Appendices

Appendix A: Important Links and Additional Resources

Applets

• Normal Distribution Calculator
• Rossman and Chance Applets
• Simulating the Central Limit Theorem

Run R Online

• Run R Online
• WebR
• RStudio Cloud

Appendix B: Average Deviance

The deviance of an observation from its mean is $x - \bar{x}$. We denote the deviation for the $i$th observation as $x_i - \bar{x}$. So the sum over all $n$ deviances is

$$\begin{align} \text{Sum of Deviances} &= \Sigma_{i=1}^n (x_i - \bar{x}) \\ &= (x_1 - \bar{x}) + (x_2 - \bar{x}) + \dots + (x_{n-1} - \bar{x}) + (x_n - \bar{x}) \\ &= x_1 - \bar{x} + x_2 - \bar{x} + \dots + x_{n-1} - \bar{x} + x_n - \bar{x} \\ &= x_1 + x_2 + \dots + x_{n-1} + x_n - \bar{x} - \bar{x} - \dots - \bar{x} - \bar{x} \\ &= (x_1 + x_2 + \dots + x_{n-1} + x_n) - (\bar{x} + \bar{x} + \dots + \bar{x} + \bar{x}) \end{align}$$

where the first half is the sum over all of the $x$ values and the term ($\bar{x}$) appears $n$ times. So we can rewrite this as

$$\text{Sum of Deviances} = \Sigma_{i=1}^n (x_i) - n\bar{x}$$ Now notice that, because $\bar{x} = \frac{\Sigma_{i=1}^n (x_i)}{n}$, we can multiply $\Sigma_{i=1}^n (x_i)$ by $\frac{n}{n}$ to get $\frac{n\Sigma_{i=1}^n (x_i)}{n} = n\bar{x}$ and rewrite the sum over the deviances as

$$\begin{align} \text{Sum of Deviances} &= n\bar{x} - n\bar{x} \\ &=0 \end{align}$$

Appendix C: Deriving a Confidence Interval

Assume we are taking a sample from a normal distribution with mean $\mu$ and standard deviation $\sigma$. We will assume the value of $\sigma$ is known to us. Then $\bar{X}$ is Normal($\mu, \sigma/\sqrt{n}$). If we standardize $\bar{X}$, we get $$Z = \frac{\bar{X}-\mu}{\sigma/\sqrt{n}}.$$

We want some interval $(a,b)$. We will start by considering $a < Z < b$, so $a < Z$ and $Z < b$ (or $b > Z$). Then

$$\begin{aligned} Z &< b\\ \frac{\bar{X}-\mu}{\sigma/\sqrt{n}} &< b\\ \bar{X}-\mu &< b\sigma/\sqrt{n} \\ \bar{X}-b\sigma/\sqrt{n} &< \mu \end{aligned}$$

and

$$\begin{aligned} a &< Z \\ a &< \frac{\bar{X}-\mu}{\sigma/\sqrt{n}} \\ a\sigma/\sqrt{n} &< \bar{X}-\mu \\ \mu &< \bar{X}-a\sigma/\sqrt{n} \end{aligned}$$

putting these together, $$\bar{X}-b\frac{\sigma}{\sqrt{n}} < \mu < \bar{X}-a\frac{\sigma}{\sqrt{n}}.$$ If we want to be 95% confident, then we want $P(a < Z < b)=0.95$: $$P\left(\bar{X}-b\frac{\sigma}{\sqrt{n}} < \mu < \bar{X}-a\frac{\sigma}{\sqrt{n}}\right) = 0.95.$$ To calculate the 95% confidence interval, we need to find $a$ and $b$ such that $P(a < Z < b)=0.95$.

We want this interval to be as narrow (small) as possible. Why? Narrower intervals are more informative. If I say I’m 95% confident that tomorrow’s high will be between -100 and 200 degrees Fahrenheit, that’s a useless interval. If I change it to between 70 and 100, that’s a little better. Changing it to between 85 and 90 is even better. This is what we mean by more informative.

It turns out that with a symmetric distribution like the normal distribution, the way to make a confidence interval as narrow as possible is to take advantage of this symmetry. Each of the plots below show a shaded area of 0.95. The narrowest interval (along the horizontal axis) is the first interval, which is shaded on $(-1.96 < Z < 1.96)$.

Using the symmetry of the normal distribution, we find that the narrowest interval uses $a = -1.96$ and $b = 1.96$, which results in the 95% confidence interval $$\left(\bar{x} - z_*\frac{\sigma}{\sqrt{n}}, \bar{x} + z_*\frac{\sigma}{\sqrt{n}}\right)$$ where $z_* = 1.96$.