The manager of an assembly process wants to determine whether or not the number
ID: 3152716 • Letter: T
Question
The manager of an assembly process wants to determine whether or not the number of defective articles manufactured depends on the day of the week the articles are produced. She collected the following information. Is there sufficient evidence to reject the hypothesis that the number of defective articles is independent of the day of the week on which they are produced? Use = 0.05.
(a) Find the test statistic. (Give your answer correct to two decimal places.)
(ii) Find the p-value. (Give your answer bounds exactly.)
_________< p < __________
Explanation / Answer
Solution:
Test statistic value = 5.1933
p-value = 0.26803
Here, we get the p-value = 0.26803 which is greater than the given level of significance so we do not reject the null hypothesis that the number of defective articles is independent of the day of the week on which they are produced.
The detailed test is given as below:
Test statistic formula = [(O – E)^2 / E]
Chi-Square Test
Observed Frequencies
Column variable
Row variable
M
Tu
W
Th
F
Total
Non-defective
95
89
93
85
91
453
Defective
11
13
7
15
7
53
Total
106
102
100
100
98
506
Expected Frequencies
Column variable
Row variable
M
Tu
W
Th
F
Total
Non-defective
94.89723
91.31621
89.52569
89.52569
87.73518
453
Defective
11.10277
10.68379
10.47431
10.47431
10.26482
53
Total
106
102
100
100
98
506
Data
Level of Significance
0.05
Number of Rows
2
Number of Columns
5
Degrees of Freedom
4
Results
Critical Value
9.487729
Chi-Square Test Statistic
5.19333
p-Value
0.26803
Do not reject the null hypothesis
Expected frequency assumption
is met.
Chi-Square Test
Observed Frequencies
Column variable
Row variable
M
Tu
W
Th
F
Total
Non-defective
95
89
93
85
91
453
Defective
11
13
7
15
7
53
Total
106
102
100
100
98
506
Expected Frequencies
Column variable
Row variable
M
Tu
W
Th
F
Total
Non-defective
94.89723
91.31621
89.52569
89.52569
87.73518
453
Defective
11.10277
10.68379
10.47431
10.47431
10.26482
53
Total
106
102
100
100
98
506
Data
Level of Significance
0.05
Number of Rows
2
Number of Columns
5
Degrees of Freedom
4
Results
Critical Value
9.487729
Chi-Square Test Statistic
5.19333
p-Value
0.26803
Do not reject the null hypothesis
Expected frequency assumption
is met.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.