Homework Answers
applying 2 way ANOVA on above data:
| Source of Variation | SS | df | MS | F |
| treatments | 25.44 | 2 | 12.72 | 7.11 |
| blocks | 37.11 | 5 | 7.42 | |
| Error | 17.89 | 10 | 1.79 | |
| Total | 80.44 | 17 |
test statistic =7.11
critical value =4.10
p value =0.0120
reject the null hypothesis: Yes
Conclusion: conclude that there is a signifcant difference between treatment means
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#16
The test statistic is
a.
6.00.
b.
2.25.
c.
3.00.
d.
2.67.
#17
The mean square due to error (MSE) is
a.
60.
b.
18.
c.
15.
d.
20.
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In a completely randomized design, seven experimental units were used for each of the five levels...
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In a completely randomized design, seven experimental units were used for each of the five levels of the factor. Complete the following ANOVA table (to 2 decimals, if necessary). Round p-value to four decimal places. If your answer is zero enter "0". F p-value Source of Variation Sum of Squares Degrees of Freedom Mean Square Treatments 300 Error Total 460 a. What hypotheses are implied in this problem? Ho: Select H. Select b. At the - .05 level of significance,...
#16
The test statistic is
a.
6.00.
b.
2.25.
c.
3.00.
d.
2.67.
#17
The mean square due to error (MSE) is
a.
60.
b.
18.
c.
15.
d.
20.
Part of an ANOVA table is shown below. Sum of Degrees Squares Freedom Mean of Source of Variation Square 180 Between Treatments Within Treatments (Error) TOTAL 480 18
For either independent-measures or repeated-measures designs comparing two treatments, the mean difference can be evaluated with either at test or an ANOVA. The two tests are related by the equation F=12. The following data are from a repeated-measures study: Person Difference Scores 3 I 4 2 3 7 M = 4.00 T = 16 SS = 14 Treatment II 7 11 6 10 M 8.50 T-34 SS = 17 3 3 Mo 4.50 SS = 27.00 Use a repeated-measures t...
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