# Confidence Interval Estimate of Mean with a Small Sample

As long as $${\sigma ^2}$$ is known, the confidence interval estimate of a 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 with its unbiased estimate $${S^2}$$, defined by

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

The previous 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 previous tutorials.

\[t = \frac{{\overline X – \mu }}{{S/\sqrt n }}\]

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

The method of constructing a confidence interval is the same as for large samples, with \[Z\] being 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 a specific example, the 90% confidence limits may be stated as $$\overline X \pm {t_{0.5}}\frac{S}{{\sqrt n }}$$.

Here $${t_{0.05}}$$ 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 MBA students scored an average 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 98% confidence.