3.2 Confidence intervals ii

Confidence intervals ii

Confidence intervals for means

In the preceding considerations ( Confidence Intervals I ), the confidence interval for the mean $\mu$ of a normal distribution was found, assuming that the value of the standard deviation $\sigma$ is known. However, in most applications, the value of the standard deviation $\sigma$ is rather unknown, although in some cases one might have a very good idea about its value.

Suppose that the underlying distribution is normal and that ${\sigma }^2$ is unknown. It is shown that given random sample $X_1,X_2,\mathrm{...},X_n$ from a normal distribution, the statistic $$T=\frac{\overline{X}-\mu }{S/\sqrt{n}}$$ has a t distribution with $r=n-1$ degrees of freedom, where $S^2$ is the usual unbiased estimator of ${\sigma }^2$ , (see, t distribution ).

Select $t_{\alpha /2}\left(n-1\right)$ so that $$P\left[T\ge t_{\alpha /2}\left(n-1\right)\right]=\alpha /2.$$ Then

$$\begin{array}{l}1-\alpha =P\left[-t_{\alpha /2}\left(n-1\right)\le \frac{\overline{X}-\mu }{S/\sqrt{n}}\le t_{\alpha /2}\left(n-1\right)\right]\\ =P\left[-t_{\alpha /2}\left(n-1\right)\frac{S}{\sqrt{n}}\le \overline{X}-\mu \le t_{\alpha /2}\left(n-1\right)\frac{S}{\sqrt{n}}\right]\\ =P\left[-\overline{X}-t_{\alpha /2}\left(n-1\right)\frac{S}{\sqrt{n}}\le -\mu \le -\overline{X}+t_{\alpha /2}\left(n-1\right)\frac{S}{\sqrt{n}}\right]\\ =P\left[\overline{X}-t_{\alpha /2}\left(n-1\right)\frac{S}{\sqrt{n}}\le -\mu \le \overline{X}+t_{\alpha /2}\left(n-1\right)\frac{S}{\sqrt{n}}\right].\end{array}$$

Thus the observations of a random sample provide a $\overline{x}$ and ${\text{s}}^{\text{2}}$ and $\overline{x}-t_{\alpha /2}\left(n-1\right)\frac{s}{\sqrt{n}},\overline{x}+t_{\alpha /2}\left(n-1\right)\frac{s}{\sqrt{n}}$ is a $100\left(1-\alpha \right)\%$ interval for $\mu$ .

Let T have a t distribution with n -1 degrees of freedom. Then, $t_{\alpha /2}\left(n-1\right)>z_{\alpha /2}$ . Consequently, the interval $\overline{x}\pm z_{\alpha /2}\sigma /\sqrt{n}$ is expected to be shorter than the interval $\overline{x}\pm t_{\alpha /2}\left(n-1\right)s/\sqrt{n}$ . After all, there gives more information, namely the value of $\sigma$ , in construction the first interval. However, the length of the second interval is very much dependent on the value of s . If the observed s is smaller than $\sigma$ , a shorter confidence interval could result by the second scheme. But on the average, $\overline{x}\pm z_{\alpha /2}\sigma /\sqrt{n}$ is the shorter of the two confidence intervals.

If it is not possible to assume that the underlying distribution is normal but $\mu$ and $\sigma$ are both unknown, approximate confidence intervals for $\mu$ can still be constructed using $$T=\frac{\overline{X}-\mu }{S/\sqrt{n}},$$ which now only has an approximate t distribution.

Generally, this approximation is quite good for many normal distributions, in particular, if the underlying distribution is symmetric, unimodal, and of the continuous type. However, if the distribution is highly skewed , there is a great danger using this approximation. In such a situation, it would be safer to use certain nonparametric method for finding a confidence interval for the median of the distribution.

Confidence interval for variances

The confidence interval for the variance ${\sigma }^2$ is based on the sample variance $$S^2=\frac{1}{n-1}{{\displaystyle \sum _{i=1}^n\left(X_i-\overline{X}\right)}}^2.$$

In order to find a confidence interval for ${\sigma }^2$ , it is used that the distribution of $\left(n-1\right)S^2/{\sigma }^2$ is ${\chi }^2\left(n-1\right)$ . The constants a and b should selected from tabularized Chi Squared Distribution with n -1 degrees of freedom such that $$P\left(a\le \frac{\left(n-1\right)S^2}{{\sigma }^2}\le b\right)=1-\alpha .$$

That is select a and b so that the probabilities in two tails are equal: $$a={\chi }_{1-\alpha /2}^2\left(n-1\right)$$ and $$b={\chi }_{\alpha /2}^2\left(n-1\right).$$ Then, solving the inequalities, we have $$1-\alpha =P\left(\frac{a}{\left(n-1\right)S^2}\le \frac{1}{{\sigma }^2}\le \frac{b}{\left(n-1\right)S^2}\right)=P\left(\frac{\left(n-1\right)S^2}{b}\le {\sigma }^2\le \frac{\left(n-1\right)S^2}{a}\right).$$

Thus the probability that the random interval $${\text{ [(n-1)S}}^{\text{2}}{\text{/b, (n-1)S}}^{\text{2}}\text{/a]}$$ contains the unknown ${\sigma }^2$ is 1- $\alpha$ . Once the values of $X_1,X_2,\mathrm{...},X_n$ are observed to be $x_1,x_2,\mathrm{...},x_n$ and $s^2$ computed, then the interval $${\text{[(n-1)S}}^{\text{2}}{\text{/b, (n-1)S}}^{\text{2}}\text{/a]}$$ is a $100\left(1-\alpha \right)\%$ confidence interval for ${\sigma }^2$ .

It follows that $$\left[\sqrt{\left(n-1\right)/bs},\sqrt{\left(n-1\right)/as}\right]$$ is a $100\left(1-\alpha \right)\%$ confidence interval for $\sigma$ , the standard deviation.

Although a and b are generally selected so that the probabilities in the two tails are equal, the resulting $100\left(1-\alpha \right)\%$ confidence interval is not the shortest that can be formed using the available data. The tables and appendixes gives solutions for a and b that yield confidence interval of minimum length for the standard deviation.