5 A researcher is investigating whether doing yoga has any effect on quality of sleep in older people. The researcher selects a random sample of 40 older people, who then complete a yoga course. Before they start the course and again at the end, the 40 people fill in a questionnaire which measures their perceived sleep quality. The higher the score, the better is the perceived quality of sleep.
The researcher uses software to produce a 90\% confidence interval for the difference in mean sleep quality (sleep quality after the course minus sleep quality before the course). The output from the software is shown below.
Z Estimate of a Mean
Confidence level □ 0.9
Sample
Result
Z Estimate of a Mean
| Mean | 0.586 |
| s | 2.14 |
| SE | 0.3384 |
| N | 40 |
| Lower limit | 0.029 |
| Upper limit | 1.143 |
| Interval | \(0.586 \pm 0.557\) |
- Explain why the confidence interval is based on the Normal distribution even though the distribution of the population of differences is not known.
- Explain whether the confidence interval suggests that the mean sleep qualities before and after completing a yoga course are different.
- In the output from the software, SE stands for 'standard error'.
- Explain what standard error is.
- Show how the standard error was calculated in this case.
- A colleague of the researcher suggests that the confidence level should have been \(95 \%\) rather than \(90 \%\).
Determine whether this would have made a difference to your answer to part (b).