Question

A cognitive psychologist working in the area of suggestion asked a sample of children to solve as many problems as they could in 10 minutes. Half of the children are told that this is a problem-solving task and the other half are told that this is just a time-filling task. The psychologist is interested in the impact of the suggestion on problem solving. What can be concluded with an α of 0.05? Below is the data for the number of problems solved.

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

a) What is the appropriate test statistic?
---Select---naz-testOne-Sample t-testIndependent-Samples t-testRelated-Samples t-test

b)
Condition 1:
---Select---the childrensuggestiontime-fillingproblem-solvingproblems solved
Condition 2:
---Select---the childrensuggestiontime-fillingproblem-solvingproblems solved

c) Obtain/compute the appropriate values to make a decision about H₀.
(Hint: Make sure to write down the null and alternative hypotheses to help solve the problem.)
critical value = ______; test statistic = _________
Decision: ---Select---Reject H0Fail to reject H0

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---natrivial effectsmall effectmedium effectlarge effect
r² = _________; ---Select---natrivial effectsmall effectmedium effectlarge effect

f) Make an interpretation based on the results.

Children that were told this is a problem-solving task solved significantly more problems than children that were told this is a time-filling task.

Children that were told this is a problem-solving task solved significantly less problems than children that were told this is a time-filling task.

The suggestions had no significant impact on problem solving.

Answer 1

a)

Independent-Samples t-test

b)

condition 1 : problem solving

condition 2: time filling

c)

Sample #1   ---->   1                  
mean of sample 1,    x̅1=   4.286                  
standard deviation of sample 1,   s1 =    1.704                  
size of sample 1,    n1=   7                  
                          
Sample #2   ---->   2                  
mean of sample 2,    x̅2=   7.286                  
standard deviation of sample 2,   s2 =    1.604                  
size of sample 2,    n2=   7                  
                          
difference in sample means =    x̅1-x̅2 =    4.2857   -   7.3   =   -3.00  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.6547                  
std error , SE =    Sp*√(1/n1+1/n2) =    0.8845                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -3.0000   -   0   ) /    0.88   =   -3.3918
                          
Degree of freedom, DF=   n1+n2-2 =    12                  
t-critical value , t* =        2.179   (excel formula =t.inv(α/2,df)              

decision: Reject Ho

d)

Degree of freedom, DF=   n1+n2-2 =    12              
t-critical value =    t α/2 =    2.1788   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.6547              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    0.8845              
margin of error, E = t*SE =    2.1788   *   0.88   =   1.93  
                      
difference of means =    x̅1-x̅2 =    4.2857   -   7.286   =   -3.0000
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    -3.0000   -   1.9271   =   -4.927
Interval Upper Limit=   (x̅1-x̅2) + E =    -3.0000   +   1.9271   =   -1.073

e)

effect size,      
cohen's d =    |( x̅1-x̅2 )/Sp | =    1.8130 (large)
      
r²=   0.489 (large)

f)

Children that were told this is a problem-solving task solved significantly less problems than children that were told this is a time-filling task.