11.4 One-Way ANOVA Examples (In Progress)
Example 1 -Compressive Strength of Concrete
Type your examples here.
- First
- Second
Example 2 – Tea Leaf Pluckings
The following data represent quantities of tea leaf pluckings (tender shoots from tea plants) from sixteen different plots of tea bushes intended for experimental use in Ceylon, a type of tea from Sri Lanka. The tea bushes are randomly divided into four different treatment groups. Each treatment group contains four tea bushes. The number of tea leaf pluckings for each tea bush is given below in Table 1.
Treatment 1 Pluckings |
Treatment 2 Pluckings |
Treatment 3 Pluckings |
Treatment 4 Pluckings |
88 | 102 | 91 | 88 |
94 | 110 | 109 | 118 |
109 | 105 | 115 | 94 |
88 | 102 | 91 | 96 |
Experimenters wish to determine if the mean number of pluckings differs among the four treatments. Test this question at the 5% significance level. Assume that these samples are drawn from normal populations with equal variances.
Solution:
There are = 4 treatment levels. Each sample contains
= 4 bushes. The alternate hypothesis is that there is a difference in the means among the four treatments, so the null hypothesis is that the mean pluckings are the same.
There is a difference in the means.
Statistic | Treatment 1 | Treatment 2 | Treatment 3 | Treatment 4 |
Sample Mean (x) | 94.75 | 104.75 | 101.50 | 99 |
(6-9) Answer shown in table below.
Statistic | Treatment 1 | Treatment 2 | Treatment 3 | Treatment 4 |
Sample Standard Deviation (s) | 9.912 | 3.775 | 12.369 | 13.115 |
(10-13) Answer shown in table below.
Statistic | Treatment 1 | Treatment 2 | Treatment 3 | Treatment 4 |
Sample Variance (s2) | 98.25 | 14.25 | 153 | 172 |
(14) sx4.218
(15) Variance between the samples = nsx2= (4)(4.218)271.17
(16) Variance within the samples = sp2=98.25+14.25+153+1724=109.38
(17) F=n sx2sp2=71.17109.380.65
(18) df1 = k − 1 = 4 − 1 = 3. df2 = k(n − 1) = 4(4 − 1) = 12
(19) P-value = 0.598
(20) Because the P-value is not less than 0.05, the level of significance, we fail to reject the null hypothesis and do not support the alternative hypothesis.
(21) The data do not support the claim that at least one of treatments has a mean number of pluckings that differs from the other treatments’ means.
Example 3 – Potato Yields
In the book, Statistical Methods for Research Workers, statistics pioneer Ronald Fisher provides the following example. In an experiment on improving potato yields, 5 varieties of potato plants were used. For each variety, 9 plants were randomly selected. Over time, the plants yielded potatoes, and the total yield in pounds was recorded for each plant. The yield for each plant is given below in Table 2.
Table 2. Potato yield of 45 plants who received one of the five treatments.
Ajax | Arran Comrade | Great
Scot |
Iron
Duke |
Kerr’s
Pink |
3.20 | 2.25 | 3.38 | 3.43 | 3.04 |
4.00 | 2.56 | 3.07 | 3.00 | 3.57 |
3.86 | 2.58 | 3.89 | 3.96 | 3.82 |
2.55 | 1.96 | 2.79 | 3.33 | 2.96 |
3.04 | 2.15 | 3.54 | 3.08 | 3.18 |
4.13 | 2.10 | 4.14 | 3.32 | 4.32 |
2.82 | 2.42 | 3.07 | 3.50 | 2.00 |
1.15 | 2.17 | 3.15 | 2.32 | 3.00 |
4.71 | 2.17 | 3.50 | 3.29 | 3.88 |
Test the claim at 1% significance that the mean yield is the same for each variety of plant. Assume these samples are drawn from normal populations with equal variances.
(22) H0:1=2=3=4=5=6=7=8. Ha: At least one mean is different.
(23-37) The statistics for each sample are below:
Variety | x | s | s2 | n |
Ajax | 3.27 | 1.059 | 1.12 | 9 |
Arran Comrade | 2.26 | 0.213 | 0.05 | 9 |
Great Scot | 3.39 | 0.429 | 0.18 | 9 |
Iron Duke | 3.25 | 0.443 | 0.20 | 9 |
Kerr’s Pink | 3.31 | 0.680 | 0.46 | 9 |
(38) The variance between the samples is: nsx2=(9)(0.47)21.988
(39) The variance within the samples is:sp2=1.1209 + 0.0452 + 0.1837 + 0.1960 + 0.462150.402
(40) The test statistic is: F=nsx2sp2=1.9880.4024.95
(41) df1 = k − 1 = 5 − 1 = 4. df2 = k(n − 1) = 5(9 − 1) = 40
(42) P-value = 0.002. The P-value is the area to the right of F = 4.95 under the F-distribution with 4 degrees of freedom in the numerator (D1 = 4) and 40 degrees of freedom in the denominator (D2 = 40).
(43) Because the P-value is less than 0.01, the level of significance, we reject the null hypothesis and support the alternative hypothesis.
(44) The data support the claim that the mean yield of at least one variety is different from the mean yield of the other varieties. We support the alternative hypothesis, so we conclude that at least one of the population means is different from the others.
(45) Arran Comrade. This variety’s sample mean is significantly smaller than the sample means of the other varieties.
Sources
Examples 1 and 2 taken from Statway College Module 5 by Carnegie Math Pathways is licensed under CC BY NC 4.0.
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