Case Study 2 (20 points) Will a flexible workweek schedule result in positive be
ID: 3333110 • Letter: C
Question
Case Study 2 (20 points) Will a flexible workweek schedule result in positive benefits for both employer and employee? Is a more rested employee, who spends less time commuting to and from work, likely to be more efficient and take less time off for sick leave and personal leave? A report on the benefits of flexible work schedules that appeared in Environmental Health looked at the records of n 11 employees who worked in a satellite office in a county healthy department in Illinois under a 4-day workweek schedule [1]. Employees worked a conventional workweek in year 1 and a 4-day workweek in year 2. Some statistics for these employees are shown in the following table Personal Leave Sick Leave Year 1 26 Year 2 Year 1 30 59 79 Year 2 37 45 56 24 19 37 20 26 4 92 65 21 62 26 73 21 34 19 63 83 35 36 26 79 1. A 4-day workweek ensures that employees will have one more day that need not be spent at work. One possible result is a reduction in the average number of personal-leave days taken by employees on a 4-day work schedule. Do the data indicate that this is the case? Use the p-value approach to testing to reach your conclusion. You may need to test whether the data are normal. A4-day workweek schedule might also have an effect on the average number of sick-leave days an employee takes. Should a directional alternative be used in this case? Why or why not? Construct a 95% confidence interval to estimate the average difference in days taken for sick leave between these 2 years. What do you conclude about the difference between the average number of sick-leave days for these two work schedules? Based on the analysis of these two variables, what can you conclude about the advantages of a 4- day workweek schedule? 2. 3. 4.Explanation / Answer
(1) here we use t-test and
with null hypotheis H0:mean1=mean2 and alternate hypothesis H1:mean1>mean2
t=(mean1-mean2)/((sp*(1/n1 +1/n2)1/2) and sp2=((n1-1)s12+(n2-1)s22)/n and with df is n=n1+n2-2
since one tailed p-value is more than alpha=0.05 , so we fail to reject H0 and conclude that there is no reduction
(2) yes, we should use directional alternate hypothesis as we want to test weather there is reduction in sick leave after implementation of 4 days week
(3) SE(difference)=((sp*(1/n1 +1/n2)1/2) =25.83*sqrt(1/11+1/11)=11.01
(1-alpha)*100% confidence interval for differene =sample difference±t(alpha/2,df)*SE(difference)
95% confidence interval =12.37±2.07*11.01=12.37±22.79=(-10.42, 35.16)
since the 0 lies in the interval, so there is no effect of policy of 4 days
(4) there is no effect of new policy of 4 days, as there is no reduction of either leave
sample mean s s2 n (n-1)s2 year1 22.36 7.86 61.7796 11 617.796 year2 18.82 11.29 127.4641 11 1274.641 difference= 3.54 189.2437 22 1892.437 sp2= 94.62185 sp= 9.727376316 t= 0.853471237 one tailed p-value= 0.201752236 two tailed p-value= 0.403504472 one tailed critical t(0.05) 1.717144335 two tailed critical t(0.05) 2.073873058