# 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

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.

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

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.

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

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

Upper Limit

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

comments

Posted in: