In: Math
Consider the accompanying data on plant growth after the application of different types of growth hormone. 1: 13 16 7 13 2: 20 12 19 16 3: 19 16 20 16 4: 7 11 18 9 5: 6 10 15 9 (a) Perform an F test at level α = 0.05. State the appropriate hypotheses. H0: μ1 ≠ μ2 ≠ μ3 ≠ μ4 ≠ μ5 Ha: all five μi's are equal H0: μ1 ≠ μ2 ≠ μ3 ≠ μ4 ≠ μ5 Ha: at least two μi's are equal H0: μ1 = μ2 = μ3 = μ4 = μ5 Ha: all five μi's are unequal H0: μ1 = μ2 = μ3 = μ4 = μ5 Ha: at least two μi's are unequal Calculate the test statistic. (Round your answer to two decimal places.) f = What can be said about the P-value for the test? P-value > 0.100 0.050 < P-value < 0.100 0.010 < P-value < 0.050 0.001 < P-value < 0.010 P-value < 0.001 State the conclusion in the problem context. Fail to reject H0. There appears to be a difference in the average growth of at least two groups. Reject H0. There appears to be a difference in the average growth of at least two groups. Reject H0. There does not appear to be a difference in the average growth. Fail to reject H0. There does not appear to be a difference in the average growth. (b) What happens when Tukey's procedure is applied? (Round your answer to two decimal places.) w = Which means differ significantly from one another? (Select all that apply.) x1. and x2. x1. and x3. x1. and x4. x1. and x5. x2. and x3. x2. and x4. x2. and x5. x3. and x4. x3. and x5. x4. and x5. There are no significant differences. Are Tukey's method and the F test in agreement? Yes No
Answer:
Consider the accompanying data on plant growth after the application of different types of growth hormone.
Type1 13 16 7 13
Type2 20 12 19 16
Type3 19 16 20 16
Type4 7 11 18 9
Type5 6 10 15 9
(a) Perform an F test at level α = 0.05.
State the appropriate hypotheses.
H0: μ1 = μ2 = μ3 = μ4 = μ5 Ha: at least two μi's are unequal
Calculate the test statistic. (Round your answer to two decimal places.)
f = 3.47
What can be said about the P-value for the test?
0.050 < P-value < 0.100
State the conclusion in the problem context.
Reject H0. There appears to be a difference in the average growth of at least two groups.
(b) What happens when Tukey's procedure is applied? (Round your answer to two decimal places.) w = Which means differ significantly from one another? (Select all that apply.)
There are no significant differences.
Are Tukey's method and the F test in agreement? No
Excel Addon Megastat used.
One factor ANOVA |
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Mean |
n |
Std. Dev |
||||
12.3 |
4 |
3.77 |
Type1 |
|||
16.8 |
4 |
3.59 |
Type2 |
|||
17.8 |
4 |
2.06 |
Type3 |
|||
11.3 |
4 |
4.79 |
Type4 |
|||
10.0 |
4 |
3.74 |
Type5 |
|||
13.6 |
20 |
4.56 |
Total |
|||
ANOVA table |
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Source |
SS |
df |
MS |
F |
p-value |
|
Treatment |
189.80 |
4 |
47.450 |
3.47 |
.0338 |
|
Error |
205.00 |
15 |
13.667 |
|||
Total |
394.80 |
19 |
||||
Post hoc analysis |
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p-values for pairwise t-tests |
||||||
Type5 |
Type4 |
Type1 |
Type2 |
Type3 |
||
10.0 |
11.3 |
12.3 |
16.8 |
17.8 |
||
Type5 |
10.0 |
|||||
Type4 |
11.3 |
.6394 |
||||
Type1 |
12.3 |
.4029 |
.7074 |
|||
Type2 |
16.8 |
.0208 |
.0527 |
.1057 |
||
Type3 |
17.8 |
.0096 |
.0252 |
.0527 |
.7074 |
|
Tukey simultaneous comparison t-values (d.f. = 15) |
||||||
Type5 |
Type4 |
Type1 |
Type2 |
Type3 |
||
10.0 |
11.3 |
12.3 |
16.8 |
17.8 |
||
Type5 |
10.0 |
|||||
Type4 |
11.3 |
0.48 |
||||
Type1 |
12.3 |
0.86 |
0.38 |
|||
Type2 |
16.8 |
2.58 |
2.10 |
1.72 |
||
Type3 |
17.8 |
2.96 |
2.49 |
2.10 |
0.38 |
|
critical values for experimentwise error rate: |
||||||
0.05 |
3.09 |
|||||
0.01 |
3.93 |