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According to a 2011 report for IFA Educational Foundation, a sample of 100 franchised businesses located in Mississippi and find that 15 are owned by minorities.

The sample proportion is the point estimator of the population proportion p.

Computing the sample proportion is,

$\begin{array}{rcl}\hat{\mathrm{p}}& =& \frac{\mathrm{X}}{\mathrm{n}}\\ & =& \frac{15}{100}\\ & =& 0.15\end{array}$

Then the level of$100\left(1-\mathrm{伪}\right)\%$confidence interval for p (proportion) is,

$\hat{\mathrm{p}}卤{\mathrm{z}}_{\frac{\mathrm{伪}}{2}}\sqrt{\frac{\hat{\mathrm{p}}\left(1-\hat{\mathrm{p}}\right)}{\mathrm{n}}}$

For a 95% confidence interval, the value of$\frac{\mathrm{伪}}{2}$is,

$\begin{array}{rcl}100\left(1-\mathrm{伪}\right)\%& =& 95\%\\ \left(1-\mathrm{伪}\right)& =& 0.95\end{array}$

For,$\mathrm{伪}=0.05\mathrm{and}\frac{\mathrm{伪}}{2}=0.025$$\mathrm{伪}=0.05\mathrm{and}\frac{\mathrm{伪}}{2}=0.025$

The 95% confidence interval is,

$\begin{array}{rcl}\hat{\mathrm{p}}卤{\mathrm{z}}_{\frac{\mathrm{伪}}{2}}\sqrt{\frac{\hat{\mathrm{p}}\left(1-\hat{\mathrm{p}}\right)}{\mathrm{n}}}& =& 0.15卤1.960\sqrt{\frac{0.15\left(1-0.15\right)}{100}}\left[\mathrm{From}\mathrm{Standard}\mathrm{Normal}\mathrm{Table}\right]\\ & =& \left(0.15卤1.960\sqrt{0.001275}\right)\\ & =& \left(0.15卤0.07\right)\\ & =& \left(0.08,0.02\right)\end{array}$

Therefore, the 95% confidence interval that the true percentage of minority-owned franchises in Mississippi is between 8% and 22%.

Also 20.5% falls in this interval.

Hence, conclude that the percentage of minority-owned franchises in Mississippi is not less than the national value.