Question

Dr. Page believes that going through a training program will decrease weekly reading. College students read a mean of 2.3 days a week with a variance of 0.49 days. Dr. Page's sample of 31 students read a mean of 1.9 days a week. What can be concluded with α = 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--- (reading) college students days in the week individuals exposed to the training program students
Sample:
---Select--- (reading) college students days in the week individuals exposed to the training program students

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.

Individuals that went through the training program did significantly more reading than college students.Individuals that went through the training program did significantly less reading than college students.    The training program has no significant effect on weekly reading.

Answer 1

a) z test (because σ ² is known)

b)

population: college students

Sample: individuals exposed to the training program

c)

Ho :   µ =   2.3                  
Ha :   µ <   2.3       (Left tail test)          
                          
Level of Significance ,    α =    0.010                  
population std dev ,    σ =    0.7000                  
Sample Size ,   n =    31   0.4900              
Sample Mean,    x̅ =   1.9000                  
                          
'   '   '                  
                          
Standard Error , SE = σ/√n =   0.7000   / √    31   =   0.1257      
Z-test statistic= (x̅ - µ )/SE = (   1.900   -   2.3   ) /    0.1257   =   -3.182
                          
critical z value, z* =       -2.326   [Excel formula =NORMSINV(α/no. of tails) ]              
                          
Decision: test stat < critical value, 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 =   0.7000   / √   31   =   0.1257
margin of error, E=Z*SE =   2.5758   *   0.1257   =   0.3238
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    1.90   -   0.323843   =   1.5762
Interval Upper Limit = x̅ + E =    1.90   -   0.323843   =   2.2238
99%   confidence interval is (   1.58   < µ <   2.22   )

e)

Cohen's d=|(mean - µ )/std dev|=   0.571 (medium)

r² = d²/(d² + 4) =    0.075 (small)

f)

Individuals that went through the training program did significantly less reading than college students.