Question

A graduate student in education has developed an mathematical intervention program for under-performing school children. To test the effectiveness of his intervention he administers a math skills assessment at the beginning of the school year to a sample of n-25 students. He then implements the program throughout the year and re-assess the students again at the end of the school year. On average there was a 3 point difference between time 1 and time 2, with a variance value of s-36. Attach File Browse My Computer QUESTION 43 Would the graduate student use a one-tailed or two-tailed test? One-tailed Two-tailed QUESTION 44 If the graduate student used an alpha of α-.05, what would be degrees of freedom and corresponding critical value used to judge significance? df-24; critical value- 1.711 df 2; critical value 9.925 df= 23; critical value = 2.069 df= 25; critical value = 2.787 QUESTION 45 What would be the test statistic value the graduate student would generate based on his data? 0.08 2.45 0.05 2.5 QUESTION 46 Based on the result of the statistical test, what would the graduate student conclude about the math intervention program? Would he reject or retain the null hypothesis? Retain null hypothesis (Ho): There is insufficient evidence in favor of the math intervention program havinga significant effect. Reject null hypothesis (Ho): The math intervention program was effective at raising scores across the school year. QUESTION 47 What is the effect size in terms of proportion of variance accounted for (r)? 2-0.21 2--0.21 d-0.5 2 0.00 QUESTION 48 Calculate a 95% confidence interval based on the math intervention data. 95% CI -0.52 to 5.48 95% CI =0.90 to 5.10 95% CI =-0.43 to 6.43 95% CI =-0.06 to 6.06

Answer 1

43)

one tail test

Ho :   µd=   0
Ha :   µd >   0

44)

Degree of freedom, DF=   n - 1 =    24  
t-critical value , t* =        1.711   [excel function: =t.inv(α,df) ]

45)

mean of difference ,    D̅ =    3.000                  
                          
std dev of difference , Sd =        6.0000                  
                          
std error , SE = Sd / √n =    6.0000   / √   25   =   1.2000      
                          
t-statistic = (D̅ - µd)/SE = (   3.0000   -   0   ) /    1.2000   =   2.5
                          

46)

test stat >critical value, reject Ho

answer: option B

48)

Degree of freedom, DF=   n - 1 =    24   and α =    0.05  
t-critical value =    t α/2,df =    2.0639   [excel function: =t.inv.2t(α/2,df) ]      
                  
std dev of difference , Sd =        6.0000          
                  
std error , SE = Sd / √n =    6.0000   / √   25   =   1.2000
margin of error, E = t*SE =    2.0639   *   1.2000   =   2.4767
                  
mean of difference ,    D̅ =   3.000          
confidence interval is                   
Interval Lower Limit= D̅ - E =   3.000   -   2.4767   =   0.5233
Interval Upper Limit= D̅ + E =   3.000   +   2.4767   =   5.4767
                  
so, confidence interval is (   0.52 < Dbar <   5.48 )  

answer: option a)