This document provides information on conducting a one-way analysis of variance (ANOVA) using SPSS. It uses an example where a farmer tests the effect of different fertilizers (biological, chemical, none) on the weight of parsley plants. The summary is: The document walks through running a one-way ANOVA in SPSS to analyze the weights of parsley plants that received different fertilizers. The ANOVA results show that fertilizer significantly affects weight. A post hoc test finds a significant difference between plants that received chemical fertilizer versus no fertilizer. The document also briefly describes two-way ANOVAs for analyzing the effects of two independent variables.

1. Data Analysis with SPSS
Analysis of variance
(ANOVA)
Surjeet Singh Dhaka
PhD Scholar
IABM, Bikaner
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2. Relationship Amongst Tests
One Independent One or More
Metric Dependent Variable
t Test
Binary
Variable
One-Way Analysis
of Variance
One Factor
N-Way Analysis
of Variance
More than
One Factor
Analysis of
Variance
Categorical:
Factorial (Nonmetric)
Analysis of
Covariance
Categorical
and Interval
Regression
Interval
Independent Variables
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3. Example
• A farmer wants to know whether the weight of parsley
plants is influenced by using a fertilizer. He selects 90
plants and randomly divides them into three groups of
30 plants each. He applies a biological fertilizer to the
first group, a chemical fertilizer to the second group and
no fertilizer at all to the third group. After a month he
weighs all plants, resulting in parsley.sav. Can we conclude
from these data that fertilizer affects weight?
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4. Hypothesis
• H0: μ1 = μ2 = μ3
(Other words the mean weights of the three groups of
plants are equal)
• Ha: at least one μi is different
(the mean weights of the three groups of plants are not
equal)
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5. Running SPSS One-Way ANOVA
• Dependent variable: weight in grams of parsley plants
(metric)
• Independent variable: Fertilizer used for parsley plants
(categorical or nonmetric)
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7. 2
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8. 3
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9. 5
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Press “Post Hoc”
In this example, I have
chosen “Scheffe”.
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10. we'll first inspect the Descriptive table.
“N” in the first column refers to the number of cases used for calculating
the descriptive statistics. These numbers being equal to our sample sizes
tells us that there are no missing values on the dependent variable.
The mean weights are the core of our output. After all, our main research
question is whether these differ for different fertilizers. On average,
parsley plants weigh some 51 grams if no fertilizer was used. Biological
fertilizer results in an average weight of some 54 grams whereas
chemical fertilizer does best with a mean weight of 57 grams.
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11. Now, we'll focus on the ANOVA table.
• The degrees of freedom (df) and F statistic are used for reporting our
results correctly.
• The p value (denoted by “Sig.”) is .028. This means that if the
population mean weights are exactly equal, we only have a 2.8%
chance of finding the differences that we observe in our sample.
• The null hypothesis is usually rejected if p < .05 so we conclude that
the mean weights of the three groups of plants are not equal.
• The weights of parsley plants are affected by the fertilizer if anythat's
used.
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We do not know which group means are different, post hoc test will indicate this
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12. Post Hoc Tests
Scheffe Multiple Comparisons test shows there is significant difference
between a pair of means: “NONE” and “CHEMICAL”, p = 0.028 (≤0.05)It
means that the differences in the weight of plants was due to chemical
fertiliser and none application . Meaning that, effect of biological fertiliser
on the weight of the plants could be attributed to chance. Whiles the
effect of chemical and none is not attributed to chance.
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13. Why Post Hoc test
Post hoc tests in the Analysis of Variance (ANOVA). Post
hoc tests are designed for situations in which the
researcher has already obtained a significant F-test with a
factor that consists of three or more means and additional
exploration of the differences among means is needed to
provide specific information on which means are
significantly different from each other.
posthoc.pdf
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14. Reporting a One-Way ANOVA
• Step 1H0: μ1 = μ2 = μ3
(Other words the mean weights of the three groups of plants are equal)
• Ha: at least one μi is different
Step 2: Significance Level
α = 0.05
Step 3: Rejection Region
Reject the null hypothesis if p-value ≤ 0.05.
Step 4: Construct the ANOVA Table (re-formatted from original SPSS
output)
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15. Reporting a One-Way ANOVA
Step 4: Construct the ANOVA Table (SPSS output)
From the output, F = 3.743 with 3 and 87 degrees of freedom. p-value
= Sig. ≈ 0.028
Step 5: Conclusion
Since p-value ≈ 0.0000 ≤ 0.05 = α, we shall reject the null hypothesis.
Step 6: State conclusion in words
The null hypothesis is usually rejected if p < .05 so we conclude that
the mean weights of the three groups of plants are not equal.
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16. Reporting a One-Way ANOVA
We do not know which group means are different,
post hoc test will indicate this
Scheffe Multiple Comparisons test shows there is
significant difference between a pair of means:
“NONE” and “CHEMICAL”, p = 0.028 (≤0.05)
All others None*Biological, Biological*Chemical,
pairs are not significant different .
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Factorial (Two or more way) ANOVA
One dependent variable- interval or ratio with a normal distribution
Two independent variables- nominal (define groups), and
independent of each other
Three hypothesis tests:
Test effect of each independent variable controlling for the effects of
the other independent variable
One: H0: Treatment type has no impact on Outcome
Two: H0: Age Group has no impact on Outcome
Three: Test interaction effect for combinations of categories
H0: Treatment and Age Group interact in affecting Outcome
Two-Way ANOVA

18. Structure of the Two-Factor Analysis of Variance

19. Two-factors considered at one time
• Factor 1 (independent variable, e.g. type of crop variety)
• Always nominal or ordinal (it defines distinct groups)
• Factor 2 (independent variable, e.g., fertilizer)
• Always nominal or ordinal (it defines distinct groups)
• Outcome (dependent variable, e.g. yield)
• Always interval or ratio
• Mean Outcomes of the
groups defined by Factor 1
and Factor 2 are being
compared.

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Factorial ANOVA asks three questions
and tests three null hypotheses
• First: Does type of crop variety have any impact on the yield?
Null: The groups defined by type of crop variety will have the
same Mean yield.
• Second: Does fertilizer has any impact on the yield?
Null: The groups defined by fertilizer will have the same Mean
yield.
• Third: Do type of crop variety and fertilizer interact in
influencing yield?
Null: No combination of type of crop variety and fertilizer
produces unusually high or unusually low mean yield.

21. Thank You
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