Question

Coaching companies claim that their courses can raise the SAT scores of high school students. Of course, students who retake the SAT without paying for coachinggenerally 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 verbalscores on the first and second tries, we have these summary statistics:


Try 1 Try 2 Gain
x s x s x s
Coached 500 92 529 97 29 59
Uncoached 506 101 527 101 21 52
Let's first ask if students who are coached significantly increased their scores.

1. You could use the information given to carry out either a two-sample t test comparing Try 1 with Try 2 for coached students or a matched pairs t test using Gain.Which is the correct test? Why?

A. Two-sample test, because we are given data for the two samples which is more accurate than the data only on the gain.
B. Matched pairs test, because we are not interested in the difference in mean between the two groups, but the gain that each student has achieved.
C. Two-sample test, because matched pairs test can be performed only if we have information on every individual.
D. Matched pairs test, because we are interested in the difference in mean between the two groups ? not the gain that each student has achieved


2. Carry out the proper test. What do you conclude?

A. There is sufficient evidence that coached student achieved a better score on the second try.
B. There is insufficient evidence that coached student achieved a better score on the second try.
C. There is no evidence that coached student achieved a better score on the second try.

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

A. 20.9 to 37.1
B. 20.8 to 37.2
C. 21.5 to 36.5
D. 20 to 38

Answer 1

ans=

(a) The appropriate test is the matched-pairs test because a student’s score on Try 1 is certainly correlated with his/her score on Try 2. Using the differences, we have xbar =  29 and s = 59.

(b) To test H0: mu=0 vs.  H1 mu > 0, we compute

t = (29-0)/((59/sqrt(427))=10.16

with df = 426. This is certainly significant, with P < 0.0005. Coached students do improve their scores on average

(a) H₀: μ₁ = μ₂ vs. H_a: μ₁ > μ₂, where μ₁ is the mean gain among all coached students and μ₂ is the mean gain among uncoached students. H0 and Ha. Using the conservative approach, df = 426 is rounded down to df = 100 in  (t table) and we obtain 0.0025 < P < 0.005. Using software, df = 534.45 and P = 0.004. There is evidence that coached students had a greater average increase.

(b) 8 ± t*(3.0235) where t* equals 2.626 (using df = 100 with (t table) ) or 2.585 (df = 534.45 with software). This gives either 0.06 to 15.94 points, or 0.184 to 15.816 points, respectively.

(c) Increasing one’s score by 0 to 16 points is not likely to make a difference in being granted admission or scholarships from any colleges.

Answer 2

Level of Significance ,    α =    0.005          
degree of freedom=   DF=n-1=   426          
't value='   tα/2=   2.822   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   59.0000   / √   427   =   2.8552
margin of error , E=t*SE =   2.8217   *   2.8552   =   8.0566
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    29.00   -   8.056638   =   20.9434
Interval Upper Limit = x̅ + E =    29.00   -   8.056638   =   37.0566
99.5%   confidence interval is (   20.9434   < µ <   37.0566   )
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please try this if above gets wrong(20.9853   < µ <   37.0147)