Question

In: Statistics and Probability

To test whether the mean time needed to mix a batch of material is the same...

To test whether the mean time needed to mix a batch of material is the same for machines produced by three manufacturers, the Jacobs Chemical Company obtained the following data on the time (in minutes) needed to mix the material. Manufacturer 1 2 3 25 30 21 31 28 20 29 33 24 27 29 23

a. Use these data to test whether the population mean times for mixing a batch of material differ for the three manufacturers. Use . Compute the values below (to 2 decimals, if necessary). Sum of Squares, Treatment Sum of Squares, Error Mean Squares, Treatment Mean Squares, Error Calculate the value of the test statistic (to 2 decimals). The -value is What is your conclusion?

b. At the level of significance, use Fisher's LSD procedure to test for the equality of the means for manufacturers and . Calculate Fisher's LSD Value (to 2 decimals). What is your conclusion about the mean time for manufacturer and the mean time for manufacturer ?

Solutions

Expert Solution

A one way ANOVA analysis is used to test whether all group means are equal. The hypothesis is defined as,

Null Hypothesis:

Alternative hypothesis; At least one mean is significantly different.

The test is performed in excel by using the following steps,

Step 1: Write the data values in excel. The screenshot is shown below

Step 2: DATA > Data Analysis > ANOVA: Single Factor > OK.  The screenshot is shown below,

Step 3: Select Input Range: All the data values column, Grouped By: Columns. The screenshot is shown below,

The result is obtained.  The screenshot is shown below,

a)

Sum of Squares, Treatment = 138.67

Sum of Squares, Error = 44

Mean Squares, Treatment = 69.33

Mean Squares, Error = 4.89

F Statistic = 14.18

P-value = 0.0017 < 0.05 at a 5 % significance level hence the null hypothesis is rejected.

b)

Fisher's LSD

The LSD value is obtained using the formula,

Since the n1 = n2 = n3 = 4, the LSD value will be the same for all pairwise comparison.

Where,

From ANOVA result summary, Mean Squares, Error (MSw) = 4.89

Difference Absolute Value LSD Conclusion
2 <    3.41 No Significant Difference
6 > 3.41 Significant Difference
8 > 3.4 Significant Difference

Group 1 and group 2 => not different

Group 1 and group 3 => significantly different

Group 2 and group 3 => significantly different


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