Interpret simulated sampling distributions and interval variation
A library record checks whether a borrowed item is returned by the due date
The population proportion for borrowed items returned by the due date is \(p=\frac{3}{4}\)
A simulation took \(12\) random samples of \(n=50\) borrowed items. The success count and sample proportion for each sample are shown
\[\widehat p=\frac{X}{n}\]
| Run | Success count | Sample proportion |
|---|---|---|
| Run 1 | \(39\) | \(\frac{39}{50}\) |
| Run 2 | \(37\) | \(\frac{37}{50}\) |
| Run 3 | \(38\) | \(\frac{19}{25}\) |
| Run 4 | \(40\) | \(\frac{4}{5}\) |
| Run 5 | \(41\) | \(\frac{41}{50}\) |
| Run 6 | \(40\) | \(\frac{4}{5}\) |
| Run 7 | \(35\) | \(\frac{7}{10}\) |
| Run 8 | \(33\) | \(\frac{33}{50}\) |
| Run 9 | \(34\) | \(\frac{17}{25}\) |
| Run 10 | \(38\) | \(\frac{19}{25}\) |
| Run 11 | \(38\) | \(\frac{19}{25}\) |
| Run 12 | \(42\) | \(\frac{21}{25}\) |
A second simulation used the same population proportion and the same number of samples, but its sample size was \(n=25\)
| Run | Simulation A, n = 50 | Simulation B, n = 25 |
|---|---|---|
| Run 1 | \(39: \frac{39}{50}\) | \(18: \frac{18}{25}\) |
| Run 2 | \(37: \frac{37}{50}\) | \(16: \frac{16}{25}\) |
| Run 3 | \(38: \frac{19}{25}\) | \(16: \frac{16}{25}\) |
| Run 4 | \(40: \frac{4}{5}\) | \(17: \frac{17}{25}\) |
| Run 5 | \(41: \frac{41}{50}\) | \(15: \frac{3}{5}\) |
| Run 6 | \(40: \frac{4}{5}\) | \(16: \frac{16}{25}\) |
| Run 7 | \(35: \frac{7}{10}\) | \(22: \frac{22}{25}\) |
| Run 8 | \(33: \frac{33}{50}\) | \(22: \frac{22}{25}\) |
| Run 9 | \(34: \frac{17}{25}\) | \(15: \frac{3}{5}\) |
| Run 10 | \(38: \frac{19}{25}\) | \(17: \frac{17}{25}\) |
| Run 11 | \(38: \frac{19}{25}\) | \(18: \frac{18}{25}\) |
| Run 12 | \(42: \frac{21}{25}\) | \(20: \frac{4}{5}\) |