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For the probability $1-\alpha $ , it is possible to find a number ${z}_{\alpha /2}$ , such that $$P\left(-{z}_{\alpha /2}\le \frac{\overline{X}-\mu}{\sigma /\sqrt{n}}\le {z}_{\alpha /2}\right)=1-\alpha .$$
For example , if $1-\alpha =0.95$ , then ${z}_{\alpha /2}={z}_{0.025}=1.96$ and if $1-\alpha =0.90$ , then ${z}_{\alpha /2}={z}_{0.05}=\mathrm{1.645.}$
Recalling that $\sigma >0$ , the following inequalities are equivalent : $$-{z}_{\alpha /2}\le \frac{\overline{X}-\mu}{\sigma /\sqrt{n}}\le {z}_{\alpha /2}$$ and $$-{z}_{\alpha /2}\left(\frac{\sigma}{\sqrt{n}}\right)\le \overline{X}-\mu \le {z}_{\alpha /2}\left(\frac{\sigma}{\sqrt{n}}\right),$$
$$-\overline{X}-{z}_{\alpha /2}\left(\frac{\sigma}{\sqrt{n}}\right)\le -\mu \le -\overline{X}+{z}_{\alpha /2}\left(\frac{\sigma}{\sqrt{n}}\right),$$ $$\overline{X}+{z}_{\alpha /2}\left(\frac{\sigma}{\sqrt{n}}\right)\ge \mu \ge \overline{X}-{z}_{\alpha /2}\left(\frac{\sigma}{\sqrt{n}}\right).$$
Thus, since the probability of the first of these is 1- $1-\alpha $ , the probability of the last must also be $1-\alpha $ , because the latter is true if and only if the former is true. That is, $$P\left[\overline{X}-{z}_{\alpha /2}\left(\frac{\sigma}{\sqrt{n}}\right)\le \mu \le -\overline{X}+{z}_{\alpha /2}\left(\frac{\sigma}{\sqrt{n}}\right)\right]=1-\alpha .$$
So the probability that the random interval $$\left[\overline{X}-{z}_{\alpha /2}\left(\frac{\sigma}{\sqrt{n}}\right),\overline{X}+{z}_{\alpha /2}\left(\frac{\sigma}{\sqrt{n}}\right)\right]$$ includes the unknown mean $\mu $ is $1-\alpha $ .
For illustration , $$\overline{x}\pm 1.96\left(\sigma /\sqrt{n}\right)$$ is a 95% confidence interval for $\mu $ .
It can be seen that the confidence interval for $\mu $ is centered at the point estimate $\overline{x}$ and is completed by subtracting and adding the quantity ${z}_{\alpha /2}\left(\sigma /\sqrt{n}\right)$ .
A shorter confidence interval indicates that there is more reliance in $\overline{x}$ as an estimate of $\mu $ . For a fixed sample size n , the length of the confidence interval can also be shortened by decreasing the confidence coefficient $1-\alpha $ . But if this is done, shorter confidence is achieved by losing some confidence.
Let $\overline{x}$ be the observed sample mean of 16 items of a random sample from the normal distribution $N\left(\mu ,{\sigma}^{2}\right)$ . A 90% confidence interval for the unknown mean $\mu $ is $$\left[\overline{x}-1.645\sqrt{\frac{23.04}{16}},\overline{x}+1.645\sqrt{\frac{23.04}{16}}\right].$$ For a particular sample this interval either does or does not contain the mean $\mu $ . However, if many such intervals were calculated, it should be true that about 90% of them contain the mean $\mu $ .
If one cannot assume that the distribution from which the sample arose is normal, one can still obtain an approximate confidence interval for $\mu $ . By the Central Limit Theorem the ratio $\left(\overline{X}-\mu \right)/\left(\sigma /\sqrt{n}\right)$ has, provided that n is large enough, the approximate normal distribution $N\left(0,1\right)$ when the underlying distribution is not normal. In this case $$P\left(-{z}_{\alpha /2}\le \frac{\overline{X}-\mu}{\sigma /\sqrt{n}}\le {z}_{\alpha /2}\right)\approx 1-\alpha ,$$ and $$\left[\overline{x}-{z}_{\alpha /2}\left(\frac{\sigma}{\sqrt{n}}\right),\overline{x}+{z}_{\alpha /2}\left(\frac{\sigma}{\sqrt{n}}\right)\right]$$ is an approximate $100\left(1-\alpha \right)\%$ confidence interval for $\mu $ . The closeness of the approximate probability $1-\alpha $ to the exact probability depends on both the underlying distribution and the sample size. When the underlying distribution is unimodal (has only one mode) and continuous, the approximation is usually quite good for even small n , such as $n=5$ . As the underlying distribution becomes less normal ( i.e. , badly skewed or discrete), a larger sample size might be required to keep reasonably accurate approximation. But, in all cases, an n of at least 30 is usually quite adequate.
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