Question

The article "College Graduates Break Even by Age 33" (USA Today, September 21,2010 ) reported that \(2.6 \%\) of college graduates were unemployed in 2008 and \(4.6 \%\) of college graduates were unemployed in \(2009 .\) Suppose that the reported percentages were based on independently selected representative samples of 500 college graduates in each of these two years. Construct and interpret a \(95 \%\) confidence interval for the difference in the proportions of college graduates who were unemployed in these 2 years.

Short answer

The \(95\%\) confidence interval for the difference in the proportions of college graduates who were unemployed in 2008 and 2009 is \((-0.00152, 0.04152)\). This means that anytime a sample is taken, there is a \(95\%\) likelihood that the difference in proportions will lie in this interval. Since 0 is contained within the interval, no significant change in the unemployment rate from 2008 to 2009 can be concluded.

Step-by-step solution

- Understand the data and problem

The problem presents two samples independently chosen from college graduates in 2008 and 2009. In 2008, the unemployment proportion (\(p_1\)) was \(2.6\%\) and in 2009, the unemployment proportion (\(p_2\)) was \(4.6\%\). The sample size for both years (n) is 500. The task is to compute a \(95\%\) confidence interval for the difference between these two proportions.

- Determine the proportions

First, change the percentage to a decimal: \(p_1 = 2.6\% = 0.026\) and \(p_2 = 4.6\% = 0.046\). Next, calculate the number of unemployed graduates for each year: \(X_1 = n * p_1 = 500 * 0.026 = 13\) and \(X_2 = n * p_2 = 500 * 0.046 = 23\). Then, compute the sample proportions: \(\hat{p_1} = X_1 / n = 13 / 500 = 0.026\) and \(\hat{p_2} = X_2 / n = 23 / 500 = 0.046\).

- Compute standard error

Then, estimate the standard error of these proportions using the formula for the standard error of a difference between two proportions: \(SE = \sqrt{\hat{p_1}*(1-\hat{p_1})/n_1 + \hat{p_2}*(1-\hat{p_2})/n_2} = \sqrt{0.026*0.974/500 + 0.046*0.954/500} = 0.01098\). This estimates the variability we might expect if we took multiple samples.

- Determine the confidence interval

Next, construct the \(95\%\) confidence interval for the difference in proportions. The formula for this is \((\hat{p_2}-\hat{p_1}) \pm Z*SE\), where \(Z = 1.96\) for a \(95\%\) confidence interval. Thus, the confidence interval is \((0.046-0.026) \pm 1.96*0.01098 = 0.020 \pm 0.02152\). Hence the confidence interval is \((-0.00152, 0.04152)\).

- Interpret the confidence interval

The interpretation of the confidence interval is that we can be \(95\%\) certain that the difference in the unemployment rates of college graduates between 2008 and 2009 is between \(-0.00152\) and \(0.04152\). The negative lower limit may seem strange, but it just means that the unemployment rate in 2008 could be slightly higher than in 2009, according to our sample data. Since 0 is contained within the confidence interval, we cannot conclude that there was a significant change in the unemployment rate from 2008 to 2009.