Air Force Training Program

An Air Force introductory course uses a system which guides students to view a videotaped lecture and then offers them a programmed instruction text. Students work independently until they complete the training and pass the test. One of the problems is that students' spaces vary from each other to complete this portion of their training. Some students are able to cover the program relatively quick, while others might take longer, and the faster students have to wait for the slower students so that the entire group could move to the next level of training together.

To solve this program, there is a proposed alternative system which involves using a computer assisted instruction. With this method, each student is assigned to a computer which guides students to work independently through the self-training portion. To compare the proposed method with the current method, a class of 122 students was assigned randomly into two groups. One group of 61 students used the current method, and the other group of 61 students used the proposed method. The time was recorded for each student and the basic results are listed in the following table:

Proposed Method Current Method

Count 61 61

Mean 64.99 70.84

Median 64.92 71.73

Range 38.76 25.34

Sample Variance 50.93 29.16

As can be seen, the mean value of the proposed method is lower than the current method; however, the range and the sample variance are higher than the current method. This is to say, although the mean completion time of proposed method is faster than the other group, but their paces vary a lot with the proposed method.

To ensure which method is more effectively, we need to compare the difference between the two population means, and we made the following hypothesis test:

H0: µ1-µ2 ≤ 0

Ha: µ1-µ2 > 0

We use µ1 to represent the population mean of completion time with the proposed method, and µ2 to represent the population mean of completion time with the current method. We develop a 0.05 level of significance. Since it is an upper-tail test, we could only reject H0 when the p-value of this test is less than 0.05, or the test statistic is higher than the critical value.

T-test Results

Test Statistic -5.21