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Data analysis with spss anova

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Surjeet Randhir Singh Dhaka
Surjeet Randhir Singh Dhaka

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.

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Data Analysis with SPSS Analysis of variance (ANOVA) Surjeet Singh Dhaka PhD Scholar IABM, Bikaner 1
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 2
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? 3
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) 4
Running SPSS One-Way ANOVA • Dependent variable: weight in grams of parsley plants (metric) • Independent variable: Fertilizer used for parsley plants (categorical or nonmetric) 5
1 6
2 7
3 4 8
5 6 7 8 9 10 11 12 13 Press “Post Hoc” In this example, I have chosen “Scheffe”. 9
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. 14 10
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. 15 16 17 We do not know which group means are different, post hoc test will indicate this 11
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. 18 12
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 13
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) 14
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. 15
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 . 16
17 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.
20 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 21

More Related Content

Data analysis with spss anova

  • 1. Data Analysis with SPSS Analysis of variance (ANOVA) Surjeet Singh Dhaka PhD Scholar IABM, Bikaner 1
  • 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 2
  • 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? 3
  • 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) 4
  • 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) 5
  • 6. 1 6
  • 7. 2 7
  • 9. 5 6 7 8 9 10 11 12 13 Press “Post Hoc” In this example, I have chosen “Scheffe”. 9
  • 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. 14 10
  • 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. 15 16 17 We do not know which group means are different, post hoc test will indicate this 11
  • 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. 18 12
  • 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 13
  • 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) 14
  • 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. 15
  • 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 . 16
  • 17. 17 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.
  • 20. 20 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.