Question

An intervention program was implemented to improve the work conditions of city bus drivers. Heart rates of 10 drivers who drove on the improved routes (intervention) showed a mean of 67.90 beats per mins and a standard deviation of 5.49 mins whereas the heart rate of 31 drivers assigned the normal routes (control) showed a mean of 66.81 mins and a standard deviation of 9.04 mins. a) At alpha = 0.05, is there evidence to conclude that the intervention program reduces the mean heart rate of bus drivers? Assume that heart rate is normally distributed in both groups with unequal variances. (Use the critical value approach and the P-value approach). b) Find a 95% confidence interval for the difference between the mean heart rate of bus drivers in the two environments.

Answer 1

$\mu_{1}$ :- Heart rates of drivers who drove on the improved routes

:-  Heart rates of drivers who drove on the normal routes

μ2

To Test :-

H0 :-  $\mu_{1} = \mu_{2}$

H1 :-  $\mu_{1} < \mu_{2}$

Test Statistic :-
$t = (\bar{X_{1}} - \bar{X_{2}}) / \sqrt{ ( S_{1}^{2} / n1) + (S_{2}^{2} / n2)}$

t = 0.4586

t (67.9 66.81)/V(5.49/10)9.042/31)

Test Criteria :-
Reject null hypothesis if $t < -t_{\alpha, DF}$


DF = 25

ta.DF to.05.25 1.708

$t < -t_{\alpha, DF} = 0.4586 > -1.708$
Result :- Fail to Reject Null Hypothesis

Decision based on P value
P - value = P ( t > 0.4586 ) = 0.6748
Reject null hypothesis if P value < $\alpha = 0.05$ level of significance
P - value = 0.6748 > 0.05 ,hence we fail to reject null hypothesis
Conclusion :- We Accept H0

Confidence interval :-
$\( \bar{X}_{1} - \bar{X}_{2}) \pm t_{\alpha /2 , DF}\sqrt{ (S_{1}^{2}/n1) + (S_{2}^{2}/n2)}$
$t_{\alpha, DF} = t_{ 0.05 , 25 } = 2.06$
$( 67.9 - 66.81 ) \pm t_{ 0.05/2 , 25 }\sqrt{ ( 5.49^{2}/ 10 ) + ( 9.04^{2}/ 31 )}$
Lower Limit = $( 67.9 - 66.81 ) - t_{ 0.05/2 , 25 }\sqrt{ ( 5.49^{2}/ 10 ) + ( 9.04^{2}/ 31 )}$
Lower Limit = -3.8055
Upper Limit = $( 67.9 - 66.81 ) + t_{ 0.05/2 , 25 }\sqrt{ ( 5.49^{2}/ 10 ) + ( 9.04^{2}/ 31 )}$
Upper Limit = 5.9855
95% Confidence interval is ( -3.8055 , 5.9855 )

Since the value 0 lies in the interval, hence we fail to reject H0

There is insufficient evidence to support the claim that the intervention program reduces the mean heart rate of bus drivers.