Question

A researcher hypothesizes that caffeine will affect the speed with which people read, but differently if people get a full night’s sleep (e.g., 8 hours) than if they do not get a full night’s sleep (e.g., 2 hours). To test this, the researcher randomly assigns 24 people into one of 6 conditions defined by fully crossing Factor A and Factor B:

Factor A: 100 mg Caffeine (a1), 50mg Caffeine (a2) or Placebo (a3) Factor B: 8 hours of sleep (b1) or 2 hours of sleep (b2)

After the amount of sleep specified by Factor B, the study participants were given a pill that contained either 100mg Caffeine, 50mg Caffeine, or no caffeine (i.e., placebo). An hour after the treatment, each participant was asked to read from a book for 1 minute; the researcher counted the number of words each participant finished reading in that time. The following are the data. (The grand mean G = 430)

X111 = 525

b1 X112 = 525 X113 = 600 X114 = 550

(Ma1b1 = 550)

X121 = 175

b2 X122 = 375 X123 = 260 X124 = 450

(Ma1b2 = 315)

Ma1 = 432.5

X211 = 450 X212 = 420 X213 = 500 X214 = 450 (Ma2b1 = 455)

X221 = 350 X222 = 480 X223 = 360 X224 = 490 (Ma2b2 = 420)

Ma2 = 437.5

X311 = 300 X311 = 450 X311 = 430 X311 = 500 (Ma3b1 = 420)

X321 = 300 X321 = 450 X321 = 430 X321 = 500 (Ma3b2 = 420)

Ma3 = 420

Mb1 = 475

Mb2 = 385

a1 a2 a3

Use an a-level of .05 for this problem:

a. What is p (the number of conditions in A)?

b. What is q (the number of conditions in B)?

c. What is n (the number of subjects in each condition)? d. What is dfwithin? What is dftotal?

e. What is SSwithin? What is SStotal? f. What is MSwithin?

For the main effect of A:

g. What is the null hypothesis (H0A)?

  

h. What is the research hypothesis (H1A)?

i. WhatisdfA,SSA,MSA,andFA?

j. What is FcritA(dfA,dfwithin) for a = .05?

k. Draw a picture of the F-distribution. Indicate where FcritA(dfA,dfwithin) is in

the picture. Indicate what part of the curve corresponds to a. Indicate where the FA is in the picture.

l. Can you reject the null hypothesis, H0A? Explain your answer.

m. Can you accept the research hypothesis, H1A? Explain your answer.

For the main effect of B:

n. What is the null hypothesis (H0B)?

o. What is the research hypothesis (H1B)?

p. WhatisdfB,SSB,MSB,andFB?

q. What is FcritB(dfB,dfwithin) for a = .05?

r. Draw a picture of the F-distribution. Indicate where FcritB(dfB,dfwithin) is in

the picture. Indicate what part of the curve corresponds to a. Indicate where the FB is in the picture.

s. Can you reject the null hypothesis, H0B? Explain your answer.

t. Can you accept the research hypothesis, H1B? Explain your answer.

For the interaction effect AxB:

u. What is the null hypothesis (H0AxB)?

v. What is the research hypothesis (H1AxB)?

w. WhatisdfAxB,SSAxB,MSAxB,andFAxB?

x. What is FcritAxB(dfAxB,dfwithin) for a = .05?

y. Draw a picture of the F-distribution. Indicate where FcritAxB(dfAxB,dfwithin) is

in the picture. Indicate what part of the curve corresponds to a. Indicate where the FAxB is in the picture.

z. Make a graph of the 6 condition means. Your x-axis should be marked with the 3 levels of A. Put the dependent variable (i.e., number of words read) on the y- axis. Draw a line graph of A for b1, i.e., plot Ma1b1, Ma2b1, and Ma3b1 and then

connect the dots. Draw a line graph of A for b2 as well. (This graph should display the AxB interaction, i.e., the extent to which the pattern across levels of A is different for b1 and b2.)

aa. Can you reject the null hypothesis, H0AxB? Explain your answer.

bb. Can you accept the research hypothesis, H1AxB? Explain your answer.

Answer 1

a) p=3

b) q=2

c) n=4

d) df within = (n-1)*p*q = 3*3*2 = 18, df total = p*q*n - 1 = 3*2*3 - 1 =23

e)SSwithin = 112100, SStotal = SSA+SSB+SSAB+SSwithin = 1300+48600+64300+112100 = 226300

f) MSwithin = SSwithin/dfwithin = 112100/18= 6227.78

g) H_0A: The speed with which the people read does not depend on the level of caffeine consumed.

h) H_1A: The speed with which the people read depends on the level of caffeine consumed

i) dfA= p-1=2, SSA= 1300, MSA= SSA/dfA = 1300/2= 650 and FA = MSA/MSwithin = 650/6227.78= 0.104

j) FcritA(dfA,dfwithin) = 3.5546

l) We cannot reject the null hypothesis since FA = 0.104 < FcritA = 3.5546.

m) We cannot accept the research hypothesis since FA = 0.104 < FcritA = 3.5546.

n)H_0B:  The speed with which the people read does not depend on the level of sleep.

o) H_1B: The speed with which the people read depends on the level of sleep.

p)  dfB= q-1=1, SSB= 48600, MSB= SSB/dfB = 48600/1= 48600 and FB = MSB/MSwithin = 48600/6227.78 = 7.80

q) FcritB(dfB,dfwithin) = 4.4139

s) We can reject the null hypothesis since FB = 7.80 > FcritB = 4.4139.

t) We can accept the research hypothesis since FB = 7.80 > FcritB = 4.4139.

u)H_0A*B: there is no interaction between level of caffeine consumed and the level of sleep.

v) H_1A*B: there is interaction between level of caffeine consumed and the level of sleep.

w)  dfAB= (p-1)(q-1)=2*1= 2, SSAB= 64300, MSAB= SSAB/dfAB = 464300/2= 232150 and FAB = MSAB/MSwithin = 232150/6227.78 = 37.28

x)FcritAB(dfAB,dfwithin) = 3.5546

aa) We can reject the null hypothesis since FAB = 37.28 > FcritAB = 3.5546.

bb) We can accept the research hypothesis since FAB = 37.28 > FcritAB = 3.5546.