Question

A sample obese adult males were randomly divided into two groups in a clinical program. In the experimental group, participants were involved in a vigorous 30-minute-per-day exercise program and also given instructions on the importance of diet. The control group only received the diet instructions. The program directors want to determine if the exercise program had an impact on LDL cholesterol levels. Below are the LDL cholesterol data after one month in the program. What can be concluded with an α of 0.05?

control	exprt
129 127 131 126 127 129 128 148 125 129	132 130 135 131 132 128 137 131 132 132

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

b)
Condition 1:
---Select---obese adult malesexprt groupLDL cholesterolone monthcontrol group
Condition 2:
---Select---obese adult malesexprt groupLDL cholesterolone monthcontrol group

c) Input the appropriate value(s) to make a decision about H₀.
(Hint: Make sure to write down the null and alternative hypotheses to help solve the problem.)
p-value =______ ; Decision: ---Select---Reject H0Fail to reject H0

d) Using the SPSS results, 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

e) Make an interpretation based on the results.

Participants that exercised with diet instructions had significantly reduced LDL cholesterol levels.

Participants that exercised with diet instructions had significantly increased LDL cholesterol levels.    

There is no significant LDL cholesterol difference between the groups.

Answer 1

a)

Independent-Samples t-test

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

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   129.90                  
standard deviation of sample 1,   s1 =    6.59                  
size of sample 1,    n1=   10                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   132.00                  
standard deviation of sample 2,   s2 =    2.49                  
size of sample 2,    n2=   10                  
                          
difference in sample means =    x̅1-x̅2 =    129.9000   -   132.0   =   -2.10  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    4.9827                  
std error , SE =    Sp*√(1/n1+1/n2) =    2.2284                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -2.1000   -   0   ) /    2.23   =   -0.942
                          
Degree of freedom, DF=   n1+n2-2 =    18                  

p-value =        0.358465   (excel function: =T.DIST.2T(t stat,df) )              
Conclusion:     p-value>α , Do not reject null hypothesis              

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

effect size,      
cohen's d =    |( x̅1-x̅2 )/Sp | =    0.421 medium effect
...............

      

There is no significant LDL cholesterol difference between the groups.

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

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