Question

To compare two programs for training industrial workers to perform a skilled job, 12 workers are included in an experiment. Of these, 5 workers are selected at random and trained by method 1; the remaining are trained by method 2. After completion of training, all the workers are subjected to a time-and-motion test that records the speed of performance of a skilled job. The following data are obtained.

Method 1 - 15 20 11 23 16

Method 2 - 23 31 13 19 23 16 22

(a) State the null hypothesis and alternative hypothesis

(b) Find the test statistic. You need to use (larger of S2 / smaller of S2) to justify split or pooled t-test

(c) If the type-I error α = 0.1, find the critical value(s) and shade the rejection region(s)

(d) Base on the type-I error α and rejection region(s), given above, what is your conclusion?

Answer 1

Suppose, random variables X and Y denote speed of performance of method 1 and method 2 respectively. Here, two different groups are used to collect data after two different methods of training. Further we do not know population standard deviation (or variance). So, we have to perform two sample t-test.

(a)

We have to test for null hypothesis

against the alternative hypothesis

(b)

Our test statistic is given by

Here,

First sample size

Second sample size

Sample mean of first sample is given by

Sample mean of second sample is given by

(c)

Level of significance

Critical values are given by

So, the rejection regions (shaded in red) are as follows.

(d)

We observe that our calculated test statistic does not lie in critical region. So, we cannot reject our null hypothesis.

Hence, based on the given data we can conclude that there no significant evidence that speeds of performance are different for these two training methods.