llowing data are from a completely randomized design. Sample mean Sample variance A 162 Source of Variation 141 Treatments 164 Error 146 Total 147 176 156 Treatment B 141 156 123 142 135 161 143 с 126 122 137 139 Compute the sum of squares between treatments. Sum of Squares 149 Compute the mean square between treatments. 125 Compute the sum of squares due to error. 133 181.2 192.4 108.4 Compute the mean square due to error. (Round your answer to two decimal places.) 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.) Degrees of Freedom Mean Square F p-value

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Q1 Probability and Statistics

 

The following data are from a completely randomized design.
Sample
mean
Sample
variance
A
162
Source
of Variation
141
Treatments
164
Error
146
Total
147
176
156
Treatment
B
141
156
123
142
135
161
143
с
126
122
137
139
(a) Compute the sum of squares between treatments.
Sum
of Squares
149
(b) Compute the mean square between treatments.
125
(c) Compute the sum of squares due to error.
133
181.2 192.4 108.4
(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.)
Degrees
of Freedom
Mean
Square
F
p-value
Transcribed Image Text:The following data are from a completely randomized design. Sample mean Sample variance A 162 Source of Variation 141 Treatments 164 Error 146 Total 147 176 156 Treatment B 141 156 123 142 135 161 143 с 126 122 137 139 (a) Compute the sum of squares between treatments. Sum of Squares 149 (b) Compute the mean square between treatments. 125 (c) Compute the sum of squares due to error. 133 181.2 192.4 108.4 (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.) Degrees of Freedom Mean Square F p-value
(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.
Ho: At least two of the population means are equal.
Ha: At least two of the population means are different.
O Ho: Not all the population means are equal.
Ha: μA = MB = MC
O Ho: MAHB = μC
Ha: MA MB MC
#
© Ho: HA = HB = HC
Ha: Not all the population means are equal.
HO: MAMB #MC
Hai HA = HB = HC
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.
O Reject Ho. There is sufficient evidence to conclude that the means for the three treatments are not equal.
Do not reject Ho. There is not sufficient evidence to conclude that the means for the three treatments are not equal.
o Reject Ho. There is not sufficient evidence to conclude that the means for the three treatments are not equal.
O Do not reject Ho. There is sufficient evidence to conclude that the means for the three treatments are not equal.
Transcribed Image Text:(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. Ho: At least two of the population means are equal. Ha: At least two of the population means are different. O Ho: Not all the population means are equal. Ha: μA = MB = MC O Ho: MAHB = μC Ha: MA MB MC # © Ho: HA = HB = HC Ha: Not all the population means are equal. HO: MAMB #MC Hai HA = HB = HC 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. O Reject Ho. There is sufficient evidence to conclude that the means for the three treatments are not equal. Do not reject Ho. There is not sufficient evidence to conclude that the means for the three treatments are not equal. o Reject Ho. There is not sufficient evidence to conclude that the means for the three treatments are not equal. O Do not reject Ho. There is sufficient evidence to conclude that the means for the three treatments are not equal.
Expert Solution
Step 1: Description of what test use:

By the given data,here data have one categorical independent variable and one quantitative dependent variable. The independent variable have at least three levels (i.e. at least three different groups or categories).

Hence, we use One way ANOVA to solve the problem. 

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