Question

A manager wishes to see if the time (in minutes) it takes for their workers to complete a certain task will change when they are allowed to wear ear buds at work. A random sample of 25 workers' times were collected before and after. Test the claim that the time to complete the task has changed at a significance level of α=0.10. For the context of this problem, μd=μafter−μbefore where the first data set represents after and the second data set represents before. Assume the population is normally distributed. Round answers to 4 decimal places.

Ho:μd=0
Ha:μd≠0

You obtain the following sample of data:

After	Before
52.7	53.3
43.1	42.7
55.9	61.2
22.5	23.9
32.5	47.6
53.5	50
50.1	50
58.5	54.2
30.1	25.8
20.4	18.7
49.3	60.5
51.4	57.6
68.6	69.7
54.2	52.9
44.5	43.9
16.2	29.5
49.6	56.1
35.5	36.7
61	58.2
33.8	45
29.6	28.8
37.6	30.7
51.3	48.8
16.6	21.7
33.5	36.7

What is the critical value for this test?  
critical value = ± ________________

What is the test statistic for this sample?  
test statistic = __________________

The test statistic is: in the critical region or not in the critical region

This test statistic leads to a decision to?

Reject the claim

Fail to reject the null

Accept the null

reject the null

Accept the claim

As such, the final conclusion is that...

__There is sufficient evidence to warrant rejection of the claim that the time to complete the task has changed.

__There is not sufficient sample evidence to support the claim that the time to complete the task has changed.

__The sample data support the claim that the time to complete the task has changed.

__There is not sufficient evidence to warrant rejection of the claim that the time to complete the task has changed.

Answer 1

Here we have given that

Claim: To check whether the time to complete the task has changed at a significance level of α=0.10.

The Hypothesis is

v/s

Now,

After	Before	Diff=after-before
52.7	53.3	-0.6
43.1	42.7	0.4
55.9	61.2	-5.3
22.5	23.9	-1.4
32.5	47.6	-15.1
53.5	50	3.5
50.1	50	0.1
58.5	54.2	4.3
30.1	25.8	4.3
20.4	18.7	1.7
49.3	60.5	-11.2
51.4	57.6	-6.2
68.6	69.7	-1.1
54.2	52.9	1.3
44.5	43.9	0.6
16.2	29.5	-13.3
49.6	56.1	-6.5
35.5	36.7	-1.2
61	58.2	2.8
33.8	45	-11.2
29.6	28.8	0.8
37.6	30.7	6.9
51.3	48.8	2.5
16.6	21.7	-5.1
33.5	36.7	-3.2

we get,

n= Number of observation =25

= sample mean of differences = -2.09  

Sd= sample standard deviation of differences =5.82

here we assume that population is normally distributed

Now, we can find the test statistic

= -1.79

Now, we find the critical value

Degrees of freedom = n-1= 25-1=24

= level of significance=0.10

we get

T-critical =1.711 ( using t -table)

Decision:

| T-statistics | > T-critical

i.e Here we Reject the Ho Null hypotheisis

Conclusion:

There is sufficient evidence to warrant rejection of the claim that the time to complete the task has changed.