(Use Excel) A recent report criticizes SAT-test-preparation providers for promis

Question

(Use Excel) A recent report criticizes SAT-test-preparation providers for promising big score gains without any hard data to back up such claims (The Wall Street Journal, May 20, 2009). Suppose eight college- bound students take a mock SAT, complete a three-month test-prep course, and then take the real SAT. Let the difference be defined as Score on Mock SAT minus Score on Real SAT. Use Table 2 Student 2 4 Score on Mock SAT Score on Real SAT 1,868 1,728 2,041 2,009 1,643 1,806 1,945 1,729 1,852 1,787 2,008 2,222 1,691 1,965 1,805 1,775 7 8 &Click; here for the Excel Data File Let the difference be defined as scores on Mock SAT -Real SAT a. Specify the competing hypotheses that determine whether completion of the test-prep course increases a student's score on the real SAT b. Assuming that the SAT scores difference is normally distributed, calculate the value of the test statistic and its associated p-value. (Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places. Compute the p-value using your unrounded test statistic value. Round "Test statistic" value to 2 decimal places and "p-value" to 4 decimal places.) Test statistic p-value C. At the 5% significance level, do the sample data support the test-prep providers' claims? Reject H0 since the p-value is less than . Reject Ho since the p-value is more than Do not reject H0 since the p-value is less than Do not reject Ho since the p-value is more than a

Explanation / Answer

Given that,
null, H0: Ud >= 0
alternate, H1: Ud < 0
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.895
since our test is left-tailed
reject Ho, if to < -1.895
we use Test Statistic  
to= d/ (S/n)
where
value of S^2 = [ di^2 – ( di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = -42
We have d = -42
pooled variance = calculate value of Sd= S^2 = sqrt [ 99496-(-336^2/8 ] / 7 = 110.44
to = d/ (S/n) = -1.08
critical Value
the value of |t | with n-1 = 7 d.f is 1.895
we got |t o| = 1.08 & |t | =1.895
make Decision
hence Value of |to | < | t | and here we do not reject Ho
p-value :left tail - Ha : ( p < -1.0756 ) = 0.15889
hence value of p0.05 < 0.15889,here we reject Ho
ANSWERS
---------------
test statistic: -1.08
critical value: reject Ho, if to < -1.895
decision: Do not Reject Ho
p-value: 0.15889

X Y X-Y (X-Y)^2 1868 1852 16 256 1728 1787 -59 3481 2041 2008 33 1089 2009 2222 -213 45369 1643 1691 -48 2304 1806 1965 -159 25281 1945 1805 140 19600 1729 1775 -46 2116 -336 99496

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