Question

A health psychologist hypothesizes that stress levels in the nation have been increasing over the years. In the late 1990s, the average score on the Birk Stress Inventory (BSI) in the nation was 17 with a standard deviation of 6.2. A current sample of 29 produces a mean BSI score of 14.8. What can be concluded with an α of 0.10?

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--- current sample the nation stress levels the years the BSI score
Sample:
---Select--- current sample the nation stress levels the years the BSI score

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 current population is significantly more stressed.The current population is significant less stressed.    The current population does not significantly differ in stress.

Answer 1

a) z test

because σ is known

b) population :the nation

Sample: current sample

c)

Ho :   µ ≥ 17                  
Ha :   µ <   17       (Left tail test)          
                          
Level of Significance ,    α =    0.100                  
population std dev ,    σ =    6.2000                  
Sample Size ,   n =    29   38.4400              
Sample Mean,    x̅ =   14.8000                  
                          
'   '   '                  
                          
Standard Error , SE = σ/√n =   6.2000   / √    29   =   1.1513      
Z-test statistic= (x̅ - µ )/SE = (   14.800   -   17   ) /    1.1513   =   -1.911
                          
critical z value, z* =       -1.282   [Excel formula =NORMSINV(α/no. of tails) ]              
                          
Decision: test stat < critical value, Reject null hypothesis                       

d)

Level of Significance ,    α =    0.1          
'   '   '          
z value=   z α/2=   1.645   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   6.2000   / √   29   =   1.1513
margin of error, E=Z*SE =   1.6449   *   1.1513   =   1.8937
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    14.80   -   1.893738   =   12.9063
Interval Upper Limit = x̅ + E =    14.80   -   1.893738   =   16.6937
90%   confidence interval is (   12.906   < µ <   16.694   )

e)

Cohen's d=|(mean - µ )/std dev|=   0.35 (medium)
  
r² = d²/(d² + 4) =    0.031 (small)

f)

The current population is significant less stressed.