Question

A neuropsychologist and physical therapist believe that patients with brain injury who begin physical therapy very early have better cognitive outcomes that patients who do not engage in early intervention. To test their research question they design a study using two groups of patients on the rehabilitation unit. One group of patients begins physical therapy within two hours of their transfer to the unit. Physical therapy involves sitting up, moving to the edge of the bed, placing feet on the floor and standing (with assistance) three times per day. The other group receives current treatment practices. After 3-weeks on the unit, a cognitive test of attention was administered. Using α = .05, determine whether the results were statistically significant.

Early Intervention	No Intervention
n = 4	n = 4
M =63	M = 57
SS = 28	SS = 20

1. Identify the independent variable.

2. Identify the dependent variable.

3. State the null and alternative hypothesis . You can use words or notation.

4. Establish the critical boundary for the research question

5. Calculate

   • Include answer for pooled variance
   • Include answer for standard error
   • Include answer for t-score

6. Summarize your results, including research notation, your decision and an explanation

7. Calculate and interpret r² (if appropriate)

8. What else could explain the change in attention between the samples (confounding or extraneous variables - think groups not just individual people)?

Answer 1

independent variable : two groups recieving Physical therapy early and late

dependent variable : cognitive outcomes

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

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

           
                          
Level of Significance ,    α =    0.05              

t-critical value , t* =        1.943   (excel function: =t.inv(α,df)      
                          
Sample #1   ---->   1                  
mean of sample 1,    x̅1=   63.00                  
standard deviation of sample 1,   s1 =    3.06                  
size of sample 1,    n1=   4                  
                          
Sample #2   ---->   2                  
mean of sample 2,    x̅2=   57.00                  
standard deviation of sample 2,   s2 =    2.58                  
size of sample 2,    n2=   4                  
                          
difference in sample means =    x̅1-x̅2 =    63.0000   -   57.0   =   6.00  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    2.8284                  
std error , SE =    Sp*√(1/n1+1/n2) =    2.0000                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   6.0000   -   0   ) /    2.00   =   3.000
                          
Degree of freedom, DF=   n1+n2-2 =    6                  
           
Decision:   | t-stat | > | critical value |, so, Reject Ho  

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

  
r²=   0.6
.......................

thanks

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