Question

Monica has performed another study on the amount of remorse experienced following a relational transgression (i.e., rule violation). This time she thinks that participants will experience more remorse following a transgression in a serious dating relationship than either a casual dating relationship or a friends with benefits relationship. The latter two groups are not expected to differ from one another.

a. Below are the descriptive statistics (mean, variance, and sample size for each group). On the next page is an ANOVA table for this study. Some of the cells already completed in this table (i.e., the SS and df). In that table, complete all the blank cells (i.e., a cell where there either is no number or X already in it). In other words, given the information provided, complete the ANOVA table (i.e., compute MS_B, MS_W, F, p, and η²). SHOW ALL YOUR WORK (on an extra sheet if necessary). NOTE: You do not have to compute anything in a cell that already has an ‘X’ in it or where a number is already provided

	Serious Dating	Casual Dating	Friends With Benefits
Mean =	18.50	12.00	11.50
Variance =	75.00	80.00	85.00
n =	50	50	50

Source                             SS                        df               MS                 F                       p                 η²

Between Groups	1825	2
Within Groups	12,000	147		X	X	X
Total	13, 825	149	X	X	X	X

b. What are Monica’s degrees of freedom (df) for this F-test?

c. Assuming α is .05, what is the critical value for her F-test?

d. Assuming α is .05, is the amount of variation among the means statistically significant? How do you know? What do you conclude about the null hypothesis? What if her α is .01?

e. Compute η² for this result. What does this number mean?

f. Are her results consistent with what Monica expected? How do you know?

Answer 1

Part (a)

Source	SS	df	MS	F	p	η2
Between Groups	1825	2	612. 5	7.5031	0.000789	0.1320
Within Groups	12,000	147	81.6327	X	X	X
Total	13, 825	149	X	X	X	X

Explanations:

MS = SS/df

F = MSB/MSW

p = P(F_{2, 147} > 7.5031) [obtained using Excel Function: FDIST(x, Deg_freedom1, Deg_freedom2)]

η² = SSB/SST

DONE

Part (b)

Degrees of freedom (df) for this F-test = (2, 147) ANSWER

[i.e., df of numerator (Between Groups), df of denominator (Within Groups) of F-ststistic.]

Part (c)

Assuming α is 0.05, critical value for the F-test = F_{2, 147, 0.05} = 3.0576 ANSWER

[obtained using Excel Function: FINV(Probability, Deg_freedom1, Deg_freedom2)]

Part (d)

Assuming α is .05, the amount of variation among the means is statistically significant. ANSWER 1

The above conclusion is derived by the condition that F (7.5031) > Fcrit (3.0576).

This can also be derived by the condition that p-value < α. ANSWER 2

The null hypothesis is rejected. ANSWER 3

If α is 0.01, null hypothesis is still stand rejected because p-value < α. ANSWER 4

Part (e)

η² = 0.1320 ANSWER 1

This means that only 13.2% of the total variation is accounted by the treatment. ANSWER 2

Part (f)

Yes ANSWER

Because, rejection of the null hypothesis implies that the means for the three groups are not the same. This, coupled with the empirical evidence that mean for ‘Serious Dating’ group is significantly higher affirms the expectation.

DONE