Question

To test whether the mean time needed to mix a batch af 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 use these data to test whether the pptation mean times for moing a batch d material difrir Compute the values belaw (to 2 decma's, necessary) Sum of Squares, Treatment |-78 Sum of Squares, Error Moan Squares, Treatment39 Moan Squares, Error the treennaatuies. Usc a-.05. 81 3 Calculate the value of the test statistc(to 2 decmals). The p-value is | between as and .10回! What your condusion? Conclude the mearn batch of material is not the same b. At the α·.05 level of significance, use FShr's LSD procedure to test for the couaityof the means for manufacturers 1 and 3 Calculate Fsher's LSD Value (to 2 decimals). What is your condusion about the mean time for manufacturer 1 and the moan time for manufacturer 3? Cannet conclude there io a diffinence il he mdan Lime for the mauictuen

Answer 1

a)The Hypothesis are

Ho : Mean1=Mean2=Mean3

Ha: Mean1$\neq$Mean2$\neq$Mean3

Using excel Add on the ANOVA table is shown below

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
1	4	87	21.75	14.25
2	4	108	27	18
3	4	84	21	12
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	85.5	2	42.75	2.898305	0.106748	4.256495
Within Groups	132.75	9	14.75
Total	218.25	11

Hence,

Sum of square, Treatment=85.5

Sum of Square, Error=132.75

Mean Square Treatment= 42.75

Mean square, Error=14.75

The teat statistics=4.26

The p valueis between 0.10 and 0.25, (The option is not shown, you can choose accordingly, P value =0.107 )

Conclusion:

Cannot conclude that the mean time needed to mix a batch of material is not the same for all manufacturers,.

Because the P value is greater than the level of significance hence we fail to reject the Null hypothesis.

b) The Fishers Lsd is calculated bt formula

Here For sample 1 and 3

MSW=14.75

DFW= Degree of freedom Within=9

t_0.005=2.26

n_A=Sample 1 size=4 also n_B=Sample 3 size

Npw by formulae

$\Rightarrow 2.26*\sqrt{14.75(\frac{1}{4}+\frac{1}{4})}=6.14$

and if we look at

Mean1-Mean3=21.75-21=0.75

which is smaller than the Fishers LSD hence We fail to reject the null hypothesis and

Conclusion:

Cannot conclude that there is a difference in the mean of Sample 1 and Sample 3.