Question

Given the following sample information, test the hypothesis that the treatment means are equal at the 0.10 significance level:

Treatment 1	Treatment 2	Treatment 3
3	9	6
2	6	3
5	5	5
1	6	5
3	8	5
1	5	4
	4	1
	7	5
	6
	4

a. State the null hypothesis and the alternative hypothesis.

H₀ : μ_{1 click to select (=,>,<)} μ₂click to select (=, <, >) μ₃ *please select correct multiple choice from brackets)

a. State the null hypothesis and the alternative hypothesis.

H₀ : μ₁

μ₂  

μ₃

H₁ : Treatment means

all the same.

b. What is the decision rule? (Round the final answer to 2 decimal places.)

Reject H₀ if F >                .

c. Compute SST, SSE, and SS total. (Round the final answers to 2 decimal places.)

SST =               

SSE =               

SS total =               

d. Complete the ANOVA table. (Round the SS, MS, and F values to 2 decimal places.)

	Source	SS	DF	MS	F	x
	Factor					x
	Error				x	x
	Total			x	x	x
* fill in all squares without x*

e. State your decision regarding the null hypothesis.

Decision: click to select (Do not reject, Reject) H_{0. *select correct multiple choice*}

f.Find the 95% confidence interval for the difference between treatment 2 and 3. (Round the final answers to 2 decimal places.)

95% confidence interval is: ____ ± _____ (fill in ___)   

We can conclude that the treatments 2 and 3 are click to select (different, the same)

Answer 1

(a)

H0: Treatments means all the same

H1: Treatments means all not the same. At least one mean is different from other two means.

(b)

=0.10

Degrees of Freedom for numerator = 3 - 1 = 2

Degrees of Freedom for denominator = 24 - 3 = 21

From Table, critical value of F = 2.575

Decision Rule:
Reject H0 if F > 2.575

(c)

Fromthe given data, the following Table is calculated:

	Treatment 1	Treatment 2	Treatment 3
n	6	10	8
Sum	15	60	34
Mean	2.5	6.0	4.25
	49	384	162
Std. Dev.	1.517	1.633	1.581
SS	11.5	24.0	17.5

(i)

(ii)

(iii)

(d)

The ANOVA Table is completed as follows:

Source	SS	DF	MS	F
Factor	46.958	2	46.958/2=23.479	23.479/2.528=9.303
Error	53.0	21	53.0/21=2.528
Total	99.958	23

(e)

Since calculated value of F = 9.303 is greater than critical value of F = 2.575:

Reject H0

(f)

For Treatments 2 & 3:

(i)

n1 = 10

1 = 6.0

s1 = 1.6330

n2 = 8

2 = 4.25

s2 = 1.5811

= 0.05

Pooled Standard Deviaion is given by:

df = 10+8 - 2 = 16

=0.05

From Table, critical values of t = 2.12

Confidence Interval:

So,

Answer is:

(0.131. 3.369)

Since all values of the confidence interval are positive:

we can conclude that treatments 2 and 3 are different.