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

\begin{center}
\begin{tabular}{ r l }
Mean & 0.586 \\
$s$ & 2.14 \\
 & 40 \\
\end{tabular}
\end{center}

Result\\
Z Estimate of a Mean

\begin{center}
\begin{tabular}{ l l }
Mean & 0.586 \\
s & 2.14 \\
SE & 0.3384 \\
N & 40 \\
Lower limit & 0.029 \\
Upper limit & 1.143 \\
Interval & $0.586 \pm 0.557$ \\
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Explain why the confidence interval is based on the Normal distribution even though the distribution of the population of differences is not known.
\item Explain whether the confidence interval suggests that the mean sleep qualities before and after completing a yoga course are different.
\item In the output from the software, SE stands for 'standard error'.
\begin{enumerate}[label=(\roman*)]
\item Explain what standard error is.
\item Show how the standard error was calculated in this case.
\end{enumerate}\item 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).
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics Major 2024 Q5 [10]}}