Question

2. Repeated-measures ANOoVA Aa Aa Suppose you are interested in studying whether temperature affects spatial reasoning abilities. You decide to test spatial reasoning using completion time scores for the paper-falding test with five peaple, repeating the test on each person with three different temperatures (40 degrees, 60 degrees, and 80 degrees Fahrenheit). In this experiment, the null hypothesis is that: O There are no differences in the mean completion times among the temporatures compared O There are no individual differences in the completion time means The completion time mean for at least one temperature is diftferent from another The factor of interest is the and the dependent variable is the The results of the study are presented in the following data table. All scares are times necessary to complete the paper-folding task, recorded in seconds Temperature Participant 40 Degrees 60 Degrees 80 Degrees Participant Totals n Class 3.22 P-10.10 A 3.42 3.46 n 5 B. 3.27 3.31 3.35 P 9.93 k 3 3.47 C 3.93 3.69 P-11.09 N-15 co search 331 PM 11/10/2019 DELL

Answer 1

Null

option A) is correct

factor of interest is temperature
dependent variable is spatial reasonling abilities

п 1 SSTreatment ΣΤ?-CF n i-1

Since you didn't provide data properly

MSSBetween MSEBetween F ratioBetween

you calculate yourself and replace following Excel formula

ANOVA
Source of Variation	SS	df	MS	F
Between	0.1619	2	0.0810	3.5582
Within	0.7308	12	0.0609
Between	0.5488	4	0.1372
Error	0.1820	8	0.0228
Total	1.6235	14

Excel formula

ANOVA
Source of Variation	SS	df	MS	F
Between	0.1619	2	=B3/C3	=D3/D6
Within	=C4*D4	12	0.0609
Between	=C5*D5	4	0.1372
Error	=B4-B5	8	=B6/C6
Total	=SUM(B3:B6)	14

Note that there first row (between ) will be wrong like SS Between MS Between, F and SS Total, you have to change accordingly

The critical value of F at 0.05, level of significance is 8.649.

if F > 8.649, then fail to reject the null hypothesis, otherwise fail to reject the null