Question

College students were randomly sorted into one of two groups. Members of each group performed a series of mental tasks while music was playing in the background. One group listened to pop music and the other to country music. It is hypothesized that music will influence the number of tasks completed. What can be concluded with an α of 0.10? The results are below:

pop	country
1 = 34.34 1 = 5.33 n1 = 16	2 = 37.05 2 = 4.37 n2 = 9

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

b)
Condition 1:
---Select---country musicmental taskpop musicmusicschool
Condition 2:
---Select---country musicmental taskpop musicmusicschool

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.

Students that listened to pop music significantly completed more tasks.

Students that listened to country music significantly completed more tasks.

    Music had no significant effect on the number of tasks completed.

Answer 1

a) Independent-Samples t-test

b)

condition 1: pop music

condition 2: country music

c)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   1                  
mean of sample 1,    x̅1=   34.340                  
standard deviation of sample 1,   s1 =    5.330                  
size of sample 1,    n1=   16                  
                          
Sample #2   ---->   2                  
mean of sample 2,    x̅2=   37.050                  
standard deviation of sample 2,   s2 =    4.370                  
size of sample 2,    n2=   9                  
                          
difference in sample means =    x̅1-x̅2 =    34.3400   -   37.1   =   -2.71  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    5.0170                  
std error , SE =    Sp*√(1/n1+1/n2) =    2.0904                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -2.7100   -   0   ) /    2.09   =   -1.2964
                          
Degree of freedom, DF=   n1+n2-2 =    23                  
t-critical value , t* = ± 1.714   (excel formula =t.inv(α/2,df)              
Decision:   | t-stat | < | critical value |, so, Fail to Reject Ho                      
  

d)

Degree of freedom, DF=   n1+n2-2 =    23              
t-critical value =    t α/2 =    1.7139   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    5.0170              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    2.0904              
margin of error, E = t*SE =    1.7139   *   2.09   =   3.58  
                      
difference of means =    x̅1-x̅2 =    34.3400   -   37.050   =   -2.7100
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    -2.7100   -   3.5827   =   -6.293
Interval Upper Limit=   (x̅1-x̅2) + E =    -2.7100   +   3.5827   =   0.873

e)

d, r² = Na

f)

    Music had no significant effect on the number of tasks completed.