Interpret simulated sampling distributions and interval variation
A transport survey records whether a commuter uses public transport for a trip
The population proportion for commuters who use public transport for the trip is \(p=\frac{1}{3}\)
A simulation took \(8\) random samples of \(n=20\) commuters. The success count and sample proportion for each sample are shown
\[\widehat p=\frac{X}{n}\]
| Run | Success count | Sample proportion |
|---|---|---|
| Run 1 | \(3\) | \(\frac{3}{20}\) |
| Run 2 | \(5\) | \(\frac{1}{4}\) |
| Run 3 | \(10\) | \(\frac{1}{2}\) |
| Run 4 | \(9\) | \(\frac{9}{20}\) |
| Run 5 | \(5\) | \(\frac{1}{4}\) |
| Run 6 | \(3\) | \(\frac{3}{20}\) |
| Run 7 | \(9\) | \(\frac{9}{20}\) |
| Run 8 | \(5\) | \(\frac{1}{4}\) |
A second simulation used the same population proportion and the same number of samples, but its sample size was \(n=10\)
| Run | Simulation A, n = 20 | Simulation B, n = 10 |
|---|---|---|
| Run 1 | \(3: \frac{3}{20}\) | \(2: \frac{1}{5}\) |
| Run 2 | \(5: \frac{1}{4}\) | \(3: \frac{3}{10}\) |
| Run 3 | \(10: \frac{1}{2}\) | \(1: \frac{1}{10}\) |
| Run 4 | \(9: \frac{9}{20}\) | \(6: \frac{3}{5}\) |
| Run 5 | \(5: \frac{1}{4}\) | \(6: \frac{3}{5}\) |
| Run 6 | \(3: \frac{3}{20}\) | \(3: \frac{3}{10}\) |
| Run 7 | \(9: \frac{9}{20}\) | \(6: \frac{3}{5}\) |
| Run 8 | \(5: \frac{1}{4}\) | \(3: \frac{3}{10}\) |