document.write( "Question 620853: Two methods of memorizing difficult material are being tested to determine if one produces better retention. A random sample of 7 pairs of students matched according to I.Q. and academic background are included in the study. Method A is randomly assigned to one student in each pair. The scores on a memorization test are listed below (by percent). Using a = .05, is there evidence to conclude a significant difference between the effectiveness of the two methods?
\n" ); document.write( "t – test: Paired Sample for Mean Method A Method B
\n" ); document.write( "Mean 61.86 65.43
\n" ); document.write( "Variance 70.4762 170.2857
\n" ); document.write( "Observations 7 7
\n" ); document.write( "t statistic -1.622
\n" ); document.write( "P-value one-tail 0.078
\n" ); document.write( "t Critical value one-tail 1.943
\n" ); document.write( "P-value two-tail 0.156
\n" ); document.write( "t Critical value two-tail 2.447 \r
\n" ); document.write( "\n" ); document.write( "Is there evidence of a difference between the effectiveness of the two methods?\r
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Algebra.Com's Answer #567483 by Cupjohn(1)\"\" \"About 
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This is a matched pair t -test.
\n" ); document.write( "So d\"bar\" is 61.86-65.43 = -3.57
\n" ); document.write( "the standard deviation is sqrt(70.4762+170.2857)=15.52
\n" ); document.write( "n = 7\r
\n" ); document.write( "\n" ); document.write( "So the t test statistic is -3.57/(15.52/sqrt(7))=-0.608591\r
\n" ); document.write( "\n" ); document.write( "Ho = \"mu\"d = 0
\n" ); document.write( "H1 = \"mu\"d is not = 0\r
\n" ); document.write( "\n" ); document.write( "test statistic is less than the critical values so we cannot reject the null, thus there is no evidence of a difference between the two methods
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