Question

Please answer asap!

2. (17 pts total) Research has shown that people find symmetrical faces more attractive than faces that are not symmetrical. For example, a psychologist selected a random sample of 12 people and used random assignment to place them in 1 of 3 conditions. Depending on their condition assignment, participants were shown one of three pictures: a face that is perfectly symmetrical, a face that is slightly asymmetrical, and a face that is highly asymmetrical. The researcher then asked them to rate the face in terms attractiveness on a scale from 1 to 7 with 7 being the most attractive. Use the provided steps (a-) to conduct a one-factor between subjects ANOVA that tests whether there are any differences in the attractiveness ratings across the three conditions. Perfectly Slightly Highly Symmetrical Asymmetrical Asymmetrical 7 5 2 5 3 6 3 3 5 5 3 M 5.75 4.25 2.75 0.96 0.96 0.50 s 3 d. If applicable, compute the effect size and write a one-sentence explanation of ght this value. Include the formula! If not applicable, write N/A.(2 pts) nebo that combine bars to be 01002 e. State the results in APA format (see the slides for examples!). (3 pts) f. For these data, I already computed the Tukey's HSD = 1.64. Given this Tukey's value, explicitly state which means are different/not different using APA format. (2 pts)

Answer 1

From the summary data, we construct the ANOVA for computing the effect size.

We are given with the data :

	ps	sa	ha
	7	5	2
	5	4	3
	6	3	3
	5	5	3
M	5.75	4.25	2.75
s	0.957427	0.957427	0.5

1. Calculate the overall mean:

Υ ΣΥ, 3

$\bar{Y}=\frac{5.75+4.25+2.75}{3}$

$\bar{Y}=\frac{12.75}{3}=4.25$

2. The sum of squares due to faces: $FSS=\sum n_{i}*(\bar{Y}_{i}-\bar{Y})^2$

$FSS=4*(5.75-4.25)^2+4*(4.25-4.25)^2+4*(2.75-4.25)^2$

  $FSS=4*(1.5)^2+4*(0)^2+4*(-1.5)^2$

  $FSS=4*2.25+4*0+4*2.25=18$

3. Error sum of squares $ESS=\sum (n_{i}-1)s_{i}^2$

$ESS=(4-1)*0.95^2+(4-1)*0.95^2+(4-1)*0.5^2$

  $ESS=3*(0.92+0.92+0.25)=6.25$

4. Total sum of squares TSS=6.25+18=24.25

We complete the ANOVA table:

Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	18	2	9	12.96	0.0022	4.2565
Within Groups	6.25	9	0.6944
Total	24.25	11

d. The effect size is given by $\eta ^{2}=\frac{Between SS}{Total SS}=\frac{18}{24.25}=0.7423$

This indicates that the model could account for 74.23% of the variability in the data.

e. From the ANOVA table, we observe that the calculated value of F >the critical value and so we reject the null hypothesis. Hence, we conclude that there are differences in attractiveness ratings across the three conditions.

f. It is given that HSD is 1.64. We shall form a table of differences to find out the significant difference.

Groups	Average	Comparison	Abs(Diff)	HSD	Significant or not
ps	5.75	ps vs. sa	1.5	1.64	No
sa	4.25	ps Vs. ha	3	1.64	Yes
ha	2.75	sa Vs. ha	1.5	1.64	No

Among the differences, it is observed that the difference between Perfectly symmetric and highly  asymmetric group is significant.