Question

10. Over the past year, the vice president for human resources at a large medical center has run a series of three-month workshops aimed at increasing worker motivation and performance. To check the effectiveness of the workshops, she selected a random sample of 35 employees from the personnel files. She collected the employee performance ratings recorded before and after workshop attendance and stored the paired ratings in the Excel data file for this assignment under the tab Perform.   

Conduct the appropriate hypothesis test and then state your findings and conclusions regarding the value of these workshops. (Significance level of 5%)

1. State your null and alternative hypothesis.
2. What is the conclusion of your test? Justify your answer, including an interpretation of the confidence interval and the p-value.
3. Assuming the cost of the training is $1000 per employee, what would be your decision regarding these workshops under the assumption that an increase in performance of at least 10% would justify the cost?

   Before	After
   59	72
   72	74
   89	62
   67	74
   81	78
   88	86
   71	81
   67	72
   78	77
   64	85
   72	80
   89	80
   87	76
   69	86
   61	84
   82	80
   82	87
   65	82
   80	76
   70	80
   76	79
   78	88
   77	83
   74	83
   63	81
   62	76
   84	79
   71	81
   68	86
   88	89
   73	75
   77	71
   83	78
   82	78
   60	94

Answer 1

Soln

	Before	After	Diff
	59	72	13
	72	74	2
	89	62	-27
	67	74	7
	81	78	-3
	88	86	-2
	71	81	10
	67	72	5
	78	77	-1
	64	85	21
	72	80	8
	89	80	-9
	87	76	-11
	69	86	17
	61	84	23
	82	80	-2
	82	87	5
	65	82	17
	80	76	-4
	70	80	10
	76	79	3
	78	88	10
	77	83	6
	74	83	9
	63	81	18
	62	76	14
	84	79	-5
	71	81	10
	68	86	18
	88	89	1
	73	75	2
	77	71	-6
	83	78	-5
	82	78	-4
	60	94	34
Mean			5.26
Std Dev			11.523

From the paired data, we calculate the Mean of Difference (After – Before) and Std Deviation of Difference using the given formula.

From the data, we calculate the mean and standard deviation

Mean (X bar) = Sum of Values /n

and

XD = 5.26

sD = 11.523

n = 35

Std Error (SE) = Std Dev / n1/2 = 1.95

a)

Null and Alternate Hypothesis

H0: µD = 0 (ie There is no difference between Performance ratings after workshop)

Ha: µD > 0 (ie Performance ratings after workshop have improved)

b)

alpha = 0.05

Test Statistic

t = XD / SE = 5.26 / 1.95 = 2.699

p-value = TDIST(2.699,35-1,1) = 0.005 (Note i have used excel to calculate the p-value using the given formula)

ZCritical = 1.96

95% CI = XD +/- SE * ZCrtical = 5.26 +/- 1.96*1.95 = {1.44, 9.07}

Interpretation of CI: There is 95% probability that true mean of Difference of Performance Rating (After Workshop – Before Workshop) lies in the interval {1.44, 9.07}

Result

Since the p-value is less than 0.05, we reject the null hypothesis.

Conclusion

Performance ratings after workshop have improved

c)

Since the upper limit of 95% ie 9.07 is less than 10, Hence the cost of training is not justified