Question

A cognitive psychologist working in the area of suggestion asked a sample of children to solve as many problems as they could in 15 minutes. Half of the children are told that this is a problem-solving test and the other half are told that this is just a time-filling task. The psychologist hypothesizes that thinking it is a time-filling task will reduce problem solving. What can the psychologist conclude with an α of 0.05? Below is the data for the number of problems solved.

problem- solving	time- filling
10 5 9 7 6 8 7	3 4 7 4 2 6 4

Compute the appropriate test statistic(s) to make a decision about H₀.
(Hint: Make sure to write down the null and alternative hypotheses to help solve the problem.)
critical value =  

d) If appropriate, compute the CI. If not appropriate, input "na" for both spaces below.
[   ,   ]

e) Compute the corresponding effect size(s) and indicate magnitude(s).
If not appropriate, input and/or select "na" below.
d =   ;  ---Select--- na trivial effect small effect medium effect large effect
r² =   ;  ---Select--- na trivial effect small effect medium effect large effect

Answer 1

For Problem solving :  

Sample mean using excel function AVERAGE(), x̅1 = 7.4286

Sample standard deviation using excel function STDEV.S, s1 = 1.7182

Sample size, n1 = 7

For Time filling :  

Sample mean using excel function AVERAGE(), x̅2 = 4.2857

Sample standard deviation using excel function STDEV.S, s2 = 1.7043

Sample size, n2 = 7

a) Null and Alternative hypothesis:

Ho : µ1 ≤ µ2

H1 : µ1 > µ2

Pooled variance :

S²p = ((n1-1)*s1² + (n2-1)*s2² )/(n1+n2-2) = ((7-1)*1.7182² + (7-1)*1.7043²) / (7+7-2) = 2.9286

Test statistic:

t = (x̅1 - x̅2) / √(s²p(1/n1 + 1/n2 ) = (7.4286 - 4.2857) / √(2.9286*(1/7 + 1/7)) = 3.4358

df = n1+n2-2 = 12

Critical value :

Right tailed critical value, t crit = ABS(T.INV(0.05, 12)) = 1.782

Decision:

t > tcrit, Reject the null hypothesis

d) 95% Confidence interval :

At α = 0.05 and df = n1+n2-2 = 12, two tailed critical value, t-crit = T.INV.2T(0.05, 12) = 2.179

Lower Bound = (x̅1 - x̅2) - t-crit*√(S²p*(1/n1 +1/n2)) = (7.4286 - 4.2857) - 2.179*√(2.9286*(1/7 + 1/7)) = 1.1498

Upper Bound = (x̅1 - x̅2) + t-crit*√(S²p*(1/n1 +1/n2)) = (7.4286 - 4.2857) + 2.179*√(2.9286*(1/7 + 1/7)) = 5.1359

1.1498 < µ1 - µ2 < 5.1359

e) Cohen's d = (x̅1- x̅2)/√S²p = 1.84

r² = d²/(d²+4) = 0.46