In: Statistics and Probability
In the production process of growing Azaleas, the plants are transplanted from a small pot to a larger pot. Lee’s green thumbs have realized that the newly transplanted azaleas seem to have better chance of growing to be large and healthy if they are transplanted immediately after the plant has had a major growth spurt. Therefore, they started looking at growth rates of the plants. They studied the growth of 10 plants for five periods.
Can you determine if one period is a better for transplanting than another period? Test at a 0.05 level of significance. Conduct a multiple comparisons test using Turkey’s method. Which week has the most growth? Which week has the least growth? Produce a main effects plot and an interval plot.
The data is as follow:-
Plant | Period-1 | Period-2 | Period-3 | Period-4 | Period-5 |
1 | 1.58 | 1.09 | 1.00 | 2.22 | 1.20 |
2 | 1.62 | 1.03 | 1.00 | 2.40 | 1.40 |
3 | 2.04 | 1.00 | 1.07 | 2.47 | 1.34 |
4 | 1.28 | 1.23 | 1.18 | 1.85 | 1.00 |
5 | 1.43 | 1.13 | 1.34 | 2.50 | 1.20 |
6 | 1.93 | 1.15 | 1.25 | 1.20 | 1.10 |
7 | 2.20 | 1.03 | 1.26 | 1.33 | 1.10 |
8 | 1.96 | 1.09 | 1.03 | 2.40 | 1.06 |
9 | 2.23 | 1.00 | 1.12 | 2.23 | 1.17 |
10 | 1.54 | 1.00 | 1.26 | 2.57 | 1.25 |
(Example interpretation of the data: the 1.58 would be interpreted as: “at the end of the first period, the plant is 58% larger than at the beginning of that same period”.)
One-way ANOVA: Period-1, Period-2, Period-3, Period-4, Period-5
Method
Null hypothesis
All means are equal
Alternative hypothesis At least one mean is different
Significance level α = 0.05
Equal variances were assumed for the analysis.
Factor Information
Factor Levels Values
Factor 5 Period-1, Period-2,
Period-3, Period-4, Period-5
Analysis of Variance
Source DF Adj SS Adj MS F-Value P-Value
Factor 4 8.557 2.13918
27.35 0.000
Error 45 3.520 0.07822
Total 49 12.077
Model Summary
S R-sq
R-sq(adj) R-sq(pred)
0.279682 70.85%
68.26% 64.02%
Means
Factor N
Mean StDev 95%
CI
Period-1 10 1.781 0.333 ( 1.603,
1.959)
Period-2 10 1.0750 0.0775 (0.8969, 1.2531)
Period-3 10 1.1510 0.1238 (0.9729, 1.3291)
Period-4 10 2.117 0.493 ( 1.939,
2.295)
Period-5 10 1.1820 0.1244 (1.0039, 1.3601)
Pooled StDev = 0.279682
p-value = 0.000 < 0.05
hence there is significant difference between 5 period
Period 1 is better than other period