Question

I have answered sections A/B. Can you double check my answers for C/D, then solve E/F. In total, I need the correct answers for C, D. E. F. Thank you! The question is below...

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Dr. Pepper wants to examine if a training program has an effect on weekly exercise. College students exercise an average of 2.2 days a week with a standard deviation of 1.7 days. Dr. Pepper's sample of 27 students exercise an average of 2.3 days a week. What can be concluded with an α of 0.10?

a) What is the appropriate test statistic?
-- z-test

b)
Population:
--- college students
Sample:
--individuals exposed to the training program

c) Obtain/compute the appropriate values to make a decision about H₀.
(Hint: Make sure to write down the null and alternative hypotheses to help solve the problem.)
critical value =  1.645; test statistic = 0.3056
Decision:  --Fail to reject H0

d) If appropriate, compute the CI. If not appropriate, input "na" for both spaces below.
[ 1.7618 ,  2.8382]

e) Compute the corresponding effect size(s) and indicate magnitude(s).
If not appropriate, input and 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

f) Make an interpretation based on the results.

1) Individuals that went through the training program did significantly more exercise than college students.

2) Individuals that went through the training program did significantly less exercise than college students.    

3) The training program has no significant effect on weekly exercise.

Answer 1

c)

Ho :   µ =   2.2  
Ha :   µ ╪   2.2   (Two tail test)

population std dev ,    σ =    1.7000
Sample Size ,   n =    27
Sample Mean,    x̅ =   2.3000
      
'   '   '
      
Standard Error , SE = σ/√n =   1.7/√27=   0.3272
Z-test statistic= (x̅ - µ )/SE =    (2.3-2.2)/0.3272=   0.3057
      
critical z value, z* =   ±   1.645

answer: correct

d)

Level of Significance ,    α =    0.1          
'   '   '          
z value=   z α/2=   1.645   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   1.7/√27=   0.3272          
margin of error, E=Z*SE =   1.6449   *   0.3272   =   0.5381
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    2.30   -   0.5381   =   1.7619
Interval Upper Limit = x̅ + E =    2.30   -   0.5381   =   2.8381

answer: correct

e)

Cohen's d=|(mean - µ )/std dev|=   0.06 (trivial effect)

r = d/√(d²+4) =   0.029 (trivial effect)

f)

3) The training program has no significant effect on weekly exercise.

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