4/5\)Reset the count and use the next group
\(\displaystyle \overline{\widehat p}=\frac{3/5+4/5}{2}=7/10\ne3/5\)Average the two stored proportions; the finite mean need not equal p.

Common mistakes

Treating one run as the centre

✕\(\displaystyle \text{centre}=\widehat p_4\)✓\(\displaystyle \text{centre}\approx p\)

One run is one observation, while the centre describes the repeated-sampling distribution

Claiming a larger sample is more variable

✕\(\displaystyle 50>10\Rightarrow\frac{p(1-p)}{50}>\frac{p(1-p)}{10}\)✓\(\displaystyle 50>10\Rightarrow\frac{p(1-p)}{50}<\frac{p(1-p)}{10}\)

For a fixed population proportion, increasing n reduces the variance of the sample proportion

Reading a count as a proportion

✕\(\displaystyle \text{centre}=14\)✓\(\displaystyle \widehat p_1=\frac{14}{25}=0.56\)

A success count must be divided by the sample size before it becomes a sample proportion

Confusing repetitions with sample size

✕Use p(1−p)/R for the variance of one sample proportion✓Use p(1−p)/n.

Each proportion is calculated from n observations; R counts repeated samples

Resetting the random stream per sample

✕Restart the same seed before every sample✓Seed once, then use fresh draws while resetting only the success count

Restarting produces repeated copies rather than the intended independent samples

Forcing nominal coverage

✕Change intervals until exactly 95% contain p.✓Apply the same declared procedure and report the observed fraction

Finite-run variability and approximation error must not be edited out

Difficulty levels

introIdentify the centre of a simulated distribution and compare two sample sizes

standardCalculate one run proportion and identify the centre, spread or shape from simulated data

challengeCount intervals containing the population proportion and interpret interval variation

examComplete a two-part interpretation with one exact calculation from the simulated table

Questions about this topic

Why is the simulation centre only near p?+

A simulation contains a finite number of random samples, so its observed centre can differ from p even though repeated sampling is centred on p.

How does sample size affect variation?+

For a fixed population proportion, the variance p(1-p)/n becomes smaller as n increases, so sample proportions are less variable

Why can confidence intervals from different samples have different endpoints?+

Each sample can have a different sample proportion, and the interval is formed around that sample proportion with its own estimated margin

Does this lesson assess execution in an external calculator or spreadsheet?+

It checks algorithm tracing and interpretation. You can implement the stated procedure in your technology, but these checked entries do not prove that you independently ran it

Can I enter a percentage for a proportion?+

These scalar and generated proportion entries request a fraction or decimal. For example, enter 0.35 or 7/20 for 35%

Can I use a calculator?+

No. This topic is practised without a calculator

Which year level and chapter is this?+

Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 14: Statistical inference

In this chapter

Populations, samples, parameters, statistics and sampling methods Allocate a stratified random sample Find a percentage of a quantity Calculate a sample proportion and list its possible values Exact sampling distribution of a sample proportion Finite events, probability assignments and complements Conditional probability from exact representations Evaluate factorial and combination expressions Convert between fractions, decimals and percentages Large-sample conditions and the sample-proportion distribution Calculate a standard score or actual value Use z-score boundaries to estimate normal-model percentages and counts Calculate the standard deviation from the formula Solve a linear equation Solve a quadratic exactly with the quadratic formula Approximate confidence intervals for a population proportion Rearrange literal equations and recognise equivalent line forms Interpret simulated sampling distributions and interval variation Design and interpret a probability simulation

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Interpret simulated sampling distributions and interval variation

Contents

Simulated sampling distributions

Write the repeated-sampling algorithm

Distinguish n from R

Build an interval for each sufficiently large sample

Compare four deliberately selected outcomes

Implement and verify the procedure in your technology

Checks to finish