Stressed-Out Bus Drivers. An intervention program designed by the Stockholm Transit District was implemented to improve the work conditions of the city's bus drivers. Improvements were evaluated by G. Evans et al., who collected physiological and psychological data for bus drivers who drove on the improved routes (intervention) and for drivers who were assigned the normal routes (control). Their findings were published in the article "Hassles on the Job: A Study of a Job Intervention with Urban Bus Drivers" (Journal of Organizational Behavior, Vol. 20, pp. 199-208). Following are data, based on the results of the study, for the heart rates, in beats per minute, of the intervention and control drivers.

a. At the $5\%$significance level, do the data provide sufficient evidence to conclude that the intervention program reduces mean heart rate of urban bus drivers in Stockholm? (Note; $x_1=67.90$, $s_1=5.49,$${\overline{x}}_2=66.81$and $s_2=9.04$.

b. Can you provide an explanation for the somewhat surprising results of the study?

c. Is the study a designed experiment or an observational study? plain your answer.

To see if the data is sufficient to infer that the intervention programme reduces the mean heart rate of Stockholm's urban bus drivers.

Let's consider ${\mu }_1$be the intervention drivers' mean heart rate and${\mu }_2$be the control drivers' mean heart rate.

Think about the null and alternative hypotheses.

Null hypothesis,$H_0:{\mu }_1={\mu }_2$

Alternative hypothesis, $H_0:{\mu }_1<{\mu }_2$

The hypothesis test is left untailed in this case.

The significance level is set at $5\%$

Assume the null hypothesis and calculate the test statistics as follows.

$t=\frac{\overline{x_1}-\overline{x_2}}{\sqrt{\frac{s_1^2}{n_1}+\frac{x_2^2}{n_2}}}$

$=\frac{67.90-66.81}{\sqrt{\frac{(5.49{)}^2}{10}+\frac{(9.04{)}^2}{31}}}$

$=\frac{1.09}{2.377013}$

$=0.46$

Ascertain the degree of freedom by performing the following calculations.

$d_f=\frac{{\left[\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right]}^2}{\frac{{\left(s_1^2/n_1\right)}^2}{n_1-1}+\frac{{\left(s_2^2{m}_2\right)}^2}{n_2-1}}$

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

$\approx 25$

To determine the $P-$value, refer to the table with $25$degrees of freedom, which is$P-$value $>0.10$.

P-value is $0.675$,when using technology.

Because the P-value is$0.675$it is more than the significance level of$\alpha =0.05$. As a result, our null hypothesis is not rejected.

To find an explanation for the study's somewhat unexpected finding.

The intended cause is an explanatory variable that explains the results. The intended effect is called a response variable, and it responds to explanatory factors.

The unexpected results could be due to some variables that are correlated with the response variable or explanatory variables.

Explain the answer to determine whether the study is a deliberate experiment or an observational study.

We measure or survey members of a sample in an observational research without attempting to influence them. In a controlled experiment, we divide people or items into groups and administer a treatment to one group while leaving the other group untreated.

Here, researchers impose treatment and control and then observe characteristics and take measurements, so our study is designed experimental study.