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4.2.1 One-sample mean

We construct confidence intervals for the population mean $\mu$ of a Normal distribution. Again we assume that our data $x_{1},\ldots,x_{n}$ are realisations of a sequence of IID random variables $X_{1},\ldots,X_{n}$ with $\operatorname{Normal}(\mu,\sigma^{2})$ distribution. As before, if the sample size is large enough the Normal assumption can be relaxed using the Central Limit Theorem. We obtain the 95% confidence interval first, before giving the formula more generally.

As we saw earlier,

\displaystyle\frac{\bar{X}-\mu}{S/\sqrt{n}}\sim t_{n-1}.

We can therefore say that

\displaystyle\Pr\left(t_{n-1}(0.025)\leq\frac{\bar{X}-\mu}{S/\sqrt{n}}\leq t_{% n-1}(0.975)\right)=0.95

where

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$t_{n-1}(0.025)$ is the 2.5% quantile of the $t_{n-1}$ -distribution, and
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$t_{n-1}(0.975)$ is the 97.5% quantile of the $t_{n-1}$ -distribution.

By the symmetry of the $t$ -distribution, $t_{n-1}(0.025)=-t_{n-1}(0.975)$ , so that we can write

\displaystyle\Pr\left(-t_{n-1}(0.975)\leq\frac{\bar{X}-\mu}{S/\sqrt{n}}\leq t_% {n-1}(0.975)\right)=0.95

Rearranging,

	$\displaystyle 0.95$	$\displaystyle=\Pr\left(-t_{n-1}(0.975)S/\sqrt{n}\leq\bar{X}-\mu\leq t_{n-1}(0.% 975)S/\sqrt{n}\right)$
		$\displaystyle=\Pr\left(-\bar{X}-t_{n-1}(0.975)S/\sqrt{n}\leq-\mu\leq-\bar{X}+t% _{n-1}(0.975)S/\sqrt{n}\right)$
		$\displaystyle=\Pr\left(\bar{X}-t_{n-1}(0.975)S/\sqrt{n}\leq\mu\leq\bar{X}+t_{n% -1}(0.975)S/\sqrt{n}\right).$

And the $95$ % confidence interval for $\mu$ is given by {mdframed}

\displaystyle\left(\bar{X}-t_{n-1}(0.975)S/\sqrt{n},\bar{X}+t_{n-1}(0.975)S/% \sqrt{n}\right).

Remark.

The end points of the confidence interval are random variables, since they are functions of the estimators $\bar{X}$ and $S$ . Different samples of the same size therefore produce different confidence intervals. In practice, $\bar{X}$ is replaced by $\bar{x}$ and $S^{2}$ is replaced by $s^{2}$ .

To extend this to obtain the more general $100(1-\alpha)\%$ confidence interval, simply replace the 97.5% quantiles of the $t$ -distribution with the $100(1-\alpha/2)$ % quantiles, {mdframed}

\displaystyle\left(\bar{X}-t_{n-1}(1-\alpha/2)S/\sqrt{n},\bar{X}+t_{n-1}(1-% \alpha/2)S/\sqrt{n}\right).

(4.1)

TheoremExample 4.2.1 Arctic sea ice

Use the sample of Arctic sea ice data from Example 4.1.1 to create a 95% confidence interval for the population mean $\mu$ of the minimum sea ice extent.

We use the formula given in equation (4.1). For a 95% confidence interval, $\alpha=0.05$ , and so we need to find the $t_{9}(0.975)$ quantile. Either from tables or using R,

⬇

> qt(0.975,9)

[1] 2.262157

so $t_{9}(0.975)=2.26$ . From previous examples, $\bar{x}=6.06$ , $s=0.911$ and $n=10$ . So using equation (4.1), the 95% confidence interval for $\mu$ is

\displaystyle\left(6.06-2.26\times 0.911/\sqrt{10},6.06+2.26\times 0.911/\sqrt% {10}\right)=(5.41,6.71)\times 10^{6}km^{2}.

Remark.

Can you create a 90% confidence interval for the Arctic sea ice data? How does it compare to the 95% confidence interval?