Confidence Interval Estimate of Mean with Small Sample

As long as {\sigma ^2} is known the confidence interval estimate of population mean can be obtained by the method discussed earlier, provided the sample is large. Even if {\sigma ^2} is not known we could replace it by its unbiased estimate {S^2}, defined by

{S^2} = \frac{{\sum {{\left( {{X_i} - \overline X } \right)}^2}}}{{n - 1}}

The pervious method of estimation fails to provide good estimations if the sample size is small (smaller than 30). In such cases, which are quite frequent, we can use the t statistics as defined in pervious tutorials.

t = \frac{{\overline X - \mu }}{{S/\sqrt n }}

Where    {S^2} = \frac{{\sum {{\left( {{X_i} - \overline X } \right)}^2}}}{{n - 1}}

The method of construction of a confidence interval is same as for the large samples, with

Z

replaced by t. The applications of this statistic, however, presume that the population is approximately bell shaped. A \left( {1 - \alpha } \right)100% confidence interval estimate for the mean may be obtained as follows.

\begin{gathered} P\left[ { - {t_{\alpha /2}} < t < {t_{\alpha /2}}} \right] = 1 - \alpha \\ P\left[ { - {t_{\alpha /2}} < \frac{{\overline X - \mu }}{{S/\sqrt n }} < {t_{\alpha /2}}} \right] = 1 - \alpha \\ P\left[ { - {t_{\alpha /2}}\frac{S}{{\sqrt n }} < \overline X - \mu < {t_{\alpha /2}}\frac{S}{{\sqrt n }}} \right] = 1 - \alpha \\ P\left[ { - \overline X - {t_{\alpha /2}}\frac{S}{{\sqrt n }} < - \mu < - \overline X + {t_{\alpha /2}}\frac{S}{{\sqrt n }}} \right] = 1 - \alpha \\ P\left[ {\overline X - {t_{\alpha /2}}\frac{S}{{\sqrt n }} < \mu < \overline X + {t_{\alpha /2}}\frac{S}{{\sqrt n }}} \right] = 1 - \alpha \\ \end{gathered}

The \left( {1 - \alpha } \right)100% confidence limits for the means of the population are, therefore, \overline X \pm {t_{\alpha /2}}\frac{S}{{\sqrt n }}, as an specific example, the 90% confidence limits may be stated as \overline X \pm {t_{0.5}}\frac{S}{{\sqrt n }}.

Where {t_{0.05}} is the value of t from the t - distribution table at 5% level of significance corresponding to the given degrees of freedom \left( {n - 1} \right). For further understanding let us consider the following practical example.

Example:
A random sample of 19 students of MBA made an average score of 60 with a standard deviation score of 15. Construct a 95% confidence interval for the mean of the entire MBA class.

Solution:
Since the sample size is smaller than 30, we will use the t - statistic to construct the required confidence interval. We are provided with

\begin{gathered} \overline X = 60 \\ S = 15 \\ n = 16\,\,\,\,\,\,\,\,\,\therefore \sqrt n = 4 \\ \end{gathered}

The degree of freedom  = n - 1 = 16 - 1 = 15
\alpha = 0.05\,\,\,\,\,\,\,\,\because \frac{\alpha }{2} = 0.025

Consulting the t - distribution table, for 15 degrees of freedom, we have, {t_{0.025}} = 2.131

Hence, using  formula  \overline X \pm {t_{\alpha /2}}\frac{S}{{\sqrt n }}, the 95% confidence limits would be.

Lower Limit

\begin{gathered} \overline X - {t_{0.025}}\frac{S}{{\sqrt n }} = 60 - 2.131\left( {\frac{{15}}{4}} \right) \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 60 - 7.99 = 52\,\,appox \\ \end{gathered}

Upper Limit

\overline X + {t_{0.025}}\frac{S}{{\sqrt n }} = 60 - 7.99 = 68\,\,appox

The true mean score \mu , therefore, lies between 52 and 68 with a 98% confidence.