Question

A large corporation is interested in determining whether the average days of sick leave taken annually is more for night-shift employees than for day-shift employees. It is assumed that the distribution of the days of sick leave is normal for both shifts and that the variances of sick leave taken are equal for both shifts. A random sample of 12 employees from the night shift yields an average sick leave of 16.4 days with a standard deviation of 2.2 days. A random sample of 15 employees from the day shift yields an average sick leave of 12.3 days with a standard deviation of 3.5 days. At a level of significance α of 0.05, can we conclude that the average sick leave for the night shift exceeds that in the day shift?

Please show work and explain

Answer 1

To Test :-

H0 :- µ1 <= µ2
H1 :- µ1 > µ2

Test Statistic :-
t = (X̅1 - X̅2) / SP √ ( ( 1 / n1) + (1 / n2))



t = ( 16.4 - 12.3) / 2.9983 √ ( ( 1 / 12) + (1 / 15 ) )
t = 3.5307

Test Criteria :-
Reject null hypothesis if t > t(α, n1 + n2 - 2)
Critical value t(α, n1 + n1 - 2) = t( 0.05 , 12 + 15 - 2) = 1.708
t > t(α, n1 + n2 - 2) = 3.5307 > 1.708
Result :- Reject Null Hypothesis

Decision based on P value
P - value = P ( t > 3.5307 ) = 0.0008
Reject null hypothesis if P value < α = 0.05 level of significance
P - value = 0.0008 < 0.05 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis

There is sufficient evidence to support the claim that the average sick leave for the night shift exceeds that in the day shift.