Question

0. A group of statistics students wanted to investigate the effect of social media on a person's self- esteem. For this purpose, they conducted a study where participants were randomly assigned to one of three groups: browsing ENews for 2 minutes and then taking a survey, browsing Facebook for 2 minutes and then taking a survey, or a control group where the participants just took the survey. The survey, based on the Rosenberg self-esteem scale, was the tool used to measure each participant's self-esteem. Each participant was then given a score based on their responses to the survey, and higher scores are considered indicative of higher self-esteem. A printout of the data analysis is shown below. The p-value is 0.0015, so the F statistic is significant. What is the conclusion of the test? Should the researchers do any follow-up analysis with the data collected in the study? If so, give a summary of the results of the follow-up analysis based on the data summary given below. In your summary, do not use numbers; instead, use phrase "show significantly lower self-esteem than" the 32 Boxplots Summary Statistics Mean SD 22.48 4.24 26.003.43 22.41 4.30 23.63 4.01 27 Cont FB ENew 27 pooled 81 Observed F Statistice 7.084 Source df SS MS F p-value Treatment 2 22763 113.81 7.084 0 0015 Error 78 1253 26 16.07 Total 80 1480.89 ENews- FB (-5.76, -1.42) ENows Control (-2.25, 2.10) FB Control: (1.35, 569)*

Answer 1

Solution

Since the F of ANOVA is significant, it is indicative of significant difference in the mean esteem scores of the three groups. So, the next logical step would be to undertake a Post-hoc analysis to identify the pair-wise significance of the three group means.

This is done below employing Tukey HSD Test.

Back-up Theory

Tukey HSD Test

Test Statistic

Q = (M_L - M_S)/√(MSw/n),

Where

M_L = Larger of the two means being compared

M_S = Smaller of the two means being compared

MSw = Mean Sum of Squares of Within (Error) in ANOVA

n = Number of observations per treatment in ANOVA (must be the same for all treatments)

Critical Value

The Studentized range statistic (q)* whose values are available in standard text books (or can be accessed from the Net).

In the Table, df = the df of MSw in ANOVA, which is (Total number of observations,N – Total number of treatments, k)

Decision

Reject null hypothesis of equal means if Qobs > qcrit.

Now, to work out the solution,

SE =√(MSw/n)

0.818082673

qcrit(78,3,0.05)

3.36* < q < 3.40*

α

0.05

NOTE: * qcrit(60,3,0.05) = 3.40 and qcrit(120,3,0.05) = 3.36

Comparison	Larger Mean	Smaller Mean	D = Difference	Q = D/SE	Significance
Cont vs FB	26	22.48	3.52	4.302744	yes
Cont vs ENew	22.5	22.41	0.07	0.085566	No
FB vs ENew	26.0	22.41	3.59	4.38831	yes

Since Cont vs ENew is not significant, these two groups can be treated as equal.

Further, FB is significantly higher than both Cont and ENew thus establishing that the

highest esteem results after browsing FB for 2 minutes. Answer

DONE