91Ó°ÊÓ

For urban bus drivers in Stockholm, calculate a 90% confidence interval for the difference in mean heart rates between the two environments.

The following is the formula for calculating degree of freedom:

$d_f=\frac{{[\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}]}^2}{\frac{{\left(x_1^2{\ln }_1\right)}^2}{{}^{n_1−1}}+\frac{{\left(\frac{2}{2}{l}_m\right)}^2}{m_2−1}}$

The confidence interval's end points for ${\mathrm{μ}}_1−{\mathrm{μ}}_2$are given by

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

Perform the following calculations to determine the degree of freedom.

localid="1651416467233" $\begin{array}{rl}& d_f=\frac{{[\frac{s_1^2}{n_1}+\frac{x_2^2}{n_2}]}^2}{\frac{{(x_1^2/m_1)}^2}{n_1−1}+\frac{{(s_2^2/m_2)}^2}{n_2−1}}\\ & =\frac{{[\frac{(5.49{)}^2}{10}+\frac{(9.04{)}^2}{31}]}^2}{\frac{(5.492/10{)}^2}{10−1}+\frac{{(9.04{)}^2/31)}^2}{31−1}}\\ & ≈25\end{array}$

From the table with $d_f=25$
$t_{\mathrm{Î\pm }/2}=1.708$

Determine the confidence interval's ends for ${\mathrm{μ}}_1−{\mathrm{μ}}_2$as followslocalid="1651416620470" $$\begin{gathered} \left.\stackrel{¯}{\left(x_1\right.}−\stackrel{¯}{x_2}\right)Â\pm t_{\mathrm{Î\pm }/2}\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=(67.90−66.81)Â\pm 1.708\sqrt{\frac{(5.49{)}^2}{10}+\frac{(9.04{)}^2}{31}} \\ =1.09Â\pm 4.06027 \\ =−2.97027\text{to5}15027 \end{gathered}$$