Question

Suppose you want to know if different dosages of caffeine affect students’ performance on a test. We randomly sampled 15 college students and randomly assigned them to three conditions.

• Group 1: Very large dose of caffeine
• Group 2: Moderate dose of caffeine
• Group 3: Placebo (no caffeine)

The table below displays these 15 students’ scores.

Group1	Group 2	Group 3
15	26	10
16	25	9
14	23	9
13	20	6
12	21	6

1. Calculate SSB, SSW, SST, MSB, MSW, and F observed value. Fill out the blanks of this table.

	X	Y		$\bar{Y}$		(	(
	1	15
	1	16
	1	14
	1	13
	1	12
	2	26
	2	25
	2	23
	2	20
	2	21
	3	10
	3	9
	3	9
	3	6
	3	6
SUM:

Y Y Yi) Y - Y) Y- Y)2

Answer 1

(Y Y bar i) 2 (Y bar i - Y bar)^2 Y bar i Y bar Y 15 (15-14) 2 15(15-14) 2 15(15-14)^2 15(15-14)^2 15 (15-14)^2 15 (15-23)^2 15 (15-23) 2 15 (15-23)^2 15 (15-23)^2 15 (15-23)^2 1|(14-15)^2 4 (14-15)^2 O(14-15)^2 1 (14-15)^2 4 (14-15)^2 9 (14-15) 2 4 (14-15)^2 0 (14-15) 2 9 (14-15) 2 4 (14-15) 2 1 15 14 1 16 14 1 1 14 14 1 1 13 14 1 1 1 12 14 1 26 23 2 64 25 23 64 23 23 64 2 20 23 64 2 21 23 64 6.25 (14-15)^2 15 (15-7.5)^2 3 10 7.5 56.25 15 (15-7.5)^2 15 (15-7.5)^2 15 (15-7.5)^2 15 (15-7.5)^2 2.25 (14-15)^2 2.25 (14-15)^2 2.25(14-15)^2 2.25 (14-15)^2 7.5 3 56.25 7.5 3 56.25 7.5 3 56.25 7.5 3 56.25 6 UM: 222.5 225 51.25 606.25 mean 14 1 23 2 2 7.5 y bar 14.83333 Lnn LnLn

One way anova We are given 14 S2.5 Group 1: X n15 Group 2 X2 23 S 6.5 n2 5 S2 =3.5 n3 Group 3: X3 8 5 Step 1- State the hypotheses Но : м H1 At least one mean is different form the other. Step 2 Find the critical value No. of group k 3 Total no. of observation N 5+5+5 15 dfN k-1 2 dfD N k 12 a 0.01 Using a and dfN 12, we find the critical value 2, dfD Feritical 0.7348 Step 3 Compute the test value a. Grand mean (X1XE1(Xx)+(X3x) Хам (14x5)+(23x5)+(8x5) _15 XGM 15

b. Sum of squares within and between group 2(X- XG)2( X2- XGM +ng(X3 XGM) S Shetween = S Sietweer 5(14 - 15)25(23 - 15)2 5(8 - 15) = 570 n1(n 1)Sn2(n2 1) S n3(n3- 1)S SSithi SSaithin = 5(5 - 1)2.5 5(5 1)6.55(5 - 1)3.5 = 50 c. Variance within and between group SStetreen A-1 570 Shetween = 285 3-1 SSwithin 50 Seithi 4.1667 - - (5-1)+(5-1)+(5-1) (1-1)+(2-1+(n3-1) d. F statistics Stetween Swithin 285 Fstat 68.3995 - 4.1667 e. Pvalue Using Fstat and dfN 2, dfD 12, we find the pvalue Pvalue 0 Step 4 Since the pvalue is less than the 0.01, we Reject Step 5 There is enough evidence to reject the null hypothesis and conclude that at least one mean is different from the others. Summary Table Sum of squares Mean square d.f. F Pvalue 285 68.3995 Between 570 2 50 12 4.1667 Within Total 620 14