Confidence Interval

The confidence level describes the uncertainty associated with a sampling method. Suppose we used the same sampling method to select different samples and to compute a different interval estimate for each sample. Some interval estimates would include the true population parameter and some would not. A 90% confidence level means that we would expect 90% of the interval estimates to include the population parameter; A 95% confidence level means that 95% of the intervals would include the parameter; and so on.

$P(x_l <= X <= x_h) = 1 - \alpha$

Choose $x_l$ and $x_h$ such that,

$P\big(X \leq x_l \big) = \frac{\alpha}{2}, \quad \textrm{and} \quad P\big(X \geq x_h) = \frac{\alpha}{2}$

or,

$F_X(x_l) = \frac{\alpha}{2}, \quad \textrm{and} \quad F_X(x_h) = 1 - \frac{\alpha}{2}$

re-writing,

$x_l = F^{-1}_X \left(\frac{\alpha}{2}\right), \quad \textrm{and} \quad x_h = F^{-1}_X\left(1-\frac{\alpha}{2}\right)$

Standard Deviation / Standard Error

sample mean

$\bar {x} \qquad \qquad \frac{s}{\sqrt{n}}$

sample proportion

$\qquad \qquad p \qquad \qquad \sqrt{\frac{p (1-p)}{n}}$

Questions

1. Given $\sigma^2 = 4$ and margin of error <= 0.25, how many samples are required?

$\begin{aligned} \big[\overline{X}-0.25, \overline{X}+0.25\big] \\ \left[\overline{X}- z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}} , \overline{X}+ z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}}\right] \\ z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}} & = 0.25 \\ z_{\frac{\alpha}{2}}=z_{0.05} = \Phi^{-1}(1-0.05) & = 1.645 \\ 1.645 \frac{2}{\sqrt{n}} & = 0.25 \end{aligned}$

$\begin{aligned} \dot{x} & = \sigma(y-x) \\ \dot{y} & = \rho x - y - xz \\ \dot{z} & = -\beta z + xy \end{aligned}$