Question

A company was thinking about instituting a policy that allowed for no limits to the number of sick days that employees could take rather than limiting the number of paid sick days. Each column represents a different sample of employees during the trial period.

A) Explain why an alpha level of 0.05 is appropriate for the test (compared to, say, 0.01 or 0.001).

B) What are we doing by setting an alpha level? Explain it for the layperson.

C) What is the null hypothesis and alternate hypothesis? (Do not worry about formatting the statements - I'll get it from what you write.)

D) Conduct the t-test at α = 0.05, and report the results and conclusion. Provide in your statement the necessary statistics to support your answer.

Limits	No Limits
35	23
21	12
26	6
24	15
17	18
23	5
37	21
22	18
16	34
38	10
23	23
41	14
27	19
24	23
32	8

Answer 1

A)

Level of significance alpha shows the probability of rejecting the true null hypothesis. That is we incorrectly conclude that number of leaves with limit and no limit are not same while actually they are same. Since this error is not very critical so we can assume alpha = 0.05.

B)

It is maximum probability that we allow to conclude that number of leaves with limit and no limit are not same while actually they are same.

C)

Following table shows the calculations:

Limits	No Limits	d=limits-no limits	(d-mean)^2
35	23	12	2.3409
21	12	9	2.1609
26	6	20	90.8209
24	15	9	2.1609
17	18	-1	131.5609
23	5	18	56.7009
37	21	16	30.5809
22	18	4	41.8609
16	34	-18	810.5409
38	10	28	307.3009
23	23	0	109.6209
41	14	27	273.2409
27	19	8	6.1009
24	23	1	89.6809
32	8	24	183.0609
Total		157	2137.7335