{"id":522,"date":"2015-07-17T08:32:14","date_gmt":"2015-07-17T12:32:14","guid":{"rendered":"http:\/\/sites.berry.edu\/vbissonnette\/?page_id=522"},"modified":"2024-06-02T10:44:06","modified_gmt":"2024-06-02T14:44:06","slug":"one-factor-anova-rm-design-solution","status":"publish","type":"page","link":"https:\/\/sites.berry.edu\/vbissonnette\/index\/stats-homework\/documentation\/one-factor-anova-rm-design\/one-factor-anova-rm-design-solution\/","title":{"rendered":"One-Factor ANOVA (RM Design) Solution"},"content":{"rendered":"

Your homework problem:<\/h3>\n

You are interested in the effects of arousal on motor performance. A random sample of subjects perform a complex motor task under 3 conditions: no caffeine (low arousal), a small dose of caffeine (moderate arousal), and a large dose of caffeine (high arousal).
\nThe dependent variable represents performance level: higher scores represent better performance.<\/p>\n

This study resulted in the following data:<\/p>\n\n\n\n\n\n\n\n\n\n\n\n
Subject<\/td>\nLow
\nDifficulty<\/td>\n		Moderate
\nDifficulty<\/td>\n		High
\nDifficulty<\/td>\n<\/tr>\n
1<\/td>\n15<\/td>\n17<\/td>\n19<\/td>\n<\/tr>\n
2<\/td>\n2<\/td>\n7<\/td>\n4<\/td>\n<\/tr>\n
3<\/td>\n11<\/td>\n12<\/td>\n6<\/td>\n<\/tr>\n
4<\/td>\n13<\/td>\n15<\/td>\n5<\/td>\n<\/tr>\n
5<\/td>\n12<\/td>\n12<\/td>\n7<\/td>\n<\/tr>\n
6<\/td>\n2<\/td>\n18<\/td>\n11<\/td>\n<\/tr>\n
7<\/td>\n8<\/td>\n12<\/td>\n5<\/td>\n<\/tr>\n
8<\/td>\n9<\/td>\n14<\/td>\n2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Did the perceived level of caffeine significantly affect the participants’ performance (alpha = .05)? If your analysis reveals a significant overall effect, then make sure to explore all possible mean differences with a post-hoc analysis (same alpha).<\/p>\n

\n

The appropriate\u00a0 test statistic for your study would be the value of the F<\/i>
\nstatistic that would result from\u00a0 a One-Factor Repeated Measures Analysis of Variance<\/u> (ANOVA):<\/p>\n

\"anova_rm1_1\"<\/a><\/p>\n

The F<\/i> statistic is equal to the ratio of the Mean Square
\nTreatment<\/i> divided by the Mean Square Error<\/i>. k<\/i> is
\nequal to the number of groups, and n<\/i> is equal to the number of
\nscores in each treatment condition (n<\/i> is also equal to the number of
\nparticipants in your study).<\/p>\n

Compute the test statistic<\/b>. When we conduct an ANOVA, we typically
\nconstruct a Source Table<\/i>
\nthat details the sources of your variance and the corresponding degrees
\nof freedom:<\/p>\n\n\n\n\n\n\n\n
Source<\/td>\nSS<\/td>\ndf<\/td>\nMS<\/td>\nF<\/td>\n<\/tr>\n
Treatment:<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
Subjects:<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
Error:<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
Total:<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

To facilitate your ANOVA, you should compute the following statistics: a)\u00a0 the sum of all scores (\u03a3X), the sum of all squared scores\u00a0 (\u03a3X\u00b2), the overall number of scores (N), and the number of\u00a0 scores in each treatment group (n). In our case, \u03a3X=238,\u00a0 \u03a3X\u00b2=2964, N=24, and n=8. Compute these statistics for yourself\u00a0 now.<\/p>\n

Also, you should compute the sum of scores (T<\/i>) for each treatment\u00a0 condition:<\/p>\n\n\n\n\n
<\/td>\nLow
\nArousal<\/td>\n		Moderate
\nArousal<\/td>\n		High
\nArousal<\/td>\n<\/tr>\n
T:<\/td>\n72<\/td>\n107<\/td>\n59<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Now, you are ready to begin the Analysis of Variance. The Sum of Squares<\/i> (SS<\/i>) Total is equal to:<\/p>\n

