# Confidence Interval Estimate of Mean

From central limit theorem we know that,

is a standard normal variate. From the discussion of 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

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 of the interval the margin of error $\alpha$ is distributed equally on the two sides of the curve as shown in the figure:

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, for $\alpha = 0.05$, $- {Z_{\alpha /2}}$ 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 mean can be obtained by replacing Z with, $\overline X - \mu /\frac{\sigma }{{\sqrt n }}$, thus

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

Which on simplification 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 on 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 students in the MS class made an average score of 60, with a standard deviation score of 15. Construct 99% confidence interval estimation for the mean score of entire class.

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

Using the formula for 99% confidence interval may be written as

The lower confidence limit is

Also the upper confidence limit is

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.