Question

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 n n x ¯ ¯ ¯ x¯ s s x ¯ ¯ ¯ x¯ s s x ¯ ¯ ¯ x¯ s s Coached 427 500 92 529 97 29 59 Uncoached 2733 506 101 527 101 21 52 Estimate a 96% confidence interval for the mean gain of all students who are coached. to at 96% confidence. Now test the hypothesis that the score gain for coached students is greater than the score gain for uncoached students. Let μ 1 μ1 be the score gain for all coached students. Let μ 2 μ2 be the score gain for uncoached students.

(a) Give the alternative hypothesis: μ 1 − μ 2 μ1−μ2 0 0 .

(b) Give the t t test statistic:

(c) Give the appropriate critical value for α= α= 5%: . The conclusion is A. There is sufficient evidence to support the claim that with coaching, the mean increase in scores is greater than without coaching. B. There is not sufficient evidence to support the claim that with coaching, the mean increase in scores is greater than without coaching.

please show your work . thank you !

Answer 1

sample size ,    n =    427
      

mean of difference ,    D̅ =    29.000
      
std dev of difference , Sd =        59.0000

sample size ,    n =    427          
Degree of freedom, DF=   n - 1 =    426   and α =    0.04  
t-critical value =    t α/2,df =    2.0601   [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 =    2.0601   *   2.8552   =   5.8819
                  
mean of difference ,    D̅ =   29.000          
confidence interval is                   
Interval Lower Limit= D̅ - E =   29.000   -   5.8819   =   23.118
Interval Upper Limit= D̅ + E =   29.000   +   5.8819   =   34.882
                  
so, confidence interval is (   23.1181   < µd <   34.8819   )  

..............

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 >   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   29.00                  
standard deviation of sample 1,   s1 =    59.00                  
size of sample 1,    n1=   427                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   21.00                  
standard deviation of sample 2,   s2 =    52.00                  
size of sample 2,    n2=   2733                  
                          
difference in sample means =    x̅1-x̅2 =    29.0000   -   21.0   =   8.00  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    52.9982                  
std error , SE =    Sp*√(1/n1+1/n2) =    2.7579                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   8.0000   -   0   ) /    2.76   =   2.901
                          
Degree of freedom, DF=   n1+n2-2 =    3158                  
  
p-value =        0.001874   [excel function: =T.DIST.RT(t stat,df) ]              
Conclusion:     p-value <α , Reject null hypothesis             

       
The conclusion is A. There is sufficient evidence to support the claim that with coaching, the mean increase in scores is greater than without coaching.

..............

THANKS

revert back for doubt

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