Question

A counselor hypothesizes that a popular new cognitive therapy increases depression. The counselor collects a sample of 28 students and gives them the cognitive therapy once a week for two months. Afterwards the students fill out a depression inventory in which their mean score was 50.73. Normal individuals in the population have a depression inventory mean of 50 with a variance of 12.96. What can be concluded with α = 0.01?

- 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 = __________; test statistic = ____________
Decision: (choose one) Reject H0 or Fail to reject H0

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

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. (choose one)

1.) The depression of students that underwent cognitive therapy is significantly higher than the population.

2.) The depression of students that underwent cognitive therapy is significantly lower than the population.  

3.) The new cognitive therapy technique does not significantly impact depression.

Answer 1

Ho :   µ =   50                  
Ha :   µ >   50       (Right tail test)          
                          
Level of Significance ,    α =    0.01                  
population std dev ,    σ =    3.6000                  
Sample Size ,   n =    28                  
Sample Mean,    x̅ =   50.7300                  
                          
'   '   '                  
                          
Standard Error , SE = σ/√n =   3.6000   / √    28   =   0.6803      
Z-test statistic= (x̅ - µ )/SE = (   50.730   -   50   ) /    0.6803   =   1.07
                          
critical z value, z* =       2.3263   [Excel formula =NORMSINV(α/no. of tails) ]              
                                
Decision:  test stat < critical , Do not reject null hypothesis

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

Level of Significance ,    α =    0.01          
'   '   '          
z value=   z α/2=   2.5758   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   3.6000   / √   28   =   0.680336
margin of error, E=Z*SE =   2.5758   *   0.68034   =   1.752430
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    50.73   -   1.752430   =   48.977570
Interval Upper Limit = x̅ + E =    50.73   -   1.752430   =   52.482430
99%   confidence interval is (   48.98   < µ <   52.48   )

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

Cohen's d=|(mean - µ )/std dev|=   0.20 small
  

r² = d²/(d² + 4) =    0.010

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

2.) The depression of students that underwent cognitive therapy is significantly lower than the population.  

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

THANKS

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