Please help Workers completed an intensive training program to reduce the number
ID: 3297513 • Letter: P
Question
Please help
Workers completed an intensive training program to reduce the number of defective parts produced. Below are data on the number of defects per shift before and after the training. 1. Describe what would happen if you made a Type I error in determining program effectiveness. What might be the management outcome? 2. Describe what would happen if you made a Type II error in determining program effectiveness. What might be the management outcome? 3. Use hypothesis testing to determine if the program was effective in reducing defects. 4. If the goal is to have 5 or fewer defects per worker per shift, has the training program met the goal? Use hypothesis testing to answer the question.Explanation / Answer
x = Defects before training
y = Defects after training
x
y
15
8
12
5
16
10
14
5
18
14
12
4
13
6
17
6
10
3
16
12
Question 1) In this situation Type I error will occur if by mistake it will be concluded that intensive training program helped to reduce the number of defective parts but actually it did not help to reduce. The management should not introduce the same training program for all the employees at present.
Question 2) In this situation Type II error will occur if by mistake it will be concluded that intensive training program did not help to reduce the number of defective parts but actually it helped to reduce. The management should introduce the same training program for all the employees.
Question 3)
Null and Alternative Hypothesis:
H0: µd = 0
H1: µd > 0
Level of significance is 0.05
Test Statistics:
t = ( d bar ) / ( sd / sqrt (n))
First we find the difference (d) then by using excel function we find the sample mean and sample standard deviation.
d bar = 7
sd = 2.1082
n = 10
t = (7/)/(2.1082/sqrt(10)) = 10.5
The test statistics is 10.5
Now we find the critical t for 9 degrees of freedom at 5% level of significance from t table as 1.833
The value of test statistics is greater than the critical value (10.5 > 1.833); we reject the null hypothesis.
At 5% level of significance there is sufficient evidence to conclude that the program was effective in reducing defects.
Question 4)
Null and Alternative Hypothesis:
H0: µd 5
H1: µd > 5
Level of significance is 0.05
Test Statistics:
t = ( d bar -5 ) / ( sd / sqrt (n))
First we find the difference (d) then by using excel function we find the sample mean and sample standard deviation.
d bar = 7
sd = 2.1082
n = 10
t = (7-5)(2.1082/sqrt(10)) = 3
The test statistics is 3
Now we find the critical t for 9 degrees of freedom at 5% level of significance from t table as 1.833
The value of test statistics is greater than the critical value (3 > 1.833); we reject the null hypothesis.
At 5% level of significance there is not sufficient evidence to conclude that the goal of 5 or fewer defects has reached by this training program. More than 5 in average got reduced.
x
y
15
8
12
5
16
10
14
5
18
14
12
4
13
6
17
6
10
3
16
12