\"anova_rm1_2\"<\/a><\/p>\n

SS<\/i> for the treatment conditions is equal to:<\/p>\n

\"anova_rm1_3\"<\/a><\/p>\n

Again, T is equal to the sum of scores in a particular treatment group.\u00a0 Now, because this is a repeated-measures design, we need to compute the SS for\u00a0 subjects. We will begin by computing the total (T) for each subject in the study:<\/p>\n\n\n\n\n\n\n\n\n\n\n\n
Subject<\/td>\nLow
\nArousal<\/td>\n		Moderate
\nArousal<\/td>\n		High
\nArousal<\/td>\n		Total<\/u><\/td>\n<\/tr>\n
1<\/td>\n15<\/td>\n17<\/td>\n19<\/td>\n51<\/strong><\/td>\n<\/tr>\n
2<\/td>\n2<\/td>\n7<\/td>\n4<\/td>\n13<\/strong><\/td>\n<\/tr>\n
3<\/td>\n11<\/td>\n12<\/td>\n6<\/td>\n29<\/strong><\/td>\n<\/tr>\n
4<\/td>\n13<\/td>\n15<\/td>\n5<\/td>\n33<\/strong><\/td>\n<\/tr>\n
5<\/td>\n12<\/td>\n12<\/td>\n7<\/td>\n31<\/strong><\/td>\n<\/tr>\n
6<\/td>\n2<\/td>\n18<\/td>\n11<\/td>\n31<\/strong><\/td>\n<\/tr>\n
7<\/td>\n8<\/td>\n12<\/td>\n5<\/td>\n25<\/strong><\/td>\n<\/tr>\n
8<\/td>\n9<\/td>\n14<\/td>\n2<\/td>\n25<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Once we have a T score for each subject, you compute SS for subjects:<\/p>\n

\"anova_rm1_4\"<\/a><\/p>\n

Now, we can find SS Error by subtraction:<\/p>\n

\"anova_rm1_5\"<\/a><\/p>\n

Add these values to your Source Table<\/i>. The df<\/i> Total is equal to\u00a0 the total number of scores minus one (N – 1). In our case this is: 24 – 1 = 23.\u00a0 The df<\/i> Treatment is equal to the number of treatment conditions\u00a0 minus one (k – 1). In our case this is: 3 – 1 = 2.\u00a0 The df<\/i> Subjects is equal to the number of subjects minus one.\u00a0 In our case this is: 8 – 1 = 7. The df<\/i> Error is equal to the number\u00a0 of treatment conditions minus one times the number of subjects minus one\u00a0 (k – 1)(n – 1). In our case, this is: 2 * 7 = 14.<\/p>\n

Each Mean Square (MS<\/i>) is equal to the SS for that source divided by\u00a0 the df<\/i> for that source. In our case, MS\u00a0<\/i> Treatment = 154.08 \/ 2 = 77.042,\u00a0 and MS<\/i> Error = 185.92 \/ 14 = 13.280.<\/p>\n

Now you are ready to compute your test statistic. F<\/i> is equal to:<\/p>\n

\"anova_rm1_6\"<\/a><\/p>\n

Here is your completed Source Table<\/i>:<\/p>\n\n\n\n\n\n\n\n
Source<\/td>\nSS<\/td>\ndf<\/td>\nMS<\/td>\nF<\/td>\n<\/tr>\n
Treatment:<\/td>\n154.08<\/td>\n2<\/td>\n77.042<\/td>\n5.80<\/td>\n<\/tr>\n
Subjects:<\/td>\n263.82<\/td>\n7<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
Error:<\/td>\n185.92<\/td>\n14<\/td>\n13.280<\/td>\n<\/td>\n<\/tr>\n
Total:<\/td>\n603.83<\/td>\n23<\/td>\n<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Conduct hypothesis test<\/b>. In our study, the F<\/i> statistic\u00a0 will have 2 and 14 df<\/i>, and alpha = .05.\u00a0 If you look up the critical value of F<\/i> in your\u00a0 table\u00a0of critical values<\/a>, you will see that if your obtained\u00a0 F<\/i> is greater than 3.74, you would conclude that perceived\u00a0 task difficulty significantly affected actual task performance\u00a0 (i.e., that there exists one or more mean differences among the treatment\u00a0 group means).<\/p>\n

