1. Analysis of variance (ANOVA) is a statistical technique used to test whether the means of three or more groups are equal. It analyzes the variations between and within groups. 2. ANOVA requires assumptions of normality, equal variances, independence, and random sampling. It uses sum of squares, mean squares and the F-test statistic to determine if group means are significantly different. 3. If the p-value is less than the significance level (often 0.05), the null hypothesis of equal group means is rejected, indicating at least one group mean is significantly different from the others.

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Statistics Practice (Sample) Test
Chapter 12: Contingency Tables, Analysis of Variance
Instructions: Read this Mini Lecture or your text, or study the tutorials online
thoroughly to be able to handle this Sample Test.
Mini Lecture:
Chapter 12: Analysis of Variance (ANOVA)
12-2: One-Way ANOVA
❖ Analysis of variance (ANOVA) is a method for testing the hypothesis that three
or more-population means are equal.
❖ For example:
H0: µ1 = µ2 = µ3 = . . . µk
H1: At least one mean is different
ANOVA methods require the F-distribution:
1. The F-distribution is not symmetric; it is skewed to the right.
2. The values of F can be 0 or positive, they cannot be negative.
3. There is a different F-distribution for each pair of degrees of freedom for the
numerator and denominator. Critical values of F are given in Table A-5
An Approach to Understanding ANOVA
1. Understand that a small P-value (such as 0.05 or less) leads to the rejection of the null
hypothesis of equal means. With a large P-value (such as greater than 0.05), fail to reject
the null hypothesis of equal means.
2. Develop an understanding of the underlying rationale by studying the example in this
section.
3. Become acquainted with the nature of the SS (sum of squares) and MS (mean square)
values and their role in determining the F test statistic, but use statistical software
packages or a calculator for finding those values.
Definition: Treatment (or factor)

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A treatment (or factor) is a property or characteristic that allows us to distinguish the
different populations from another. Use Technology for ANOVA calculations if possible
Assumptions
1. The populations have approximately normal distributions.
2. The populations have the same variance 2 (or standard deviation  ).
3. The samples are simple random samples.
4. The samples are independent of each other.
5. The different samples are from populations that are categorized in only one way.
Procedure for testing:
H0: µ1 = µ2 = µ3 = . . .
1. Use Technology to obtain results.
2. Identify the P-value from the display.
3. Form a conclusion based on these criteria:
❖ If P-value  , reject the null hypothesis of equal means.
❖ If P-value >  , fail to reject the null hypothesis of equal means.
Example:

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Statistics, Sample Test
Chapters 12: Analysis of Variance
Name: ______________________________ Student ID: _______________________
Instructions: Print out the following pages that contain the Sample Test, solve all
problems, show your work completely.
1. (One-Way ANOVA) Given the readability scores summarized in the following table
and a significance level of  = 0.05, Use Technology to test the claim that the three
samples come from populations with means that are not all the same.
Clancy Rowling Tolstoy
N 12 12 12
x 70.73 80.75 66.15
S 11.33 4.68 7.86
2. (One-Way ANOVA) Polar Tree Weights: Weights (kg) of poplar trees were
obtained from trees planted in a Rich and Moist region. The trees were given
different treatments identified in the accompanying table. Use a 0.05 significance
level to test the claim that the four treatment categories yield poplar trees with the
same mean weight.
No Treatment Fertilizer Irrigation Fertilizer & Irrigation
1.21 0.94 0.07 0.85
0.57 0.87 0.66 1.78
0.56 0.46 0.10 1.47
0.13 0.58 0.82 2.25
1.30 1.03 0.94 1.64
3. (One-Way ANOVA) Polar Tree Weights: Weights (kg) of poplar trees were
obtained from trees planted in a Sandy and Dry region. The trees were given
different treatments identified in the accompanying table. Use a 0.05 significance
level to test the claim that the four treatment categories yield poplar trees with the
same mean weight.

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No Treatment Fertilizer Irrigation Fertilizer & Irrigation
0.24 0.92 0.96 1.07
1.69 0.07 1.43 1.63
1.23 0.56 1.26 1.39
0.99 1.74 1.57 0.49
1.80 1.13 0.75 0.95