Question

Refer to Exercise \(10.81\) and obtain a \(90%\) confidence interval for the difference, \(\mu_{1}-\mu_{2}\), between the mean ages at arrest of East German prisoners with chronic PTSD and remitted PTSD.

Short answer

The difference between the mean age at arrest of East German prisoners with chronic PTSD and remitted PTSD is somewhere between \(0.235\) to \(7.166\).

Step-by-step solution

Step 1. Concept Used

The formula to calculate degree of freedom is given by,

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

Step 2. Calucalation

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 \(7.166\).