Since the obtained F<\/i> (5.80) is greater than the critical value for\u00a0 F<\/i> (3.74), you would conclude that there was<\/i> a significant\u00a0 overall effect of perceived task difficulty on performance.<\/p>\n

Compute effect size<\/b>. We will compute the overall effect size two\u00a0 ways. First let’s compute Overall Eta Squared<\/i>, which is very simple:<\/p>\n

\"anova_rm1_7\"<\/a><\/p>\n

Eta Squared<\/i> is an estimate of the\u00a0 proportion of variance in your\u00a0 dependent variable that can be accounted for by variation in your independent\u00a0 variable.\u00a0However, computing effect size for a repeated-measures design is more\u00a0 complicated than it is for the completely-randomized design.<\/p>\n

Some would argue that in this design, subject variance is a systematic\u00a0 source of variance that is not related to your hypothesis test.\u00a0 They would suggest that we should disregard\u00a0 subject variance and consider only treatment and error variance\u00a0 in our analysis. Thus, it would be more appropriate to compute effect size\u00a0 after we have removed subject variance from consideration.<\/p>\n

If we follow this approach, we would compute Partial Eta Squared<\/i>:<\/p>\n

\"anova_rm1_9\"<\/a><\/p>\n

Computing\u00a0Omega Squared<\/em> -an unbiased estimate of effect size for the repeated-measures design is quite a journey. Here is the formula:<\/p>\n

\"\"<\/a><\/p>\n

Yowza! We typically do not teach this formula in introductory courses.\u00a0 If you are feeling ambitious and you apply this formula to our data, you will find that\u00a0Omega Squared<\/em> for our results was .1988 ~ .20.\u00a0 Computing\u00a0Partial Omega Squared<\/em> is much easier, since the computations are based on the\u00a0F<\/em> statistic that you computed in your ANOVA:<\/p>\n

\"anova_rm1_10\"<\/a><\/p>\n

Post-Hoc testing<\/b>. The F<\/i> statistic provides us with an\u00a0 omnibus test of whether or not our independent variable affected our\u00a0 dependent variable. It does not, however, help us to understand the exact\u00a0 nature of this effect.<\/p>\n

In our case, we know that there exists at least one significant mean\u00a0 difference among our treatment groups, but we have no idea at this point\u00a0 which particular mean difference might be significant. So, we should\u00a0 follow up our ANOVA with a post-hoc test<\/u>. We will examine all\u00a0pair-wise comparisons among our treatment groups and figure out which\u00a0 treatment group mean differences are statistically significant.<\/p>\n

We will employ Tukey’s Honestly Significant Difference <\/i> (HSD<\/i>) test (see note below). This will reveal to us what would constitute a statistically significant mean difference between any two treatment group means, controlling alpha over the course of all of these comparisons. In our study, an Honestly Significant Difference would be equal to: Unchanged: We will employ Tukey’s Honestly Significant Difference <\/i> (HSD<\/i>) test. This will reveal to us what would constitute a statistically significant mean difference between any two treatment group means, controlling alpha over the course of all of these comparisons. In our study, an Honestly Significant Difference would be equal to:
\n\"anova_rm1_11\"<\/a>
\nwhere q<\/i> is equal to the Studentized Range Statistic<\/i> and n <\/i> is equal to the number of subjects. In our case, we have 8 subjects. When alpha is equal to .05, you have 3 treatment groups, and you have 14 df <\/i> within-groups, when you examine your table of critical values, you will find that the critical value of the Studentized Range Statistic<\/a> for this study is equal to 3.701. So, the value of a significant mean difference in this study would be equal to: Unchanged: where q<\/i> is equal to the Studentized Range Statistic<\/i> and n <\/i> is equal to the number of subjects. In our case, we have 8 subjects. When alpha is equal to .05, you have 3 treatment groups, and you have 14 df <\/i> within-groups, when you examine your table of critical values, you will find that the critical value of the Studentized Range Statistic<\/a> for this study is equal to 3.701. So, the value of a significant mean difference in this study would be equal to:
\n\"anova_rm1_12\"<\/a>
\nNow, create a small table of the treatment group mean differences. Each\u00a0 value in this table is equal to the mean of the row group minus\u00a0 the mean of the column group:<\/p>\n\n\n\n\n\n\n\n
Treatment Group:<\/td>\n1<\/td>\n2<\/td>\n3<\/td>\n<\/tr>\n
1. Low:<\/td>\n<\/td>\n-4.38<\/td>\n1.63<\/td>\n<\/tr>\n
2. Mod Diff:<\/td>\n<\/td>\n<\/td>\n6.00<\/b><\/td>\n<\/tr>\n
3. High Diff:<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

