Refer to Exercise $10.85$and determine a$98\%$confidence interval for the difference between the mean dopamine activities of psychotic and nonpsychotic patients.

Given in the question that, we need to determine a $98\%$confidence interval for the difference between the mean dopamine activities of psychotic and nonpsychotic patients.

The formula for determining degree of freedom is as follows:

$d_f=\frac{{\left[\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right]}^2}{\frac{{\left(s_1^2/n_1\right)}^2}{n_1-1}+\frac{{\left(s_2^2/n_2\right)}^2}{n_2-1}}$

For ${\mu }_1-{\mu }_2$ the confidence interval ends are as follows:

$\left(\overline{x_1}-\overline{x_2}\right)\pm t_{\alpha /2}\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

Ascertain the degree of freedom by performing the following calculations.

$d_f=\frac{{\left[\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right]}^2}{\frac{{\left(s_1^2/n_1\right)}^2}{n_1-1}+\frac{{\left(s_2^2/n_2\right)}^2}{n_2-1}}$

$=\frac{{\left[\frac{(0.00514{)}^2}{10}+\frac{(0.00470{)}^2}{15}\right]}^2}{\frac{{\left((0.00514{)}^2/10\right)}^2}{10-1}+\frac{{\left((0.00470{)}^2/15\right)}^2}{15-1}}$

$\approx 18$

From the table with $d_f=18$,

$t_{\alpha /2}=2.552$

The ends of the confidence interval for ${\mu }_1-{\mu }_2$should be determined as follows.

$\left(\overline{x_1}-\overline{x_2}\right)\pm t_{\alpha /2}\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=(0.02426-0.01642)$$\pm 2.552\sqrt{\frac{(0.00514{)}^2}{10}+\frac{(0.00470{)}^2}{15}}$

$=0.00784\pm 2.552(0.002028454256)$

$=0.00784\pm 0.005176615261$

$=0.00265$to $0.01300$