Question

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
	23	33	15
	29	31	14
	27	36	18
	25	32	17

a. Use these data to test whether the population mean times for mixing a batch of material differ for the three manufacturers. Use a=.05.

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 p -value is - Select your answer -less than .01between .01 and .025between .025 and .05between .05 and .10greater than .10Item 6

What is your conclusion?

- Select your answer -Conclude the mean time needed to mix a batch of material is not the same for all manufacturersDo not reject the assumption that mean time needed to mix a batch of material is the same for all manufacturersItem 7

b. At the a=.05 level of significance, use Fisher's LSD procedure to test for the equality of the means for manufacturers 1 and 3.

Calculate Fisher's LSD Value (to 2 decimals).

What is your conclusion about the mean time for manufacturer 1 and the mean time 3 for manufacturer ?

- Select your answer -These manufacturers have different mean timesCannot conclude there is a difference in the mean time for these manufacturers

Answer 1

a.

Let Ti be the total time for group i, ni be number of observations of group i.

Let G be the total time of all observations and N be total number of observations.

ΣX2 is sum of squares of all observations.

T1 = 104, T2 = 132 , T3 = 64

G = 104 + 132 + 64 = 300

ΣX2 = 2724 + 4370 + 1034 = 8128

Sum of Squares, Total = ΣX2 - G2/N = 8128 - 300^2/12 = 628

Sum of Squares, Treatment, SSTR = ΣT^2/n - G^2/N = (104^2 /4 + 132^2 /4 + 64^2 /4 ) - 300^2/12 = 584

Sum of Squares, Error, SSE = 628 - 584 = 44

DF for Treatment = Number of groups - 1 = 3 - 1 = 2

DF for Error = Number of observations - Number of groups = (4 * 3) - 3 = 9

Mean Squares, Treatment, MSTR = SSTR / DF for Treatment = 584 / 2 = 292

Mean Squares, Error, MSE = SSE / DF for Error = 44 / 9 = 4.89

F = MSTR / MSE = 292 / 4.89 = 59.7137

DF for F statistic = DF for Treatment , DF for Error = 2,9

Critical value of F at 0.01 significance level and df = 2,9 is 8.02

Since, F is greater than the critical value,

The p -value is less than .01

Reject the null hypothesis. Conclude the mean time needed to mix a batch of material is not the same for all manufacturers.

b.

Critical value of t at 0.05 significance level and df = 9 is 1.83

Fisher's LSD Value = t where n is number of obervations in each group

= 1.83

= 2.86

Mean time for manufacturer 1 = 104 / 4 = 26

Mean time for manufacturer 3 = 64 / 4 = 16

Mean difference = 26 - 16 = 10

Since the mean time difference for manufacturer 1 and 3 are greater than critical Fisher's LSD Value, we conclude that

These manufacturers have different mean times.