Question

In: Statistics and Probability

The following data are from a completely randomized design.

You may need to use the appropriate technology to answer this question.

The following data are from a completely randomized design.

Treatment
A B C
162 142 126
142 157 123
166 123 138
144 142 140
148 137 150
168 157 127
Sample
mean
155 143 134
Sample
variance
135.6 166.0 108.4

(a)

Compute the sum of squares between treatments.

(b)

Compute the mean square between treatments.

(c)

Compute the sum of squares due to error.

(d)

Compute the mean square due to error. (Round your answer to two decimal places.)

(e)

Set up the ANOVA table for this problem. (Round your values for MSE and F to two decimal places, and your p-value to four decimal places.)

Source
of Variation
Sum
of Squares
Degrees
of Freedom
Mean
Square
F p-value
Treatments
Error
Total

(f)

At the α = 0.05 level of significance, test whether the means for the three treatments are equal.

State the null and alternative hypotheses.

H0: μA = μB = μC
Ha: μAμBμCH0: μAμBμC
Ha: μA = μB = μC    H0: At least two of the population means are equal.
Ha: At least two of the population means are different.H0: μA = μB = μC
Ha: Not all the population means are equal.H0: Not all the population means are equal.
Ha: μA = μB = μC

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

Find the p-value. (Round your answer to four decimal places.)

p-value =

State your conclusion.

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

Solutions

Expert Solution

The statistical software output for this problem is :

(a)

The sum of squares between treatments = 1232

(b)

The mean square between treatments = 666

(c)

The sum of squares due to error = 2050

(d)

The mean square due to error = 136.67

(e)

ANOVA table

Source DF SS MS F-Stat P-value
Columns 2 1332 666 4.87 0.0234
Error 15 2050 136.67
Total 17 3382

Ha: μA = μB = μC    H0: At least two of the population means are equal.


Test statistics = 4.87

P-value = 0.0234


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


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