91Ó°ÊÓ

The formula to calculate degree of freedom is given by,

\(d_{f}=\frac{[\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}]^{2}}{\frac{(s_{1}^{2}/n_{2})^{2}}{n_{1}-1}+\frac{(s_{2}^{2}/n_{2})^{2}}{n_{2}-1}}\)

Determine the degree of freedom as follows.

\(d_{f}=\frac{[\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}]^{2}}{\frac{(s_{1}^{2}/n_{2})^{2}}{n_{1}-1}+\frac{(s_{2}^{2}/n_{2})^{2}}{n_{2}-1}}\)

\(d_{f}=\frac{[\frac{(9.2)^{2}}{32}+\frac{(5.7)^{2}}{20}]^{2}}{\frac{((9.2)^{2}/32)^{2}}{32-1}+\frac{((5.7)^{2}/20)^{2}}{20-1}}\)

\(\approx 49\)

From the table with \(d_{f}=49\)

\(t_{\alpha /2}=t_{0.10/2}\)

\(=t_{0.05}\)

\(=1.677\)

Determine the ends points of the confidence interval for \(\mu_{1}-\mu_{2}\) as follows.

\(\left ( \bar{x_{1}}-\bar{x_{2}} \right )\pm t_{\alpha/2}\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}=(25.8-22.1)\pm 1.677\sqrt{\frac{(9.2)^{2}}{32}+\frac{(5.7)^{2}}{20}}\)

\(=3.7\pm 1.677(2.066277)\)

\(=3.7\pm 3.465147\)

\(=0.235\) to \(7.166\)

Therefore, we can be \(90%\) confident that the difference between the mean age at arrest of East German prisoners with chronic PTSD and remitted PTSD is somewhere between \(=0.235\) to \(7.166\).