Question

In order to compare two computer software packages, a manager has 15 individuals use each software package to perform a standard set of tasks typical of those encountered in the office. Of course, in carrying out the comparison the manager was careful to use individuals who did not have an established preference of skill with either type of software, and 15 individuals were randomly selected to use software A first while the other 15 used software B first. The time required to perform the standard set of tasks, to the nearest minute, is reported in Table 1. Test the null hypothesis that there is no difference between the mean time required to perform the standard tasks by the two software packages, using the 5% level of significance.

Table:1

Individual	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15
Software A	12	16	15	13	16	10	15	17	14	12	13	14	10	11	15
Software B	10	17	18	16	19	12	17	15	17	14	17	11	18	15	13

P/s: Please show the answer in full worksheet and full method, don't make it simple

Answer 1

Suppose, random variables X and Y denote time taken by individuals for software packages A and B respectively.

Here, two different groups of users are used to collect data in case of two different software packages. Further we do not know population standard deviation (or variance). So, we have to perform two sample t-test.

We have to test for null hypothesis

against the alternative hypothesis

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

Pooled sample variance is given by

Degrees of freedom

[Using R-code 'pt(-1.911558,28)+1-pt(1.911558,28)']

Level of significance

We reject our null hypothesis if

Here, we observe that

So, we cannot reject our null hypothesis.

Hence, based on the given data we can conclude that there is no significant evidence that there is difference between the mean time required to perform the standard tasks by the two software packages.