Data Analysis with SPSS

Analysis of variance

(ANOVA)

Surjeet Singh Dhaka

PhD Scholar

IABM, Bikaner

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Relationship Amongst Tests

One Independent One or More

Metric Dependent Variable

t Test

Binary

Variable

One-Way Analysisof Variance

One Factor

N-Way Analysisof Variance

More thanOne Factor

Analysis ofVariance

Categorical:Factorial (Nonmetric)

Analysis ofCovariance

Categoricaland Interval

Regression

Interval

Independent Variables

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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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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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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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2

7

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4

8

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Press “Post Hoc”

In this example, I have

chosen “Scheffe”.

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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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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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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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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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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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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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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

Structure of the Two-Factor Analysis of Variance

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.

Thank You

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