Question

A physical therapist developed a new yoga regimen specifically for treating knee pain. He collected a random sample of chronic knee pain patients from the hospital that underwent the new regimen for 20 days. For each day of the study, the patients are asked to rate their knee pain on a scale from 0-10; no pain to extreme pain, respectively. Chronic knee pain sufferers typically report a 5 on the pain scale. Below are the average knee pain scores for the patients over the study. What can be concluded with an α of 0.01?

id	pain
11 12 13 14 15 16 17 18 19 20 21 1 2 3 4 5 6 7 8 9 10	0 9 4 7 8 6 6 3 6 6 7 5 5 8 3 3 4 8 9 7 9

b)
Population: (Choose one)
1)pain 2)patients on the regimen 3)the hospital 4)chronic knee pain sufferer 5) 20 days

Sample:
1)pain 2)patients on the regimen 3)the hospital 4)chronic knee pain sufferer 5) 20 days

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: Reject H0 or Fail 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 =____________ ; (Choose one) 1)na 2)trivial effect 3)small effect 4)medium effect 5)large effect
r² =____________ ; (Choose one) 1)na 2)trivial effect 3)small effect 4)medium effect 5)large effect

e) Make an interpretation based on the results.  (Choose one)

1)The new yoga regimen significantly worsens knee pain.

2)The new yoga regimen significantly improves knee pain.   

3) The new yoga regimen is not significantly effective at treating knee pain.

Answer 1

one sample t test

b) population: All chronic knee pain sufferer

sample: 20 days

c)

Ho :   µ =   5  
Ha :   µ >   5   (Right tail test)
          
Level of Significance ,    α =    0.010  
sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   2.3934  
Sample Size ,   n =    21  
Sample Mean,    x̅ = ΣX/n =    5.8571  
          
degree of freedom=   DF=n-1=   20  
          
Standard Error , SE = s/√n =   2.3934/√21=   0.5223  
t-test statistic= (x̅ - µ )/SE =    (5.8571-5)/0.5223=   1.6411  
          

          
p-Value   =   0.0582   [Excel formula =t.dist(t-stat,df) ]
Decision:   p-value>α, Do not reject null hypothesis       

d)

Cohen's d= nA
  

r² = Na

e) 3) The new yoga regimen is not significantly effective at treating knee pain.