document.write( "Question 1187196: A report in a research journal states that the average weight loss of people on a certain drug is 35 lbs with a margin of error of ±3 lbs with confidence level CL = 95%.\r
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document.write( "(a) According to this information, the mean weight loss of people on this drug, 𝜇, could be as low as ___ lbs.\r
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document.write( "(b) If the study is repeated, how large should the sample size be so that the margin of error would be less than 1.5 lbs? (Assume 𝑆= 8 lbs.)
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document.write( "ANSWER:
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Algebra.Com's Answer #819198 by Boreal(15235) You can put this solution on YOUR website! 32 pounds, the mean- the margin of error\r \n" ); document.write( "\n" ); document.write( "margin of error is t*s/sqrt(n)=3 pounds. \n" ); document.write( "check to see if what the original sample size was with s=8 \n" ); document.write( "t (df unknown, 0.975) is about 2 for unknown sample size*8/sqrt(n) and has to equal 3 \n" ); document.write( "16/sqrt(n)=3; sqrt n=(16/3) n=256/9 or about 28.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "error=t*s/sqrt(n); \n" ); document.write( "1.5 pounds=2*8/sqrt(n), guessing the value of t \n" ); document.write( "square both sides \n" ); document.write( "2.25=256/(n) \n" ); document.write( "n=256/2.25, or 114 rounded up. \n" ); document.write( "that is the answer given the t with df=113 is very close to 2.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |