Question

1-) A researcher who thinks that Sanav anxiety reads the effect on the scores obtained from the success tests, wants to decrease the anxiety levels with a therapy to be applied to students. In this regard, students want to have an idea about the effectiveness of the method applied by measuring their anxiety levels before and after therapy. The lowest score on the scale of fight is 10, the highest score. 10 full points show the point where the loss is highest. Pre-therapy anxiety scores of eight students were 5, 7, 8, 9, 0, 1,2,4; The therapy sonrass anxiety scores were determined as 3, 2, 7.8, 2, 1.3.3, respectively.
a. )According to both cases, we dry the research hypothesis and test it. We comment on the effectiveness of therapy.
b.) We calculate and interpret confidence intervals and effect size values.

Answer 1

A)

Ho :   µd=   0                  
Ha :   µd >   0                  
                          
Level of Significance ,    α =    0.05       claim:µd=0          
                          
sample size ,    n =    8                  
                          
mean of sample 1,    x̅1=   4.500                  
                          
mean of sample 2,    x̅2=   3.625                  
                          
mean of difference ,    D̅ =ΣDi / n =   0.875                  
                          
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    2.1002                  
                          
std error , SE = Sd / √n =    2.1002   / √   8   =   0.7425      
                          
t-statistic = (D̅ - µd)/SE = (   0.875   -   0   ) /    0.7425   =   1.178
                          
Degree of freedom, DF=   n - 1 =    7                  

                          
p-value =        0.138566   [excel function: =t.dist.rt(t-stat,df) ]               
Conclusion:     p-value>α , Do not reject null hypothesis      

There is not sufficient evidence that anxiety levels is different after treatments.

b)

sample size ,    n =    8          
Degree of freedom, DF=   n - 1 =    7   and α =    0.05  
t-critical value =    t α/2,df =    2.3646   [excel function: =t.inv.2t(α/2,df) ]      
                  
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    2.1002          
                  
std error , SE = Sd / √n =    2.1002   / √   8   =   0.7425
margin of error, E = t*SE =    2.3646   *   0.7425   =   1.7558
                  
mean of difference ,    D̅ =   0.875          
confidence interval is                   
Interval Lower Limit= D̅ - E =   0.875   -   1.7558   =   -0.881
Interval Upper Limit= D̅ + E =   0.875   +   1.7558   =   2.631
                  
so, confidence interval is (   -0.8808   < µd <   2.6308   )  

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