Psych 248: Worksheet: Introduction to One-way ANOVA Analysis

This document presents a complete solution to a Psych 248 lab worksheet focusing on the introduction to one-way ANOVA. The assignment explores the application of ANOVA in analyzing data from two different datasets. The first dataset examines the relationship between attachment style (anxious, avoidant, and secure) and sleep disturbance, where students are guided through hypothesis formulation, variable identification, ANOVA implementation using SPSS, and interpretation of results, including post-hoc tests. The second dataset focuses on the effect of different therapies and medications on weight change in participants with anorexia nervosa. The solution provides a step-by-step analysis, including checking assumptions, running ANOVA tests, and interpreting F-values and significance, ultimately determining the effectiveness of different interventions. The document also includes the null and alternative hypotheses, independent and dependent variables, and conclusions based on the statistical analysis. The document also includes a bibliography of resources used.

Psych 248 - Lab 10 Worksheet: Introduction to One-way ANOVA
One-Way ANOVA is most often used to determine whether there are any statistical differences between the
means of three or more independent groups. [NOTE: Although ANOVA can be used with two groups, the
Independent Samples t-test is more often used in that situation.] It is an omnibus test statistic and cannot tell
you which specific groups were significantly different from each other; it only tells you that at least two groups
(out of the three or more) were different. To determine which of the independent means differ from each other,
follow-up tests (post hoc) must be conducted.
Data Set 1: Attachment Style and Sleep Disturbance. An attachment researcher conducted a study to
examine the effects of anxious, avoidant, and secure attachment styles on the physiology of sleep. The
investigator hypothesized that children with anxious (and perhaps avoidant) attachment styles experience more
sleep disturbances than children with secure attachment styles because they feel responsible for monitoring the
external environment. The sleep patterns of 30 children were monitored, 10 in each style group. They measured
the percentage of time that each child spent in deep (delta) sleep. It was hypothesized that children who are
insecurely attached to their primary caregivers will spend a lower percentage of time in deep (delta) sleep as
compared to their secure counterparts.
SPSS data file Attach.sav ( ) Double Click Big E here. Look at the data file—both in data view and
variable view windows.
State the Null Hypothesis (H0) and the Alternative Hypothesis (Ha) for this study.
1. H0: There is no significant difference in percentage of time in deep sleep among children with different
attachment style.
2. Ha: Children having unsecured attachment to their primary caregivers spend lower percentage of time in
deep sleep than their secure counterparts.
3. What is the (Factor) independent variable? Attachment Style
4. What is the dependent variable? Percentage of time in deep (delta) sleep for each child
Using ANOVA, Compare the Means of the three Independent Groups
 Go to Analyze  Compare Means  One-Way ANOVA. Select the Dependent variable. Select the
Factor.
 Click on Post Hoc… Under ‘Equal Variances Assumed’ choose Tukey and Scheffe. Continue. Click on Options… Choose ‘Descriptive, Homogeneity of variance test, Means plot. Continue.
5. Did the homogeneity of variance test indicate that these groups differ significantly in variance (Can you
reject the null hypothesis of variance equality)? No, the groups do not differ significantly in variance.
6. What is the F-value? 41.425 Is it significant? ____Yes______ Did you reject H0?______No_______
7. Looking at the Post-hoc tests, which groups differ significantly from each other? The pair of attachment
groups (secure, anxious) and (secure, avoidant) differ significantly.
Data Set II, Part 1: This study is simulated (made up) data. For Part 1, Pretend this was a study that only
looked at 3 different types of therapy and the effect on weight change for participants with anorexia nervosa.
[So, use only the Therapy_Type and WeightChange variables!] We want to know if the three therapy type
groups differed significantly in terms of weight change.

