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Homework answers / question archive / A supervisor of a manufacturing plant is interested in relating the average number of defects produced per day to two factors: the operator working the machine and the machine itself
A supervisor of a manufacturing plant is interested in relating the average number of defects produced per day to two factors: the operator working the machine and the machine itself. The supervisor randomly assigns each operator to use each machine for three days and records the number of defects produced per day. Is there sufficient evidence to conclude that there is a significant difference among the average number of defects produced per day for the different machines?
| Operator A | Operator B | |
|---|---|---|
| Machine A | 9 | 8 |
| 7 | 9 | |
| 7 | 3 | |
| 6 | 5 | |
| Machine B | 5 | 5 |
| 2 | 1 | |
| 2 | 4 | |
| 2 | 7 | |
| Machine C | 5 | 0 |
| 6 | 7 | |
| 9 | 0 | |
| 2 | 0 |
| Source of Variation | SSSS | dfdf | MSMS |
|---|---|---|---|
| Total | 203.6250 | 23 | |
| Machine | 54.2500 | 2 | 27.1250 |
| Operator | 7.0417 | 1 | 7.0417 |
| Interaction | 27.5833 | 2 | 13.7917 |
| Within | 114.7500 | 18 | 6.3750 |
Step 1 of 2:
Find the value of the test statistic for testing the difference among the average number of defects produced per day for the different machines. Round your answer to two decimal places, if necessary.
Step 2 of 2:
Make the decision to reject or fail to reject the null hypothesis of equal average number of defects produced per day for the different machines and state the conclusion in terms of the original problem. Be sure to test for interaction first. Use α=0.05
Answer:
The Hypothesis:
H0: The average number of defects produced by the 3 machines are equal.
Ha: The average number of defects produced by the 3 machines are not all equal.
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(1) Here Machine is the Main Effect A and Operator is the Main effect B.
We are finding the test statistic for testing the average means for the defects produced my machines only per day.
F Machines = MS Machines / MS error = 27.125/6.375 = 4.25
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(2) The p value for Fmachine = 4.25 with dfmachine = 2 and df error = 18 is 0.0308
The Rejection Rule: If p value is < , Reject H0.
The Decision: Since p value (0.0308) is < 0.05, Reject H0.
The Conclusion: Reject H0. There is sufficient evidence at = 0.05 to warrant rejection of the claim that the average number of defects produced by the 3 machines are all equal.
