Question

A school psychologist believes that a popular new hypnosis technique reduces depression. The psychologist collects a sample of 20 students and gives them the hypnosis once a week for two months. Afterwards the students fill out a depression inventory in which their mean score was 54.5. Normal individuals in the population have a depression inventory mean of 50 with a standard deviation of 3.5. What can be concluded with an α of 0.01?

a) What is the appropriate test statistic?
---Select--- na z-test one-sample t-test independent-samples t-test related-samples t-test

b)
Population:
---Select--- normal individuals Kyolic Pills two months students receiving hypnosis new hypnosis technique
Sample:
---Select--- normal individuals Kyolic Pills two months students receiving hypnosis new hypnosis technique

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:  ---Select--- Reject H0 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 =  ;   ---Select--- na trivial effect small effect medium effect large effect
r² =  ;   ---Select--- na trivial effect small effect medium effect large effect

f) Make an interpretation based on the results.

The depression of students that underwent hypnosis is significantly higher than the population.The depression of students that underwent hypnosis is significantly lower than the population.    The new hypnosis technique does not significantly impact depression.

Answer 1

a) Z test

because σ is known

b) population: normal individuals

Sample: students receiving hypnosis

c)

Ho :   µ ≤ 50                  
Ha :   µ > 50   
Level of Significance ,    α =    0.010                  
population std dev ,    σ =    3.5000                  
Sample Size ,   n =    20   12.2500              
Sample Mean,    x̅ =   54.5000                  
                          
'   '   '                  
                          
Standard Error , SE = σ/√n =   3.5000   / √    20   =   0.7826      
Z-test statistic= (x̅ - µ )/SE = (   54.500   -   50   ) /    0.7826   =   5.750
                          
critical z value, z* =       2.326   [Excel formula =NORMSINV(α/no. of tails) ]              
                          
Decision: test stat > 2326, reject null hypothesis                       

d)

Level of Significance ,    α =    0.01          
'   '   '          
z value=   z α/2=   2.576   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   3.5000   / √   20   =   0.7826
margin of error, E=Z*SE =   2.5758   *   0.7826   =   2.0159
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    54.50   -   2.015905   =   52.4841
Interval Upper Limit = x̅ + E =    54.50   -   2.015905   =   56.5159
99%   confidence interval is (   52.48   < µ <   56.52   )

e)

Cohen's d=|(mean - µ )/std dev|=   1.2857 (large)
  
r² = d²/(d² + 4) =    0.292 ( small)

f)
The depression of students that underwent hypnosis is significantly higher than the population