You are conducting a study to see if students do better when

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 of75 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?

Solution

A)

Formulating the null and alternative hypotheses,              
              
Ho:   u1 - u2   =   0  
Ha:   u1 - u2   =/   0  
At level of significance =    0.05          
As we can see, this is a    two   tailed test.      
Calculating the means of each group,              
              
X1 =    75          
X2 =    86          
              
Calculating the standard deviations of each group,              
              
s1 =    10.95445115          
s2 =    10          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    12          
n2 = sample size of group 2 =    12          
Thus, df = n1 + n2 - 2 =    22          
Also, sD =    4.281744193          
              
Thus, the t statistic will be              
              
t = [X1 - X2 - uD]/sD =    -2.569046516   [ANSWER, T VALUE]

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b)

Calculating the means of each group,              
              
X1 =    75          
X2 =    86          
              
Calculating the standard deviations of each group,              
              
s1 =    10.95445115          
s2 =    10          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    6          
n2 = sample size of group 2 =    6          

Also, sD =    6.055300708          
              
Thus, the t statistic will be              
              
t = [X1 - X2 - uD]/sD =    -1.816590212   [ANSWER]

For part A, the P value is

p =    0.01750305

For part B, the P value is

p =    0.099328388

Hence, Part A would be significant, but part B wouldn\'t. [ANSWER]

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Hi! If you use another method/formula in calculating the degrees of freedom in this t-test, please resubmit this question together with the formula/method you use in determining the degrees of freedom. That way we can continue helping you! Thanks!

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