Question

A school psychologist is interested in the effect a popular new cognitive therapy on anxiety. The psychologist collects a sample of 13 students and gives them the cognitive therapy once a week for two months. Afterwards the students fill out an anxiety inventory in which their average score was 46.21. Normal individuals in the population have an anxiety inventory average of 49 with a standard deviation of 2.7. What can be concluded with α = 0.10?

a) What is the appropriate test statistic?'

1. na 2. z-test 3. one-sample test 4. independent-samples t-test 5. related-samples t-test

b1)
Population: (choose one of the following)

1. normal individuals 2. new hypnosis technique 3. two months 4. students receiving hypnosis

b2)
Sample: (choose one of the following)

1. normal individuals 2. new hypnosis technique 3. two months 4. students receiving hypnosis

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)*** 1. Reject H0 or 2. 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 population has significantly lower anxiety than students that underwent cognitive therapy.

2.) The population has significantly higher anxiety than students that underwent cognitive therapy.  

3.) The new cognitive therapy technique does not significantly effect anxiety.

Answer 1

z-test

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

poulation : 1. normal individuals

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

sample: . students receiving hypnosis

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

Ho :   µ =   49                  
Ha :   µ <   49       (Left tail test)          
                          
Level of Significance ,    α =    0.10                  
population std dev ,    σ =    2.7000                  
Sample Size ,   n =    13                  
Sample Mean,    x̅ =   46.2100                  
                          
'   '   '                  
                          
Standard Error , SE = σ/√n =   2.7000   / √    13   =   0.7488      
Z-test statistic= (x̅ - µ )/SE = (   46.210   -   49   ) /    0.7488   =   -3.73
                          
critical z value, z* =       -1.2816   [Excel formula =NORMSINV(α/no. of tails) ]              
                             
Decision:  test stat < critical value ,  Reject null hypothesis (lower tail test)
............

  
Level of Significance ,    α =    0.1          
'   '   '          
z value=   z α/2=   1.6449   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   2.7000   / √   13   =   0.748845
margin of error, E=Z*SE =   1.6449   *   0.74885   =   1.231741
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    46.21   -   1.231741   =   44.978259
Interval Upper Limit = x̅ + E =    46.21   -   1.231741   =   47.441741
90%   confidence interval is (   44.98   < µ <   47.44   )

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

Cohen's d=|(mean - µ )/std dev|= |-1.033| = 1.03 LARGE
  

r² = d²/(d² + 4) =    0.211 SMALL

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

THANKS

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