The ABC Corporation wants to test for a difference in two methods of training ne
ID: 3259353 • Letter: T
Question
The ABC Corporation wants to test for a difference in two methods of training new employees. A sample of 15 new employees is given a three-day seminar (Method A) and a second sample of 12 new employees is given a two-day videocassette course (Method B). Each group is tested at the end of training and the results are: for Method A, the mean is 47.73 and the standard deviation is 4.42, foe Method B, the mean is 56.5 and the standard deviation is 4.27. At a 5% level of significance, can the company assume that there is No difference in the 2 methods of training? What is your conclusion?
Explanation / Answer
Solution:
Z Test for Differences in Two Means
Data
Hypothesized Difference
0
Level of Significance
0.05
Population 1 Sample
Sample Size
15
Sample Mean
47.73
Population Standard Deviation
4.42
Population 2 Sample
Sample Size
12
Sample Mean
56.5
Population Standard Deviation
4.27
Intermediate Calculations
Difference in Sample Means
-8.77
Standard Error of the Difference in Means
1.6798
Z Test Statistic
-5.2208
Two-Tail Test
Lower Critical Value
-1.9600
Upper Critical Value
1.9600
p-Value
0.0000
Reject the null hypothesis
From the above results, we will reject the null hypothesis since p-value< 0.05 and hence we can say that the company cannot assume that there is No difference in the 2 methods of training. Therefore we will conclude that there is insufficient evidence to support the claim and that there is a difference in the 2 methods of training
Z Test for Differences in Two Means
Data
Hypothesized Difference
0
Level of Significance
0.05
Population 1 Sample
Sample Size
15
Sample Mean
47.73
Population Standard Deviation
4.42
Population 2 Sample
Sample Size
12
Sample Mean
56.5
Population Standard Deviation
4.27
Intermediate Calculations
Difference in Sample Means
-8.77
Standard Error of the Difference in Means
1.6798
Z Test Statistic
-5.2208
Two-Tail Test
Lower Critical Value
-1.9600
Upper Critical Value
1.9600
p-Value
0.0000
Reject the null hypothesis
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