Question

A manufacturing company has been selected to assemble a small but important component that will be used during the construction of numerous infrastructure projects. The company anticipates the need to assemble several million components over the next several years. Company engineers select three potential assembly methods: Method A, Method B, and Method C. Management would like to select the method that produces the fewest number parts per 10,000 parts produced that do not meet specifications. It may also be possible that there is no statistical difference between the three methods in which case the lowest cost method will be selected for production. While all parts are checked before leaving the factory, the best method will reduce the number of parts that need to be recycled back into the production process.

To test each method, six batches of 10,000 components are produced using each of the three methods. The number of components out of specification are recorded in the Microsoft Excel Online file below. Analyze the data to determine if there is any difference in the mean number of components that are out of specificaion among the three methods. After conducting the analysis report the findings to the management team.

Batch	Method A	Method B	Method C
1	164	143	123
2	142	156	121
3	165	129	139
4	145	143	141
5	147	135	152
6	173	152	128

1. Compute the sum of squares between treatments (assembly methods).

2. Compute the mean square between treatments (to 1 decimal if necessary).

3. Compute the sum of squares due to error.

4. Compute the mean square due to error (to 1 decimal if necessary).

5. Set up the ANOVA table for this problem. Round all sum of squares to the nearest whole number. Round all Mean Squares to one decimal place. Round F to two decimal places.

   Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F
   Treatments (Methods)
   Error
   Total

6. At the ? = 0.05 level of significance, test whether the means for the three methods are equal.

   Calculate the value of the test statistic (to 2 decimals).

   The p-value is (to 4 decimals):

   What is your conclusion for management?

   _________Conclude that not all method means are equal.Conclude that all method means are equal.

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Answer 1

1. Compute the sum of squares between treatments (assembly methods).

   1468

2. Compute the mean square between treatments (to 1 decimal if necessary).

   734

3. Compute the sum of squares due to error.

   2066

4. Compute the mean square due to error (to 1 decimal if necessary).

   137.7

5. Set up the ANOVA table for this problem. Round all sum of squares to the nearest whole number.

Source	SS	df	MS	F	p-value
Treatment	1,468	2	734.0	5.33	.0178
Error	2,066	15	137.7
Total	3,534	17

At the ? = 0.05 level of significance, test whether the means for the three methods are equal.

Calculate the value of the test statistic (to 2 decimals).

5.33

The p-value is (to 4 decimals):

0.0178

What is your conclusion for management?

Conclude that not all method means are equal.

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