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

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, $$\alpha = 0.05$$, $$ – {Z_{\alpha /2}}$$ 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 the 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 with $$\overline X – \mu /\frac{\sigma }{{\sqrt n }}$$, we have $$P\left[ { – 1.64 < \frac{{\overline X – \mu }}{{\sigma /\sqrt n }} < 1.64} \right] = 0.90$$

Upon simplification, this 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 along 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 MS students achieved an average score of 60 with a standard deviation score of 15. Construct a 99% confidence interval estimation for the mean score of the entire class.

 

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

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
\[\begin{gathered} \overline X – 2.57\frac{\sigma }{{\sqrt n }} = 60 – 2.57\left( {\frac{{15}}{8}} \right) \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 60 – 4.82 = 55.18 \approx 55 \\ \end{gathered}\]

 

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.