From: Andrew Lorimer <andrew@lorimer.id.au>
Date: Wed, 28 Aug 2019 04:03:45 +0000 (+1000)
Subject: [spec] confidence intervals/limits
X-Git-Tag: yr12~51
X-Git-Url: https://git.lorimer.id.au/notes.git/diff_plain/93eb70a4067f707438d4162bcef4943141871e43?ds=inline

[spec] confidence intervals/limits
---

diff --git a/spec/statistics.pdf b/spec/statistics.pdf
index 9f5622d..cd372e0 100644
Binary files a/spec/statistics.pdf and b/spec/statistics.pdf differ
diff --git a/spec/statistics.tex b/spec/statistics.tex
index 2228636..ee414c6 100644
--- a/spec/statistics.tex
+++ b/spec/statistics.tex
@@ -144,4 +144,51 @@
 
   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 \(\Pr(-k < Z < k) = \frac{C}{100}\)
+
+
 \end{document}