SOLUTION: I really need help with this problem: You are conducting a study to see if students do better when they study all at once or in intervals. One group of 12 participants took a te

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Question 1050926: I really need help with this problem:
You are conducting a study to see if students do better when they study all at once or in intervals. One group of 12 participants took a test after studying for one hour continuously. The other group of 12 participants took a test after studying for three twenty minute sessions.
The first group had a mean score of 75 and a variance of 120. The second group had a mean score of 86 and a variance of 100.
a. What is the calculated t value? Are the mean test scores of these two groups significantly different at the .05 level?
b. What would the t value be if there were only 6 participants in each group? Would the scores be significant at the .05 level?
Thank you so much for your help!

Found 2 solutions by Boreal, stanbon:
Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website!
t=(y1-y2)/spsqrt{[((1/n1)+(1/n2)]}
y1=75;y2=86
sp1^2=120;sp2^2=100
n1=n2=12
s2p=(N1−1)s1^2+(N2−1)s2^2/(N1+N2−2) (using pooled variance)
y1-y2=-11
sp^2=11(120)+11(100)/(22)=1320+1100/22=110
sp=10.49
10.49*sqrt(2/12)=10.49*sqrt(1/6)=4.28
t=-11/4.28-2.57
Critical value for t df=22 for 0.975 is +/-2.074
This exceeds the critical value; therefore, we can consider the two groups as having different means at the 5% significance level.
At 6 participants, sp^2=5(120)+5(100)/10=110; sp=10.49;sp*sqrt(1/3)-6.06
t=-11/6.06=1.82. This would no longer be significant at the 0.05 level, for the critical value would be larger as well, +/-2.23.

Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
One group of 12 participants took a test after studying for one hour continuously. The other group of 12 participants took a test after studying for three twenty minute sessions.
The first group had a mean score of 75 and a variance of 120.
The second group had a mean score of 86 and a variance of 100.
a. What is the calculated t value?
Are the mean test scores of these two groups significantly different at the .05 level?
t(75-86) = (-11)/sqrt[(120/12)+(100/12)] = -2.5690
p(value) = 2*P(t< -0.6 when df = 12-1) = 2*tcdf(-100,-2.5690,11)
= 2*0.013 = 0.026
Since the p-value is < 5%, reject equality; the scores are significantly
different
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b. What would the t value be if there were only 6 participants in each group? Would the scores be significant at the .05 level?
I'll leave that to you.
Cheers,
Stan H.
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