# Confidence Interval Estimate of the Mean

From central limit theorem we know that
$Z = \frac{{\overline X – \mu }}{{\frac{\sigma }{{\sqrt n }}}}$

is a standard normal variate. From the discussion of an introduction to interval estimation we know that P (–1.96 < Z< 1.96) = 0.95 has the least possible range. With this inequality we can construct a 95% confidence interval estimate of the population mean$\mu$ if we replace Z by $\overline X – \mu /\frac{\sigma }{{\sqrt n }}$

In general if the margin of error in our interval estimate is $\alpha$ (this is also referred to as the level of significance), then the level of confidence would be $\left( {1 – \alpha } \right)$ or $\left( {1 – \alpha } \right) \times 100\%$. For example, if $\alpha = 0.05$ or 5% then $\left( {1 – \alpha } \right) = 0.95$ or 95% would be the level of confidence. To have the shortest interval the margin of error $\alpha$ is distributed equally on the two sides of the curve, as shown in the figure below: Thus, for $\alpha = 0.05$, the curve has the margin of error $\alpha /2$= (0.025) on the left tail and the same area on the right tail. The value Z which gives an area $\alpha /2$ on the left tail is denoted by $– {Z_{\alpha /2}}$, and that which gives an area equal to $\alpha /2$ on the right tail is denoted by ${Z_{\alpha /2}}$. The area enclosed between $– {Z_{\alpha /2}}$ and ${Z_{\alpha /2}}$ would therefore be $\left( {1 – \alpha } \right)$. In our example, $\alpha = 0.05$, $– {Z_{\alpha /2}}$ and ${Z_{\alpha /2}}$ have values -1.96 and 1.96 respectively and $\left( {1 – \alpha } \right) = 0.95$.

From the above discussion we can write the following probability statement: $P\left[ { – {Z_{\alpha /2}} < Z < {Z_{\alpha /2}}} \right] = \left( {1 – \alpha } \right)$. The $\left( {1 – \alpha } \right)$ 100% confidence interval estimate of the mean can be obtained by replacing Z with $\overline X – \mu /\frac{\sigma }{{\sqrt n }}$, thus
$\begin{gathered} P\left[ { – {Z_{\alpha /2}} < \frac{{\overline X – \mu }}{{\sigma /\sqrt n }} < {Z_{\alpha /2}}} \right] = 1 – \alpha \\ P\left[ { – {Z_{\alpha /2}}\frac{\sigma }{{\sqrt n }} < \overline X – \mu < {Z_{\alpha /2}}\frac{\sigma }{{\sqrt n }}} \right] = 1 – \alpha \\ P\left[ { – \overline X – {Z_{\alpha /2}}\frac{\sigma }{{\sqrt n }} < – \mu < – \overline X + {Z_{\alpha /2}}\frac{\sigma }{{\sqrt n }}} \right] = 1 – \alpha \\ P\left[ {\overline X + {Z_{\alpha /2}}\frac{\sigma }{{\sqrt n }} > \mu > \overline X – {Z_{\alpha /2}}\frac{\sigma }{{\sqrt n }}} \right] = 1 – \alpha \\ P\left[ {\overline X – {Z_{\alpha /2}}\frac{\sigma }{{\sqrt n }} < \mu < \overline X + {Z_{\alpha /2}}\frac{\sigma }{{\sqrt n }}} \right] = 1 – \alpha \\ \end{gathered}$

The $\left( {1 – \alpha } \right)$ 100% lower and upper confidence limits are, therefore, $\overline X \mp {Z_{\alpha /2}}\frac{\sigma }{{\sqrt n }}$.

As a specific example, let $\alpha = 0.10$, so that $\left( {1 – \alpha } \right) = 0.90$ or 90%. The probability statement may thus be written as
$P\left[ { – {Z_{0.05}} < Z < {Z_{0.05}}} \right] = 0.90$ [Since $\alpha = 0.1$, $\therefore \frac{\alpha }{2} = 0.05$]

Consulting the normal table, ${Z_{0.05}} = 1.64$. Hence we can write P [–1.64 <Z< 1.64] = 0.90.

Replacing Z with $\overline X – \mu /\frac{\sigma }{{\sqrt n }}$, we have $P\left[ { – 1.64 < \frac{{\overline X – \mu }}{{\sigma /\sqrt n }} < 1.64} \right] = 0.90$

Upon simplification, this reduces to $P\left[ {\overline X – 1.64\frac{\sigma }{{\sqrt n }} < \mu < \overline X + 1.64\frac{\sigma }{{\sqrt n }}} \right] = 0.90$.

Therefore, the 90% confidence limits are: $\overline X \mp 1.64\frac{\sigma }{{\sqrt n }}$.

$\sigma$ is usually not known; therefore, for large samples it can be replaced by S (the sample standard deviation) which may be calculated from the sample by using the following formula: ${S^2} = \frac{{\sum {{\left( {{X_i} – \overline X } \right)}^2}}}{{n – 1}}$.

Proceeding along the same lines, the 95 and 99 percent confidence limits may be stated as
$\overline X \mp 1.96\frac{\sigma }{{\sqrt n }}$ for 95%
$\overline X \mp 2.57\frac{\sigma }{{\sqrt n }}$ for 99%

Example: A random sample of 64 MS students achieved an average score of 60 with a standard deviation score of 15. Construct a 99% confidence interval estimation for the mean score of the entire class.

Solution: We have the following data: $\overline X = 60,\,\,\,\,n = 64,\,\,\,\sigma = 15$

The formula for 99% confidence interval may be written as

$P\left[ {\overline X – 2.57\frac{\sigma }{{\sqrt n }} < \mu < \overline X + 2.57\frac{\sigma }{{\sqrt n }}} \right]$

The lower confidence limit is
$\begin{gathered} \overline X – 2.57\frac{\sigma }{{\sqrt n }} = 60 – 2.57\left( {\frac{{15}}{8}} \right) \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 60 – 4.82 = 55.18 \approx 55 \\ \end{gathered}$

The upper confidence limit is
$\begin{gathered} \overline X + 2.57\frac{\sigma }{{\sqrt n }} = 60 + 2.57\left( {\frac{{15}}{8}} \right) \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 60 + 4.82 = 64.82 \approx 65 \\ \end{gathered}$

Hence, the 99% confidence interval estimate for the mean score will be $55 < \mu < 65$, i.e., the mean score is between 55 and 65.