Question

The following data are from a completely randomized design.

	Treatment	Treatment	Treatment
	A	B	C
	32	45	35
	30	44	38
	30	45	37
	26	47	38
	32	49	42
Sample mean	30	46	38
Sample variance	6	4	6.5

A) At the  = .05 level of significance, can we reject the null hypothesis that the means of the three treatments are equal?

Compute the values below (to 1 decimal, 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 Selectless than .01between .01 and .025between .025 and .05between .05 and .10greater than .10Item 6

What is your conclusion?
SelectConclude that not all treatment means are equalDo not reject the assumption that the treatment means are equalItem 7

B) Calculate the value of Fisher's LSD (to 2 decimals).


Use Fisher's LSD procedure to test whether there is a significant difference between the means for treatments A and B, treatments A and C, and treatments B and C. Use  = .05.

Difference	Absolute Value	Conclusion
A -  B		SelectSignificant differenceNo significant differenceItem 10
A -  C		SelectSignificant differenceNo significant differenceItem 12
B -  C		SelectSignificant differenceNo significant differenceItem 14

Use Fisher's LSD procedure to develop a 95% confidence interval estimate of the difference between the means of treatments A and B (to 2 decimals). Since treatment B has the larger mean, the confidence interval for the difference between the means of treatments A and B ( A -  B) should be reported with negative values.
( , )

Answer 1

(a)

Following table shows the group totals:

	a, G1	b, G2	c, G3
	32	45	35
	30	44	38
	30	45	37
	26	47	38
	32	49	42
Total	150	230	190

And following table shows the grand total and total of square of observations:

	G	G^2
	32	1024
	30	900
	30	900
	26	676
	32	1024
	45	2025
	44	1936
	45	2025
	47	2209
	35	1225
	38	1444
	37	1369
	38	1444
	42	1764
Total	521	19965

So we have

Now

Now

Now

Since there are 3 different groups so we have k=3. Therefore degree of freedoms are:

-------------

Now

F test statistics is

P-value:

p-value less than .01

Conclusion:

Conclude that not all treatment means are equal

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The fisher LSD procedure to test wheter there is a signficant difference between the means:

Critical value of t for level of significance 0.05 and degree of freedom of error is dfW = 12 is 2.179.

And we have

Since sample sizes are same so LSD will be

Absolute difference between sample mean A and sample mean B is : 16

Since absoulte difference is greater than LSD so it is siginficant difference between groups A and B.

Absolute difference between sample mean A and sample mean C is : 8

Since absoulte difference is greater than LSD so it is siginficant difference between groups A and C.

Absolute difference between sample mean B and sample mean C is : 8

Since absoulte difference is greater than LSD so it is siginficant difference between groups B and C.

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The 95% required confidence interval is: