Confidence Intervals for the Mean
Unknown Population Variance \(\sigma^2\)
Student \(t\) distribution
The (Student) \(t\)-distribution was developed by William Sealy Gosset, a statistician working at the Guinness Brewery in Dublin.
🍺 Guinness relied on small samples for quality control and \(\sigma\) was not known.
🍺 He developed a new distribution that accounted for the extra variability introduced by replacing \(\sigma\) with \(s\).
🍺 Since Guiness prohibited employees from publishing under their real name, he used the pseudonym “Student”.
Standard Normal vs \(t\)-Distribution
Rows of the t-table
rows:
the degrees of freedom \(\nu\)
Rows of the t-table
rows:
the degrees of freedom \(\nu\)
Rows of the t-table
rows:
the degrees of freedom \(\nu\)
Columns of the t-table
\(\phantom{\Pr(t_{2}>1.886) = 0.1}\)
Columns of the t-table
\(\phantom{\Pr(t_{2}>1.886) = 0.1}\)
Columns of the t-table
\(\Pr(t_{2}>1.886) = 0.1\)
Columns of the t-table
\(\Pr(t_{2}>2.920) = 0.05\)
Columns of the t-table
\(\Pr(t_{2}>4.303) = 0.025\)
Columns of the t-table
\(\Pr(t_{2}>6.965) = 0.01\)
Columns of the t-table
\(\Pr(t_{2}>9.925) = 0.005\)
Probablity Ranges
\(\Pr(t_{10}>2) = ??\) We know:
\(\Pr(t_{10}>2.228) = 0.025\)
Probablity Ranges
\(\Pr(t_{10}>2) = ??\) We know:
\(\Pr(t_{10}>1.812) = 0.05\)
Probablity Ranges
Visually we can see that
\(\textcolor{purple}{\text{Area in Purple}} < \textcolor{orange}{\text{Area in Orange} = 0.05}\) \(\textcolor{purple}{\text{Area in Purple}} > \textcolor{deepskyblue}{\text{Area in Blue} = 0.025}\)
In probability statements
\(\textcolor{purple}{\Pr(t_{10}>2)} < \textcolor{orange}{\Pr(t_{10}>1.812) = 0.05}\) \(\textcolor{purple}{\Pr(t_{10}>2)} < \textcolor{deepskyblue}{\Pr(t_{10}>2.228) = 0.025}\)
Hence …
\[\textcolor{deepskyblue}{0.025} < \textcolor{purple}{\Pr(t_{10}>2)} < \textcolor{orange}{0.05}\]
Example 1: with tables and R
Example 2
\(t_{\nu, \alpha/2}\) using the tables
Since 49 is not on our table, we round down to the nearest integer (in this case 38).
We need to find the \(t_{\nu, \alpha/2}\) that satisfies:
\[ \begin{align} \Pr(t_{50} > t_{\nu, \alpha/2}) &= 0.05\\ \implies t_{\nu, \alpha/2} &\approx 2.024 \end{align} \]
Important
Always round down1 to the nearest available df in the table to ensure the confidence level is at least as high as desired. Rounding down gives a larger \(t_{\nu, \alpha/2}\) than needed and therefore produces a more conservative confidence interval.
\(t_{\nu, \alpha/2}\) Comparison
Using the tables with conservative \(\nu\) = 38:
\[ \begin{align} \Pr(t_{50} > t_{\nu, \alpha/2}) &= 0.05\\ \implies t_{\nu, \alpha/2} &\approx 2.024\\ [2.7019&,3.6981]\\ \textcolor{green}{\text{Width}} &= \textcolor{green}{0.9962} \end{align} \]
Using exact R calculations (\(\nu = 49\)) \[ \begin{align} t_{\nu, \alpha/2} &= 2.0095752 \\ [2.7055&,3.6945]\\ \textcolor{purple}{\text{Width}} &= \textcolor{purple}{0.989} \end{align} \]
\(t_{\nu, \alpha/2}\) Comparison
If we incorrectly round up (to \(\infty\)) \[ \begin{align} \Pr(t_{50} > t_{\nu, \alpha/2}) &= 0.05\\ \implies t_{\nu, \alpha/2} &\approx 1.96\\ [2.7177&,3.6823]\\ \textcolor{red}{\text{Width}} &= \textcolor{red}{0.9646} \end{align} \]
Using R with the \(\nu = 50-1 = 49\)
\[ \begin{align} t_{\nu, \alpha/2} &= 2.0095752 \\ [2.7055&,3.6945]\\ \textcolor{purple}{\text{Width}} &= \textcolor{purple}{0.989} \end{align} \]
Code
library(car)
par(mfrow=c(2,2)) ; par(mar=c(2,4,2,0)+0.1)
# install once if needed
# install.packages("vioplot")
boxplot(data, main = "Boxplot")
hist(data, main = "Histogram", col = "skyblue", border = "white")
# qqnorm(data); qqline(data)
this = qqPlot(data, main = "QQ-plot")
vioplot(data, main = "Violin plot", col = "skyblue", horizontal = FALSE)
Plots for assessing normality. This is necessary to determine whether using a t-distribution for constructing the confidence interval is appropriate.