Question

1. In the following examples, indicate whether you would perform a t-test of independent means or a dependent means (5 POINTS)
   1. Two groups of older adults were given different levels of a treatment for arthritis pain. Which treatment was more effective?
   2. A health care provider wanted to know whether his patients who received additional in-home care recovered more quickly than those who received the standard amount of in-home care
   3. One group of 85 year olds were provided an exercise program for agility and they were tested two times over a 10-month period.
   4. A group of adolescent boys were offered interpersonal skills therapy and then tested in June and December to see if there was any impact on family harmony
   5. One group of adult men was given instructions for reducing their high blood pressure whereas another group was not given any instructions. The blood pressure of men in both groups was measured before and after

2. Using the data below, compute the appropriate t-test (do it manually) and write a conclusion as to whether there as a change in a group of families’ satisfaction level (the higher the score the more satisfied) with their use of service centers following a social service intervention. (15 POINTS)

Before Intervention	After Intervention
1.3	6.5
2.5	8.7
2.3	9.8
8.1	10.2
5.0	7.9
7.0	6.5
7.5	8.7
5.2	7.9
4.4	8.7
7.6	9.1
9.0	8.4
7.6	6.4
4.5	7.2
1.1	5.8
5.6	6.9
6.2	5.9
7.0	7.6
6.9	7.8
5.6	7.3
5.2	4.6

Answer 1

a)

Independent

b)

Independent

c)

Dependent

d)

Dependent

e)

Independent

----------------------

µd= Mean difference in  a group of families’ satisfaction level before and after.

Ho :   µd=   0                  
Ha :   µd <   0                  
                          
Level of Significance ,    α =    0.05       claim:µd=0          
                          
sample size ,    n =    20                  
                          
mean of sample 1,    x̅1=   5.480                  
                          
mean of sample 2,    x̅2=   7.595                  
                          
mean of difference ,    D̅ =ΣDi / n =   -2.115                  
                          
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    2.4299                  
                          
std error , SE = Sd / √n =    2.4299   / √   20   =   0.5433      
                          
t-statistic = (D̅ - µd)/SE = (   -2.115   -   0   ) /    0.5433   =   -3.893
                          
Degree of freedom, DF=   n - 1 =    19                  

                          
p-value =        0.000490   [excel function: =t.dist(t-stat,df) ]              
Conclusion:     p-value <α , Reject null hypothesis                      

Hence there is significant change in a group of families’ satisfaction level after the intervention.

Thanks in advance!

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