Question

(1 point) Coaching companies claim that their courses can raise the SAT scores of high school students. But students who retake the SAT without paying for coaching also usually raise their scores. A random sample of students who took the SAT twice found 427 who were coached and 2733 who were uncoached. Starting with their verbal scores on the first and second tries, we have these summary statistics:

                              Try 1      Try 2      Gain

	nn	x¯¯¯x¯	ss	x¯¯¯x¯	ss	x¯¯¯x¯	ss
Coached	427	500	92	529	97	29	59
Uncoached	2733	506	101	527	101	21	52

Estimate a 99% confidence interval for the mean gain of all students who are coached.

___________________ to
at 99% confidence.
Now test the hypothesis that the score gain for coached students is greater than the score gain for uncoached students. Let μ1 be the score gain for all coached students. Let μ2 be the score gain for uncoached students.

(a) Give the alternative hypothesis: μ1−μ2_______________ 0

(b) Give the tt test statistic:

(c) Give the appropriate critical value for α=5%:

Answer 1

sample size ,    n =    427          
Degree of freedom, DF=   n - 1 =    426   and α =    0.01  
t-critical value =    t α/2,df =    2.5874   [excel function: =t.inv.2t(α/2,df) ]      
                  
std dev of difference , Sd =        59.0000          
                  
std error , SE = Sd / √n =    59.0000   / √   427   =   2.8552
margin of error, E = t*SE =    2.5874   *   2.8552   =   7.3876
                  
mean of difference ,    D̅ =   29.000          
confidence interval is                   
Interval Lower Limit= D̅ - E =   29.000   -   7.3876   =   21.6124
Interval Upper Limit= D̅ + E =   29.000   +   7.3876   =   36.3876
                  
so, confidence interval is (   21.61   < µd <   36.39   )  

============

a)

Ho :   µ1 - µ2 =   0
Ha :   µ1-µ2 >   0

b)

Sample #1   ---->   1          
mean of sample 1,    x̅1=   29.00          
standard deviation of sample 1,   s1 =    59          
size of sample 1,    n1=   427          
                  
Sample #2   ---->   2          
mean of sample 2,    x̅2=   21.000          
standard deviation of sample 2,   s2 =    52.00          
size of sample 2,    n2=   2733          
                  
difference in sample means = x̅1-x̅2 =    29.000   -   21.0000   =   8.0000
                  
std error , SE =    √(s1²/n1+s2²/n2) =    3.0235          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   8.0000   /   3.0235   ) =   2.6459

c)

t-critical value , t* =        1.6477   (excel function: =t.inv.rt(α,df)
Decision:   | t-stat | > | critical value |, so, Reject Ho