Question

In a 1996 survey of 1000 american citizens, 300 respondents claimed to be fluent in a second language. Find 94% confidence interval for the true proportion of citizens who are not fluent in a second language.

Answer 1

You have an estimated proportion of \(\hat{p}=\dfrac{700}{1000}=0.7\) of citizens that are not fluent and a sample size of \(n=1000\). The confidence interval for a proportion is given by
\[\left(\hat{p}-Z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}},\,\hat{p}+Z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right)\]
where \(Z_{\alpha/2}\) denotes the critical value for a \((1-\alpha)\%\) confidence level. The critical value for a 94% CI is approximately \(1.88\).