Question

In this study, researchers were interested in knowing if the amount of coffee (none, one cup, 4 cups) and the time coffee was ingested (am v. pm) affected student exam performance (exams were in the evening). This study took place over a 3-day period. Students were randomly assigned to either ingest coffee in the morning (AM) or the evening (PM). On day one, students were given no coffee; on day two, students were given one cup of coffee; and on day three, students were given four cups of coffee. Below are the data from this experiment. Exam scores could range from 0-100, higher scores indicate better exam performance.

Participant	Amount of Coffee	Time coffee was ingested	Exam Score
1	None	AM	75
	One cup	AM	84
	4 cups	AM	70
2	None	PM	70
	One cup	PM	84
	4 cups	PM	50
3	None	AM	60
	One cup	AM	90
	4 cups	AM	78
4	None	PM	70
	One cup	PM	80
	4 cups	PM	62
5	None	AM	70
	One cup	AM	87
	4 cups	AM	69
6	None	PM	62
	One cup	PM	79
	4 cups	PM	46
7	None	AM	75
	One cup	AM	96
	4 cups	AM	81
8	None	PM	68
	One cup	PM	92
	4 cups	PM	63

1. What type of design is this experiment? (between-subjects, within-subjects, mixed) Be sure to also write the factorial notation.
2. What are the IV(s)? Be sure to indicate the levels of and the type of variable (between-subjects, within-subjects).
3. Draw a block diagram for your variables. Be sure to include cell means and marginal means.

Answer 1

The Factorial Design: Two-way Analysis of Variance Two factors are simultaneously evaluated at two or more levels of two factors. Data frame of the model: Participant None One cup Four_cups Participant None One_cup Four_cups 70 78 69 81 84 90 87 96 70 70 62 68 75 60 70 75 2 4 6 8 84 80 79 92 50 62 46 63 7 Participant Nonme One cup Four_cups 70 75 84 90 78 60

70 75 70 70 62 68 87 96 84 80 79 92 69 81 50 62 46 63 7 4 6 75 84 70 60 90 78| 70 87 69l 「70 84 501 | 70 80 62l [75 84 70] |60 90 78| AM: 70 87 691 PM2 79 46 Score 75 96 81 I 70 84 50 70 80 62 62 79 46 68 92 63 75 96 81J 68 92 63 ccols (AM) 3 n: rows (AM)i-0,1..rows (AM)-1 j:= 0 , 1 ..c_1

[70 1 89.25 74.5 SSJE:- J -SSJEM333.75 TL λη ccols (PM)3 n :=rows ( PM j0,1..c-1 に0, 1 . . rows (PM) -| 83.75| n, 55.25j i,j ( j=0 i=0

SSE:-SSJE+SSTU-700.25 Т. ._X =68.833 SSTA:= Ž Ž ,-) 2001.6667 SSJA +SSTA-3.147.10 j:0,1..c-1 i-0,1..rows (Score)-1 L -Score cols (Score) = 3 rows(Score) n Σ (L.) Grand Mean: GM:--0 二73.375 . SST-= Σ Σ (Li",-GM) =364 1.625 j-0 i0 i.ji

SSS! 3[64.875j SSA SST-(SSJA+SSTA)-495.0417 SSB-SST-sss 1 2127.25 SSAB-SST (SSA + SSB + SSE) 319.0833 SSE-SSJE + SSTU 700.25 Number of levels of factor A 4 Number. of levels.of.factor B...c-3 Number of values (replicates) n'4 i 0,1..r-1

Number of values in the entire experiment n-r. c. n, k-0.1 ..n'-1 48 Degrees of Freedoms: dfr-(n-1) Mean sum of squares SSE SSA SSB SSAB MISA : MSAB : MSE :二 AB Two-way Analysis of Variance summary table: SourceDegr Sum of squares ees of Mean Squares (Variance) freedom MSA = 8.4834 SSA-495.0417 MSA-165.0139 Fstatー

SSB- 2127.25 MSB- 1063.625 MSB taMSE 二54.6812 AB Error Total SSAB-319.0833 MSAB 53.1806 MSAB MSE SSE-700.25 dfE_ 36 MSE-19.4514 F -2.734 stat- SST- 3641.625 Testing for factor and interaction effects: Null hypothesis: Alternative hypothesis: , Not all μ are equal F test for interaction effect: A test of hypothesis of no interaction of factors A and B Fstat_ MSAB stat: MSE 2.34 Fis the upper tail critical value from an F distribution with (r-1)*(c-1) and rc(n'-1) degrees of freedom Q:二0.05

Upper-tail critical value of F distribution: Fa qF ((1-a), dfAB,dfE) 2.3638 P1-pFFstat dAB dfe)-0.0272 We reject the null hypothesis at α:-0.05, level of significance. Because Estat 2.734 F 2.364 and. P<a You.conclude that there is sufficient evidence of an interaction effect between Exam score and Number of cups of coffee. F test for Factor A: A test of hypothesis of no difference due to factor A MSA 8.4834 FstatーMSE Fais the upper tail critical value from an F distribution with r-1 and rc(n'-1) degrees of freedom 0.05 Upper-tail critical value of F distribution: Fa F (1-a),dfA,dfE) 2.8663

P-1-pF (Fstat,dfa,dfe)0.0002 We reject the null hypothesis at a 0.05, level of significance. Because Fstt 8.4834 >F,-2.866 and P<α You conclude that there is sufficient evidence of a difference in the mean exam scores among three categories of coffee intake by the students. F test for Factor B: A test of hypothesis of no difference due to factor B MSB= 54.6812 Fais the upper tail critical value from an F distribution with r-1 and rc(n'-1) degrees of freedom. Upper-tail critical value of F distribution: α:-0.05 Reject the null hypothesis at the alpha level of significance for Fstat 54.6812 > ,-3.259 , P〈α and conclude that there is evidence of a difference in the mean exam scores among the three categories of coffee intake by the students.