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ID: 451998 • Letter: H
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home / study / math / statistics and probability / questions and answers / a manager is thinking of providing, on a regular ... Question A manager is thinking of providing, on a regular basis, in-house training for employees preparing for an inventory management certification exam. In the past, some employees received the in-house training before taking the exam, while others did not. Independent random samples taken from the company’s records provided the following exam scores for 10 workers who did not receive in-house training and 8 workers who did receive training. (The manager is confident that the distributions of both populations’ exam scores are approximately normal.) No Training 76 80 60 91 73 77 82 68 75 86 Training 80 66 71 79 94 74 83 78 Make sure to state the hypothesis null and alternative. Highligh the important results. Explain why you reject Ho indicating the values you have calculated. Using a = .05, test for any difference between the average test scores for the two populations of employees.
Explanation / Answer
We are interested to check about the effectiveness of Training with the help of sample means.
Therefore the Null hypothesis, H0: Training is not effective and there is no difference between the average scores of the two samples, say X = Y with the alternate hypothesis that X > Y (Training is effective) where X represents mean for Training and Y represents Mean for No Training.
Keeping in view the sample size (small), for testing the difference between means of given samples (independent), we are required to calculate the value of "t" ={ (X-Y)/S}{SQRT(n1n2/(n1+n2))} where S is combined standard deviation calculated as SQRT[{sigma(xi-X)2 + sigma(yi-Y)2}/(n1+n2-2)]
Calculated value of "t" will be compared with the tabulated value to give the final conclusion.
The calculated value of "t" is 0.485 which is less than the tabulated values of "t" with degree of freedom=16(10+8-2) is 1.746. Therefore Null Hypothsis is accepted or concluded that there is no significant difference between the two sample averages.
Training(x) No Training(y) x-X (xi-X)2 y-Y (yi-Y)2 1 80 76 1.875 3.515625 -0.2 0.04 2 66 80 -12.125 147.0156 3.8 14.44 3 71 60 -7.125 50.76563 -16.2 262.44 4 79 91 0.875 0.765625 14.8 219.04 5 94 73 15.875 252.0156 -3.2 10.24 6 74 77 -4.125 17.01563 0.8 0.64 7 83 82 4.875 23.76563 5.8 33.64 8 78 68 -0.125 0.015625 -8.2 67.24 9 75 -1.2 1.44 10 80 3.8 14.44 TOTAL 625 762 494.875 623.6 Average 78.125 76.2 X=78.125 Y=76.2 "X - Y" = 1.925 S' = 8.36 SQRT(n1n2/(n1+n2) 2.11 Calculated "t"= 0.485855Related Questions
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