Question

# Develop the analysis of variance computations for the following completely randomized design. At α = 0.05,...

Develop the analysis of variance computations for the following completely randomized design. At α = 0.05, is there a significant difference between the treatment means?

Treatment
A B C
136 108 91
119 115 81
113 125 85
106 105 102
130 108 88
115 109 117
129 96 109
112 115 121
104 99
85 107

xj

120 107 100

sj2

110.29 119.56 186.22

State the null and alternative hypotheses.

H0: At least two of the population means are equal.
Ha: At least two of the population means are different.

H0: μAμBμC

Ha: μA = μB = μC

H0: μA = μB = μC
Ha: μAμBμC

H0: Not all the population means are equal.

Ha: μA = μB = μC

H0: μA = μB = μC
Ha: Not all the population means are equal.

Find the value of the test statistic. (Round your answer to two decimal places.)

test statistic =

p-value =

Do not reject H0. There is sufficient evidence to conclude that the means of the three treatments are not equal.

Do not reject H0. There is not sufficient evidence to conclude that the means of the three treatments are not equal.

Reject H0. There is sufficient evidence to conclude that the means of the three treatments are not equal.

Reject H0. There is not sufficient evidence to conclude that the means of the three treatments are not equal.

H0: μA = μB = μC
Ha: Not all the population means are equal.

___________________________________

 Anova: Single Factor SUMMARY Groups Count Sum Average Variance A 8 960 120 110.2857 B 10 1070 107 119.5556 C 10 1000 100 186.2222 ANOVA Source of Variation SS df MS F P-value Between Groups 1800.714 2 900.3571 6.39 0.0057 Within Groups 3524 25 140.96 Total 5324.714 27

In Excel = DATA> DATA ANALYSIS >ANOVA(SINGLE FACTOR) > SELECT THE DATA >OK

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test statistic = 6.39

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p-value = 0.0057

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Reject H0. There is sufficient evidence to conclude that the means of the three treatments are not equal

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