Confidence Intervals for Distributions
[!example]
12 measurements are taken of the pH level of a solution with a known pH of 5 to ensure the proper calibration of a pH meter. measurements are independent observations from a normal distribution with a population standard deviation of 0.3:
Known:
- Find the 95% confidence interval for
Sinceis known, we use normal distribution instead of T-distribution
\left[ \frac{51.2}{12} - 1.95 \frac{0.3}{\sqrt{ 12 }}, \frac{61.2}{12} + 1.96 \frac{0.3}{\sqrt{ 12 }} \right] = [4.93, 5.27]
\begin{align}
0.05 & > \frac{\sigma}{\sqrt{ n }} \
n & > \left( \frac{0.3}{0.05} \right)^2 \
n & > 36
\end
\oline{x} \pm t^* \frac{s}{\sqrt{ n }}
\begin{align}
s^2 & = \frac{1}{n - 1} \left[ \sum x_{i}^2 - \frac{\left( \sum x_{i}^2 \right)^2}{n} \right] \
& = \frac{1}{11} \left( 313.42 - \frac{61.2^2}{12} \right) \
& = \dots
\end
\begin{align}
X & \sim \text{# adults taking supplements more than 4 times a week} \
X & \sim \op{Bin}(n, p) \
X & \sim \mathcal{N}(n\theta, n\theta(1 - \theta))
\end
\hat{\theta} \pm z^* \sqrt{ \frac{\hat{\theta}(1 - \hat{\theta})}{n} }
\begin{align}
0.25 & > z^*\sqrt{ \frac{\hat{\theta}(1 - \hat{\theta})}{n} } \
0.025 & > 1.96 \sqrt{ \frac{0.5}{n} } \
& \vdots \
n & > 1537
\end
\hat{\theta}(1 - \hat{\theta})