In: Statistics and Probability
The following data are given for a two-factor ANOVA with two treatments and three blocks.
Treatment | ||
Block | 1 | 2 |
A | 43 | 35 |
B | 31 | 20 |
C | 40 | 36 |
Using the 0.05 significance level conduct a test of hypothesis
to determine whether the block or the treatment means
differ.
State the null and alternate hypotheses for treatments.
State the decision rule for treatments. (Round your answer to 1 decimal place.)
State the null and alternate hypotheses for blocks. (Round your answer to 1 decimal place.)
Also, state the decision rule for blocks.
d & e. Compute SST, SSB, SS total, and SSE and complete an ANOVA table. (Round your SS, MS values to 3 decimal places and F value to 2 decimal places.)
Give your decision regarding the two sets of hypotheses.
a) H0t: There is no significance difference among the treatments
H1t: Not all treatment means are equal
Let the los be alpha = 0.05
b)
H0b: There is no significance difference among the blocks
H1b: Not all block means are equal
Let the los be alpha = 0.05
c) From the given data
Treatment | ||||
Block | 1 | 2 | Ti | Ti^2/2 |
A | 43 | 35 | 78 | 3042.000 |
B | 31 | 20 | 51 | 1300.500 |
C | 40 | 36 | 76 | 2888.000 |
Tj | 114 | 91 | 205 | 7230.500 |
Tj^2/3 | 4332.000 | 2760.333 | 7092.333 |
ANOVA Table:
From the ANOVA table, F-Ratio of Rows = 18.35, F criitcal value value = 19
since F-Ratio < F critical value so we accept H0b,
Thus we conclude that there is no significance difference among the blocks
From the ANOVA table, F-Ratio of columns = 14.30, F criitcal value value = 118.513
since F-Ratio < F critical value so we accept H0t,
Thus we conclude that there is no significance difference among the treatments