Question

A researcher would like to examine how the chemical tryptophan, contained in foods such as turkey, can affect mental alertness. A sample of n = 8 college students is obtained, and each student’s performance on a familiar video game (total points earned in the game) is measured before and after eating a traditional Thanksgiving dinner including roast turkey. The following table are the scores for each participant before and after the meal:

Participants: 1,2,3,4,5,6,7,8,

Before Meal (X1): 220,245,215,260,300,280,250,310

After Meal (X2) 210,220,195,265,275,290,220,285

a.For a two-tailed test, what is the null hypothesis using statistical notation?

b. For a two-tailed test, what is the alternative hypothesis using statistical notation?  

c. What is the sum of the difference scores (D)?

d. What is the value of MD?

e. What is the value for Sum of Squares for the difference scores (SSD)?

f. What is the sample variance for the difference scores (sD 2 )?

g. What is estimated standard error (sMD)?

h. For a two-tailed test with α = .05, what is the value/s for tcrit?

i. What is the value for tobt?  

j. Do these data indicate a significant difference between the treatments at the .05 level of significance? NOTE: simply writing that the effect is significant (e.g., only writing “reject the null”) without showing any work/calculations in parts a-e will result in point zero points.

k. Compute estimated Cohen’s d to measure the size of the treatment effect.

Answer 1

Before	After	Difference
220	210	10
245	220	25
215	195	20
260	265	-5
300	275	25
280	290	-10
250	220	30
310	285	25

Null and Alternative hypothesis:  

a) Ho : µd = 0

b) H1 : µd ≠ 0

c) Sum of difference scores:

∑d =    120

d) Mean , x̅d = Ʃd/n = 120/8 = 15

e) ∑d² = 3400  

n = 8  

Sum of square difference, SSD = (Ʃd² - (Ʃd)²/n) =    √[(3400-(120)²/8)/(8-1)] = 1600

f) Variance, sd² = (SSD)/(n-1) = (1600)/(8-1) = 228.5714

g) Standard error sMD = √(sd²/n) = 5.3452

h) df = n-1 = 7

Critical value :  

Two tailed critical value, t-crit = T.INV.2T(0.05, 7) =    2.365

Reject Ho if t < -2.365 or if t > 2.365  

i) Test statistic:  

t = (x̅d)/(sMD) = (15)/(5.3452) = 2.81

j) As t = 2.81 > 2.365, we reject the null hypothesis.

There is a significant difference.

k) r² = t²/(t²+df) = 2.8062²/(2.8062²+7) = 0.5294