We find that the mean difference between the moderate-difficult group and the high-difficulty group was significant.\u00a0 The other two mean differences were not significant.<\/p>\n

Note.\u00a0 Using Tukey’s\u00a0HSD<\/em> test in this example assumes that your data demonstrate sphericity.\u00a0 Some would argue that we should not simply assume that sphericity is present, and that using Tukey’s test would not be appropriate for the repeated-measures design.\u00a0 However, since this example is specifically for illustrating hand computations and very basic statistical concepts, this test is the most feasible for us (and it’s results will be trustworthy if the sphericity assumption is met – you will see that the Bonferroni test resulted in the same outcome).<\/p>\n

The most conservative and least controversial approach to post-hoc testing for this design involves computing a separate repeated-measures\u00a0T<\/em> test for each pairwise comparison.\u00a0 Then, we adjust the p<\/em> value for each test according the number of tests being conducted (i.e., we multiply each p<\/em> value by 3).\u00a0 This is called a Bonferroni adjustment.\u00a0 This would be easy to do if you are using a computer, and this is why that this approach is used by the Stats Homework<\/em> software package.\u00a0 However, this approach would be a challenge to conduct by hand since you would have to compute three separate T<\/em> tests and then have a special table for looking up the critical values of T<\/em>.<\/p>\n

\n

Return to One-Factor RM ANOVA<\/a><\/p>\n

Return to Table of Contents<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Your homework problem: You are interested in the effects of arousal on motor performance. A random sample of subjects perform a complex motor task under 3 conditions: no caffeine (low […]<\/p>\n","protected":false},"author":34,"featured_media":0,"parent":517,"menu_order":0,"comment_status":"open","ping_status":"open","template":"","meta":{"site-container-style":"default","site-container-layout":"default","site-sidebar-layout":"default","site-transparent-header":"default","disable-article-header":"default","disable-site-header":"default","disable-site-footer":"default","disable-content-area-spacing":"default","footnotes":""},"class_list":["post-522","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/sites.berry.edu\/vbissonnette\/wp-json\/wp\/v2\/pages\/522","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/sites.berry.edu\/vbissonnette\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/sites.berry.edu\/vbissonnette\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/sites.berry.edu\/vbissonnette\/wp-json\/wp\/v2\/users\/34"}],"replies":[{"embeddable":true,"href":"https:\/\/sites.berry.edu\/vbissonnette\/wp-json\/wp\/v2\/comments?post=522"}],"version-history":[{"count":16,"href":"https:\/\/sites.berry.edu\/vbissonnette\/wp-json\/wp\/v2\/pages\/522\/revisions"}],"predecessor-version":[{"id":2292,"href":"https:\/\/sites.berry.edu\/vbissonnette\/wp-json\/wp\/v2\/pages\/522\/revisions\/2292"}],"up":[{"embeddable":true,"href":"https:\/\/sites.berry.edu\/vbissonnette\/wp-json\/wp\/v2\/pages\/517"}],"wp:attachment":[{"href":"https:\/\/sites.berry.edu\/vbissonnette\/wp-json\/wp\/v2\/media?parent=522"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}