SPSS data file called Anorexia2.sav ( ) Double Click the Big E here. Look at the data file—both
in data view and variable view windows.
State the Null Hypothesis (H0) and the Alternative Hypothesis (Ha) for this study.
8. H0: There is no significant change in weight due to different therapy.
9. Ha: The three therapy type groups differ significantly in terms of weight change.
10. What is the (Factor) independent variable? _Therapy Type__
11. What is the dependent variable?__Weight Change_____
Check the Assumption for ANOVA: Independence of Observations
12. Looking at the Data View, are there different participants in each therapy type group, with no participant
being in more than one group? ________Yes, each therapy group has different participants._____
Check the Assumption for ANOVA: Normal distributions (Shapiro-Wilk test)
 Go to Analyze  Descriptives  Explore. Select WeightChange as the Dependent variable. Select
Therapy_Type as the Factor.
 Click on Plots… Under ‘Boxplots’ choose None. Choose Histogram. Choose Normality plots with tests
Continue. Ok.
13. Based on the Sig (p-value) for the Shapiro-Wilk test for each therapy group, is it possible to reject the null
(the distribution is equal to normal) for any of the groups? ______No, the null hypothesis assuming
normality of the groups cannot be rejected._________
Check the Assumption for ANOVA: Homogeneity of variance (Levene’s test)
 Go to Analyze  Compare Means  One-Way ANOVA. Select WeightChange as the Dependent
variable. Select Therapy_Type as the Factor. Click on Options… Choose ‘Descriptive, Homogeneity of variance test. Continue.
14. Based on the Sig (p-value) for the Levene’s Statistic “Based on Mean,” is it possible to reject the null
(the group variances are equal to each other)? _No, the p-value shows that the group variances are equal
to each other._____
Using ANOVA, Compare the Means of the three Independent Groups
15. What is the F-value? ___0.617_________ Is it significant? _No____________ Should you reject H0?
______No____
16. Should Post hoc tests be run based on this F statistic? Why or why not? Since the null hypothesis is
accepted, it indicates that there is no difference in weight change due to different therapies. Hence Post
Hoc test is not required.
OPTIONAL: Data Set II, Part 2: For Part 2, Pretend this was a study that only looked at 3 different
medication conditions (Placebo, SSRI, Zinc) and the effect on weight change for participants with anorexia
nervosa. We want to know if the three Medication groups differed significantly in terms of weight change.
[Use only Medication and WeightChange.]
Check the Assumptions for ANOVA and Compare the Means of the three Independent Groups, Post hoc tests
 Go to Analyze  Descriptives  Explore. Select WeightChange as the Dependent variable. Select
Medication as the Factor. Click on Plots… Under ‘Boxplots’ choose None. Choose Histogram. Choose

Normality plots with tests Continue. Ok.
 Go to Analyze  Compare Means  One-Way ANOVA. Select WeightChange as the Dependent
variable. Select Medication as the Factor. Click on Post Hoc… Under ‘Equal Variances Assumed’
choose Tukey. Continue. Click on Options… Choose ‘Descriptive, Homogeneity of variance test.
Continue. OK.
17. Based on the Sig (p-value) for the Shapiro-Wilk test for each medication group, is it possible to reject the
null (the distribution is equal to normal) for any of the groups? _____The p-value for Zinc is 0.029<0.05.
Hence the normality assumption for Zinc is violated at 5% level.__________________________
18. Based on the Sig (p-value) for the Levene’s Statistic “Based on Mean,” is it possible to reject the null (the
group variances are equal to each other)? _No, the null hypothesis for equal variances cannot be rejected._
19. What is the F-value? ___34.063____ Is it significant? __Yes_________Should you reject H0?
_____Yes________
20. Based on the Tukey tests, which Medication condition was most effective? Zinc is found to be most
effective among the three medication groups.

Bibliography
Elliott, A.C. and Woodward, W.A., 2014. IBM SPSS by example: a practical guide to statistical data
analysis. Sage Publications.
Goos, P. and Meintrup, D., 2016. Statistics with JMP: Hypothesis Tests, ANOVA and Regression. John
Wiley & Sons.
Larson-Hall, J., 2015. A guide to doing statistics in second language research using SPSS and R.
Routledge.
Levine, G., 2013. A guide to SPSS for analysis of variance. Psychology Press.