Question

Find a 90% confidence interval for the difference between the mean heart rates of urban bus drivers in Stockholm in the two environments

Short answer

The difference in mean heart rates between the two environments for urban bus drivers in Stockholm spans between $-2.97027$ and$5.15027.$

Step-by-step solution

Given information

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

:calculation

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 ${\mu }_1-{\mu }_2$are given by

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

Perform the following calculations to determine the degree of freedom.

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

From the table with $d_f=25$
$t_{\alpha /2}=1.708$

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