Question

The director of the Wisconsin Department of Business Licensing is looking for ways to improve employee productivity. Specifically, she would like to see an improvement in the percentage of applications that employees process correctly. The director randomly selects 50 employees and gather data on the percentage of applications each one correctly processed last month. On the recommendation of a consultant, the director has these 50 employees complete a 3-day workshop on Proactive Synergy Restructuring Techniques. At the end of the month following the training, the director collects the application processing data for the same 50 employees.

Help the director analyze these data by conducting a hypothesis test. From a statistical point of view, what can you tell the director?

employee	score1	score2
1	93	91
2	94	96
3	98	100
4	94	96
5	92	94
6	95	97
7	98	100
8	96	98
9	94	96
10	98	100
11	96	98
12	91	93
13	96	98
14	94	97
15	92	90
16	98	100
17	97	99
18	96	98
19	98	99
20	90	92
21	96	95
22	90	92
23	90	93
24	94	96
25	98	96
26	96	98
27	92	94
28	96	93
29	92	94
30	96	94
31	95	97
32	90	92
33	96	98
34	96	98
35	94	95
36	96	98
37	94	96
38	98	97
39	93	95
40	97	99
41	92	91
42	95	97
43	99	98
44	91	93
45	93	95
46	95	97
47	92	95
48	96	98
49	93	95
50	97	98

Answer 1

First enter Data into EXCEL

We have to find the sample mean.

Excel command is =AVERAGE(Select data)

Now we have to find sample standard deviation.

Excel command is =STDEV(Select data)

n1 = 50

n2 = 50

s1 = 2.50

s2 = 2.61

Null and alternative hypothesis is

H0 :u1 = u2

Vs

H1 :u1 ≠ u2

Level of significance = 0.05

Before doing this test we have to check population variances are equal or not.

Null and alternative hypothesis is

Vs

Test statistic is

F = Larger variance / Smaller variance = 6.80 / 6.24 = 1.090

Degrees of freedoms

Degrees of freedom for numerator = n1 - 1 = 50 - 1 = 49

Degrees of freedom for denominator = n2 - 1 = 50 - 1 = 49

Critical value = 1.607                ( using f-table )

F test statistic < critical value , we fail to reject null hypothesis.

Conclusion: Population variances are equal.

So we have to use pooled variance.

Formula

         d.f = n1 + n2 – 2 =98

p-value = 0.010 ( using t table )

p-value < α       Reject H0