91Ó°ÊÓ

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 ${\mathrm{μ}}_1-{\mathrm{μ}}_2$ the confidence interval ends are as follows:

$\left(\stackrel{¯}{x_1}-\stackrel{¯}{x_2}\right)Â\pm t_{\mathrm{Î\pm }/2}\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

Ascertain the degree of freedom by performing the following calculations.

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

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

$≈18$

From the table with $d_f=18$,

$t_{\mathrm{Î\pm }/2}=2.552$

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

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

$=0.00784Â\pm 2.552(0.002028454256)$

$=0.00784Â\pm 0.005176615261$

$=0.00265$to $0.01300$