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 specification among the three methods. After conducting the analysis report the findings to the management team.a. Compute the sum of squares between treatments (assembly methods).b. Compute the mean square between treatments (to 1 decimal if necessary).c. Compute the sum of squares due to error.d. Compute the mean square due to error (to 1 decimal if necessary).

Answer 1

To analyze the data and determine any difference in the mean number of components out of specification among the three assembly methods, an ANOVA should be performed. Several statistics need to be calculated including the sum of squares between treatments, mean square between treatments, sum of squares due to error, and mean square due to error.

Step-by-step explanation:

To test the three assembly methods (Method A, Method B, and Method C), six batches of 10,000 components are produced using each method. The number of components that do not meet specifications are recorded. In order to analyze the data and determine if there is any difference in the mean number of components that are out of specification among the three methods, we need to perform an analysis of variance (ANOVA). This involves calculating several statistics.

Part a:

To compute the sum of squares between treatments (assembly methods), we first need to calculate the grand mean. This is the average number of components that do not meet specifications across all treatments. Then, for each treatment, we calculate the sum of squares by subtracting the treatment mean from the grand mean and squaring the result. These individual sum of squares values are then summed to get the sum of squares between treatments.

Part b:

The mean square between treatments is calculated by dividing the sum of squares between treatments by the degrees of freedom between treatments.

Part c:

To compute the sum of squares due to error, we need to calculate the sum of squares within treatments. This is done by subtracting each observation from its respective treatment mean, squaring the result, and summing these individual sum of squares values.

Part d:

The mean square due to error is calculated by dividing the sum of squares due to error by the degrees of freedom within treatments.

Answer 2

Check the explanation

Step-by-step explanation:

a) Compute the sum of squares between treatments.

SS between treatments =1488

b) Compute the mean square between treatments.

MSS between Vestments = 744

c) Compute the sum of squares due to error.

SS dne to error = 2030

d) Compute the mean square due 1m error (to 1 decimal).

MSS due to error = 135.3

e) Set up the ANOVA table for this problem.

ANOVA

Source of Variation SS df MS P-volue F crit

Between Groups 1488 744 5.50 0.0162 3.68232

Within Groups 2030 15 135.3333

Total 3518 17

Al the = .05 level of significance test whether the means for the three treatments are equal. Calculate the value of the VA statistic ( o 2 decimals).

Test Statistics F = 5.50

The p-value M 0.0162

What is your conclusion? Reject the null hypothesis. There is a significant difference between the treatments effects. The p-value is between 0.01 and 0.025. Conclude that the means for all the treatments are not equal.