Question

Experimenter bias refers to the phenomenon that data tends to comes out in the desired direction even for the most conscientious experimenters. A social psychologist hypothesizes the opposite phenomenon. The psychologist tells a sample of students that they will be the experimenters in a study, and are then told that all subjects in the study will be given nicotine one hour before solving arithmetic problems; in reality none of the subjects were given nicotine. However, half of the experimenters are told that nicotine will lead to better performance and the other half are told nothing. The experimenters are then asked to score the arithmetic problems from the subjects. Below are the scores they gave. What can be concluded with an α of 0.01?

told nothing	told nicotine
12 14 13 8 14 10 20 10	18 15 21 12 17 14 21 24

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 H0 Fail 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--- na trivial effect small effect medium effect large effect
r² =  ;  ---Select--- na trivial effect small effect medium effect large effect

f) Make an interpretation based on the results.

Experimenters that were told nothing gave significantly higher scores than experimenters that expected a good performance.Experimenters that expected a good performance gave significantly higher scores than experimenters that were told nothing.    There was no significant score difference between experimenters that expected a good performance and those that were told nothing.

Answer 1

C)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 <   0                  
                          
Level of Significance ,    α =    0.01                  
                          
Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   12.63                  
standard deviation of sample 1,   s1 =    3.66                  
size of sample 1,    n1=   8                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   17.75                  
standard deviation of sample 2,   s2 =    4.06                  
size of sample 2,    n2=   8                  
                          
difference in sample means =    x̅1-x̅2 =    12.6250   -   17.8   =   -5.13  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    3.8672                  
std error , SE =    Sp*√(1/n1+1/n2) =    1.9336                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -5.1250   -   0   ) /    1.93   =   -2.65
                          
Degree of freedom, DF=   n1+n2-2 =    14                  
t-critical value , t* =        -2.6245   (excel function: =t.inv(α,df)              
Decision:   | t-stat | > | critical value |, so, Reject Ho                 

.............

D)

Degree of freedom, DF=   n1+n2-2 =    14              
t-critical value =    t α/2 =    2.9768   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    3.8672              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    1.9336              
margin of error, E = t*SE =    2.9768   *   1.9336   =   5.7560  
                      
difference of means =    x̅1-x̅2 =    12.6250   -   17.750   =   -5.1250
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    -5.1250   -   5.7560   =   -10.8810
Interval Upper Limit=   (x̅1-x̅2) + E =    -5.1250   +   5.7560   =   0.6310
CI (-10.881 , 0.631)

................

E)

effect size,      
cohen's d =    |( x̅1-x̅2 )/Sp | =    1.325
LARGE

r² = d²/(d² + 4) =    0.305
LARGE

..........

F)

Experimenters that expected a good performance gave significantly higher scores than experimenters that were told nothing.

..............

THANKS

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