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


estimation-mean

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 statementP\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

\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 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 toP\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

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}

Also 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.