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 Score on Mock SAT Score on Real SAT 1 1,823 1,898 2 1,799 1,750 3 2,100 2,097 4 2,103 2,224 5 1,582 1,569 6 1,857 1,959 7 1,934 1,805 8 1,691 1,748 PictureClick 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.

Answer 1

Mock SAT	Real SAT	Difference
1823	1898	-75
1799	1750	49
2100	2097	3
2103	2224	-121
1582	1569	13
1857	1959	-102
1934	1805	129
1691	1748	-57

Sample mean of difference calculated using excel function AVERAGE(), x̅_D = -20.1250
Sample standard deviation of difference calculated using excel function STDEV.S(), s_D =  84.4299
Sample size, n =   8
α =    0.05

a) Null and Alternative hypothesis: Mock SAT- Real SAT <0
H_o : µ_D >= 0      
H₁ : µ_D < 0      
          
b) Test statistic:          
t = (x̅_D)/(s_D/√n) = -0.6742 = -0.67
df =   7      
          
p-value :          
Left tailed p-value = T.DIST(-0.6742 , 7, 1 ) = 0.2609      
          
c) Decision:          

Do not reject H0 since the p-value is more than α.