Question

Application Exercise:
The U.S. state of Idaho hired an education firm to implement a program to help students in science, technology, engineering, and math (STEM). The following year, the average math SAT portion was 487 for a sample of 13 students in Idaho. The national math SAT average is 502 with a standard deviation of 82. State officials believe the program had the opposite effect. 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--- national math SAT scores education firm students in STEM Idaho State officials math SAT scores in Idaho
Sample:
---Select--- national math SAT scores education firm students in STEM Idaho State officials math SAT scores in Idaho

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.

Students from the nation had significantly lower math SAT scores than students from Idaho.

Students from the nation had significantly higher math SAT score than students from Idaho.     

The program did not have a significant effect.

Answer 1

a)

z-test

b)

Population

national math SAT scores education firm students in STEM

Sample

Idaho State officials math SAT scores in Idaho

c)

Ho :   µ =   502                  
Ha :   µ <   502       (Left tail test)          
                          
Level of Significance ,    α =    0.01                  
population std dev ,    σ =    82.0000                  
Sample Size ,   n =    13                  
Sample Mean,    x̅ =   487.0000                  
                          
'   '   '                  
                          
Standard Error , SE = σ/√n =   82.0000   / √    13   =   22.7427      
Z-test statistic= (x̅ - µ )/SE = (   487.000   -   502   ) /    22.7427   =   -0.66
                          
critical z value, z* =       -2.3263   [Excel formula =NORMSINV(α/no. of tails) ]              
                          
p-Value   =   0.2548   [ Excel formula =NORMSDIST(z) ]              
Decision:   p-value>α, Do not reject null hypothesis                       
Conclusion: There is not enough evidence that SAT score has decreased

d)

Level of Significance ,    α =    0.01          
'   '   '          
z value=   z α/2=   2.5758   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   82.0000   / √   13   =   22.742708
margin of error, E=Z*SE =   2.5758   *   22.74271   =   58.581334
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    487.00   -   58.581334   =   428.418666
Interval Upper Limit = x̅ + E =    487.00   -   58.581334   =   545.581334
99%   confidence interval is (   428.42   < µ <   545.58   )

e)

Cohen's d=|(mean - µ )/std dev|=   0.183 --- small effect

r² = d²/(d² + 4) =    0.008 ----trivial effect
  

f)

The program did not have a significant effect.

THANKS

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