Margin of Error

Confidence Level of the Confidence Interval

The confidence level for the confidence interval for θ is based on the sampling distribution of the statistic $\widehat{\mathrm{\theta }}.$ In statistical theory, pivotal quantities are used to derive confidence intervals. A pivotal quantity is a function of the sample data, the unknown parameter θ, and no other unknown quantities. The pivotal quantity also has a distribution that does not depend on θ. To illustrate how this works, assume ${\mathrm{X}}_1,,{\mathrm{X}}_n$ represent a random sample from a normal distribution with unknown mean μ and known variance σ2 It is known that the pivotal quantity

$$Z=\frac{\overline{\mathrm{X}}-\mu }{\sigma /\sqrt{n}}~N\left(0,1\right),$$

in other words, the pivotal quantity Z follows a standard normal distribution. To find a 95% confidence interval for μ, one would start with the following probability statement:

$$\begin{array}{l}P\left(-1.96<Z<1.96\right)=\\ P\left(-1.96<\frac{\overline{X}-\mu }{\sigma /\sqrt{n}}<1.96\right)=0.95.\end{array}$$

Some algebraic manipulation of the inequality in order to isolate μ produces the 95% confidence interval, $\overline{X}\pm 1.96(\sigma /\sqrt{n}).$

If a 99% confidence interval for μ were desired using the same assumptions to construct a 95% confidence interval for μ, the starting probability statement would be

$$\begin{array}{l}P\left(-2.576<Z<2.576\right)=\\ P\left(-2.576<\frac{\overline{X}-\mu }{\sigma /\sqrt{n}}<2.576\right)=0.99.\end{array}$$

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