Question

A behavioral psychologist wanted to know if making a public commitment to lose weight will make a difference in how much weight is actually lost. He obtained 14 volunteers who wanted to lose weight and randomly assigned them to one of two groups. The “commitment” group met together and each participant stood and announced an intent to lose weight and signed a pledge to adhere to the diet and exercise regimen. The “control” was placed on the same diet and exercise regimen, but did not stand to announce their intent to lose weight and did not sign a pledge to adhere to the regimen. The commitment group participants lost 13, 11, 13, 12, 10, 9,and 8 pounds; the control group participants lost 10, 6, 7, 5, 9, 6, and 8 pounds. Did making a commitment to adhere to the diet and exercise regimen make a significant difference in weight loss?

• What is the IV? How many levels?
• What is the DV?
• Construct a frequency distribution with that data above for each group
• What are the means of the 2 groups?
• What is the df?
• What type of t test needs to be conducted?
• What is the calculated t statistic?
• Is this a one tail or 2 tail test?
• Are the findings significant at alpha .05? Explain what this means.

Answer 1

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          

two tail test

Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   10.86                  
standard deviation of sample 1,   s1 =    1.95                  
size of sample 1,    n1=   7                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   7.29                  
standard deviation of sample 2,   s2 =    1.80                  
size of sample 2,    n2=   7                  
                          
difference in sample means =    x̅1-x̅2 =    10.8571   -   7.3   =   3.57  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.8772                  
std error , SE =    Sp*√(1/n1+1/n2) =    1.0034                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   3.5714   -   0   ) /    1.00   =   3.5593
                          
Degree of freedom, DF=   n1+n2-2 =    12                  
  
p-value =        0.003929   (excel function: =T.DIST.2T(t stat,df) )              
Conclusion:     p-value <α , Reject null hypothesis  

result is significant

it means that making a commitment to adhere to the diet and exercise regimen make a significant difference in weight loss

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