If \(X\) is randomly distributed with mean \(\mu\) and sd \(\sigma\), then with an adequate sample size \(n\) the distribution of the sample mean \(\overline{X}\) is approximately normal with mean \(E(\overline{X})\) and \(\operatorname{sd}(\overline{X}) = \frac{\sigma}{\sqrt{n}}\).
+ \section{Confidence intervals}
+
+ \begin{itemize}
+ \item \textbf{Point estimate:} single-valued estimate of the population mean from the value of the sample mean \(\overline{x}\)
+ \item \textbf{Interval estimate:} confidence interval for population mean \(\mu\)
+ \end{itemize}
+
+ \subsection*{95% confidence interval}
+
+ \[ \left( \overline{x} \pm 1.96 \dfrac{\sigma}{\sqrt{n}} \]
+
+ where: \\
+ \(\mu\) is the population mean (unknown) \\
+ \(\overline{x}\) is the sample mean \\
+ \(\sigma\) is the population sd \\
+ \(n\) is the sample size from which \(\overline{x}\) was calculated
+
+ Always express \(z\) as +ve. Express confidence \textit{interval} as ordered pair.
+
+ \colorbox{cas}{\textbf{On CAS}}
+
+ Menu \(\rightarrow\) Stats \(\rightarrow\) Calc \(\rightarrow\) Interval \\
+ Set Type = One-Sample Z Int, Variable
+
+ \subsection*{Interpretation of confidence intervals}
+
+ 95% confidence interval \(\implies\) 95% of samples will contain population mean \(\mu\).
+
+ \subsection*{Margin of error}
+
+ For 95% confidence interval for \(\mu\), margin of error \(M\) is:
+
+ \begin{align*}
+ M &= 1.96 \times \dfrac{\sigma}{\sqrt{n}} \\
+ \implies n &= \left( \dfrac{1.96 \sigma}{M} \right)^2
+ \end{align*}
+
+ \subsection*{General case}
+
+ A confidence interval of \(C\)% is given by
+
+ A 95% confidence interval for \(\mu\) will have \(M\) when
+ \[ \overline{x} \pm k \dfrac{\sigma}{\sqrt{n}} \]
+
+ where \(k\) is such that \(\Pr(-k < Z < k) = \frac{C}{100}\)
+
+
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