Answer :
1) Critical Value: 2.1967
2) Test Statistic (t): 1.8402
3) (A) There is not sufficient sample evidence to support the claim that the time to complete the task has Increased.
Step 1: Calculate the Differences
We calculate the differences between the "After" and "Before" times for each worker:
Before After Difference (After - Before)
53.7 49.3 -4.4
54.0 43.1 -10.9
43.8 50.3 6.5
41.4 69.5 28.1
37.7 60.9 23.2
37.1 58.6 21.5
31.1 66.4 35.3
18.8 24.7 5.9
44.0 44.7 0.7
44.3 49.8 5.5
38.2 45.4 7.2
48.0 56.2 8.2
65.7 46.4 -19.3
49.3 70.1 20.8
78.5 64.9 -13.6
57.8 63.9 6.1
48.5 52.0 3.5
30.7 51.7 21.0
39.9 38.1 -1.8
59.9 39.6 -20.3
34.2 38.1 3.9
Step 2: Calculate the Mean Difference
The mean difference (d) is calculated as follows:
d = Σ(Differences) / n
= 127.1 / 21
= 6.0524
Step 3: Calculate the Standard Deviation of the Differences
The standard deviation (sd) is calculated as follows:
First, we calculate each (di - d)²:
- (-4.4 - 6.0524)² = (-10.4524)² = 109.1744
- (-10.9 - 6.0524)² = (-16.9524)² = 288.0344
- (6.5 - 6.0524)² = (0.4476)² = 0.2000
- (28.1 - 6.0524)² = (22.0476)² = 485.2344
- (23.2 - 6.0524)² = (17.1476)² = 293.9344
- (21.5 - 6.0524)² = (15.4476)² = 238.7344
- (35.3 - 6.0524)² = (29.2476)² = 854.9344
- (5.9 - 6.0524)² = (-0.1524)² = 0.0232
- (0.7 - 6.0524)² = (-5.3524)² = 28.5856
- (5.5 - 6.0524)² = (-0.5524)² = 0.3049
- (7.2 - 6.0524)² = (1.1476)² = 1.3184
- (8.2 - 6.0524)² = (2.1476)² = 4.6336
- (-19.3 - 6.0524)² = (-25.3524)² = 642.0344
- (20.8 - 6.0524)² = (14.7476)² = 217.3716
- (-13.6 - 6.0524)² = (-19.6524)² = 386.4484
- (6.1 - 6.0524)² = (0.0476)² = 0.0023
- (3.5 - 6.0524)² = (-2.5524)² = 6.5184
- (21.0 - 6.0524)² = (14.9476)² = 223.0244
- (-1.8 - 6.0524)² = (-7.8524)² = 61.3196
- (-20.3 - 6.0524)² = (-26.3524)² = 694.8484
- (3.9 - 6.0524)² = (-2.1524)² = 4.6456
Sum of Squares of Differences:
Σ(di - d)² = 5298.9389
Standard Deviation of Differences:
sd = √(Σ(di - d)² / (n-1))
= √(5298.9389 / 20)
= √264.9469
= 15.0716
Step 4: Calculate the Test Statistic
The test statistic is calculated as follows:
t = d / (sd / √n)
= 6.0524 / (15.0716 / √21)
= 6.0524 / 3.2872
= 1.8402
Step 5: Determine the Critical Value
For a one-tailed t-test at α = 0.02 with n - 1 = 20 degrees of freedom, we find the critical value:
Critical Value:
t0.02, 20 ≈ 2.1967
Since the test statistic (1.8402) is less than the critical value (2.1967), we fail to reject the null hypothesis at the α = 0.02 significance level. Therefore, there is not enough evidence to support the claim that the time to complete the task has increased when workers are allowed to wear earbuds.
Full Question:
A manager wishes to see if the time (in minutes) it takes for their workers to complete a certain task will Increase when they are allowed to wear ear buds at work. A random sample of 21 workers' times were collected before and after. Test the claim that the time to complete the task has increased at a significance level of a=0.02. For the context of this problem, 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.
H=0
You obtain the following sample of data:
See table attached.
1) What is the critical value for this test?
critical value =
2) What is the test statistic for this sample?
test statistic =
3) This test statistic leads to a decision to?
As such, the final conclusion is that....?
(A) There is not sufficient sample evidence to support the claim that the time to complete the task has Increased.
(B) The sample data support the claim that the time to complete the task has increased.
(C) There is not sufficient evidence to warrant rejection of the claim that the time to complete the task has increased.
(D) There is sufficient evidence to warrant rejection of the claim that the time to complete the task has increased.
