document.write( "Question 1036783: In a test of the effectiveness of garlic for lowering​ cholesterol, 45 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes in their levels of LDL cholesterol​ (in mg/dL) have a mean of 3.4 and a standard deviation of 18.3.\r
\n" ); document.write( "\n" ); document.write( " Construct a 95​% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL​ cholesterol?
\n" ); document.write( "What is the confidence interval estimate of the population mean μ​?\r
\n" ); document.write( "\n" ); document.write( "Thank You for your help.
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Algebra.Com's Answer #651517 by Boreal(15235)\"\" \"About 
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This is a paired t-test
\n" ); document.write( "Ho:mean =0
\n" ); document.write( "Ha:mean ne0.
\n" ); document.write( "alpha=0.05
\n" ); document.write( "test stat is t, df=44 critical value abs(t)>2.01
\n" ); document.write( "t is mean over SE
\n" ); document.write( "calculation: SE is 18.3/sqrt (45)=2.73; mean/SE=1.25
\n" ); document.write( "Fail to reject Ho
\n" ); document.write( "CI is mean +/- t*SE=3.4 +/-2.01*2.73 and the product is 5.49.
\n" ); document.write( "The interval is (-2.09,8.89). The interval contains 0, so 0, or no change, is a plausible value, and there is no convincing evidence that garlic affects LDL cholesterol levels.
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