Concept used:

Let \(\mu_{1}\) be the mean heart rate of intervention drivers and \(\mu_{2}\) be the mean heart rate of control drivers.

Consider null and Alternative hypothesis.

Null hypothesis, \(H_{0} : \mu_{1} = \mu_{2}\).

Alternative hypothesis, \(H_{0} : \mu_{1} < \mu_{2}\)

Here, the hypothesis test is left tailed.

Level of significance is 5%.

Determine the test statistics under Null hypothesis as follows.

\(d_{f}=\frac{\frac{\left[\left(\frac{x_{1}^{2}}{n_{1}}\right)+\left(\frac{x_{2}^{2}}{n_{2}}\right)\right]^{2}}{\left(\frac{x_{1}^{2}}{n_{1}}\right)^{2}}}{n_{1}}-1+\frac{\left(\frac{x_{2}^{2}}{n_{2}}\right)^{2}}{n_{2}}-1\)

\(=\frac{\frac{\left[\left(\frac{5.49^{2}}{10}\right)+\left(\frac{9.04^{2}}{31}\right)\right]^{2}}{\frac{5.49^{2}}{10^{2}}}}{10}-1+\frac{\left(\frac{9.04^{2}}{31}\right)^{2}}{31}-1\approx25\)

Referring the table with 25 degree of freedom, to determine the P-value that is P - value > 0.10.

Using technology, P-value is 0.675 .

Since the P-value is 0.675 which is greater than level of significance \(\alpha = 0.05\). So we are fail to reject our null hypothesis.

Therefore, the given data does not provide sufficient evidence to conclude that the intervention program reduces mean heart rate of urban drivers in Stockholm.