Refer to Exercise \(10.82\) and determine a \(98%\) confidence interval for the difference, \(mu_{1}-\mu_{2}\), between the mean sizes of malignant and benign phyllodes tumors.

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.

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

\(d_{f}=\frac{[\frac{(38.5)^{2}}{52}+\frac{(26.6)^{2}}{116}]^{2}}{\frac{((38.5)^{2}/52)^{2}}{52-1}+\frac{((26.6)^{2}/116)^{2}}{116-1}}\)

\(\approx 73\)

From the table with \(d_{f}=73\)

\(t_{\alpha/2}\)

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

\(\left ( \bar{x_{1}}-\bar{x_{2}} \right )\pm t_{\alpha/2}\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}=(48.5-33.7)\pm 2.379\sqrt{\frac{(38.5)^{2}}{52}+\frac{(26.6)^{2}}{116}}\)

\(=14.8\pm 2.379(5.8826)\)

\(=14.8\pm 13.9947\)

\(=0.805\) to \(28.795\)

Therefore, we can be \(98%\) confident that the difference between the mean sizes of malignant and benign phyllodes tumors is somewhere between \(=0.805\) to \(28.795\).