### Chapter 6. Parameter Estimation and Confidence Intervals: Estimation

### Constructing a 95% Confidence Interval for the Population Mean

To construct a #95\%# confidence interval for the population mean #\mu# we will need to make use of the *sampling distribution of the sample mean*.

For the purpose of this example, let's assume that the conditions for normality have been met and the sample mean #\bar{X}# (approximately) has the #N(\mu,\sigma/\sqrt{n})# distribution.

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Constructing a 95% Confidence Interval for the Population Mean

Let #Z# be the *standardized *version of #\bar{X}#, that is:

\[Z = \cfrac{\bar{X} - \mu}{\sigma_{\bar{X}}} =\cfrac{\bar{X} - \mu}{\sigma/\sqrt{n}}\]

Then #Z# has the *Standard Normal Distribution*, that is #Z\sim N(0,1)#.

It can be shown#^{1}# that the middle #95\%# of the *Standard Normal Distribution *falls between #-1.96# and #1.96#:

\[\mathbb{P}(-1.96 \leq Z \leq 1.96) = 0.95\]

Substitituting #Z# with #\cfrac{\bar{X} - \mu}{\sigma/\sqrt{n}}# gives:

\[\mathbb{P}(-1.96 \leq \cfrac{\bar{X} - \mu}{\sigma/\sqrt{n}} \leq 1.96) = 0.95\]

With the help of a little algebra#^{2}#, this can be rewritten as:

\[\mathbb{P}(\mu -1.96\cdot\cfrac{\sigma}{\sqrt{n}} \leq \bar{X} \leq \mu + 1.96\cdot\cfrac{\sigma}{\sqrt{n}}) = 0.95\]

This result is mathematically equivalent to:

\[\mathbb{P}(\bar{X} -1.96\cdot\cfrac{\sigma}{\sqrt{n}} \leq \mu \leq \bar{X} + 1.96\cdot\cfrac{\sigma}{\sqrt{n}}) = 0.95\]

In words, there is a #95\%# probability that we will draw a random sample such that the interval

\[CI_{\mu,95\%}=(L,U)=(\bar{X} -1.96\cdot\cfrac{\sigma}{\sqrt{n}},\,\,\,\, \bar{X} + 1.96\cdot\cfrac{\sigma}{\sqrt{n}})\]

will contain the true value of the population mean #\mu#.

After the sample is drawn and the values of lower bound #L# and the upper bound #U# are computed, the interval #(L,U)# is called a #95\%# *confidence interval* for #\mu#.

The expression #1.96 \cdot \cfrac{\sigma}{\sqrt{n}}# is called the *margin of error*, while #95\%# is called the *confidence level*.

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