Question

# Use Excel) A recent report criticizes SAT-test-preparation providers for promising big score gains without any hard...

 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 Score on Mock SAT Score on Real SAT 1 1,868 1,852 2 1,728 1,787 3 2,041 2,008 4 2,009 2,222 5 1,643 1,691 6 1,806 1,965 7 1,945 1,805 8 1,729 1,775
 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. H0: μD = 0; HA: μD ≠ 0 H0: μD ≥ 0; HA: μD < 0 H0: μD ≤ 0; HA: μD > 0
 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 α. Reject H0 since the p-value is more than α. Do not reject H0 since the p-value is less than α. Do not reject H0 since the p-value is more than α.

The table given below ,

 Student Score on Mock SAT(X) Score on Real SAT(Y) di=X-Y di^2 1 1868 1852 16 256 2 1728 1787 -59 3481 3 2041 2008 33 1089 4 2009 2222 -213 45369 5 1643 1691 -48 2304 6 1806 1965 -159 25281 7 1945 1805 140 19600 8 1729 1775 -46 2116 Sum -336 99496

From table ,  Let , a) Hypothesis: Vs b) The test statistic is , p-value= The Excel function is , =TDIST(1.08,7,1)

c) Decision : Here , p-value=0.1580 > Therefore , Do not reject H0 since the p-value is more than α.

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