11.4 One-Way ANOVA Examples (In Progress)

Example 1 -Compressive Strength of Concrete

Type your examples here.

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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.

 

Table 1. Tea leaf pluckings of 16 tea bushes who received one of the four treatments.
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 k =  4 treatment levels. Each sample contains n = 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.

H_o: \mu_1 = \mu_2 = \mu_3 = \mu_4

H_a: 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.

License

Icon for the Creative Commons Attribution 4.0 International License

Introduction to Statistics for Engineers Copyright © by Vikki Maurer & Jeff Crabill & Linn-Benton Community College is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

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