document.write( "Question 1081971: A pizza shop owner wishes to find the 95% confidence interval of the true mean cost of a large plain pizza. How large should the sample be if she wishes to be accurate to within $0.13? A previous study showed that the standard deviation of the price was $0.4. \n" ); document.write( "
Algebra.Com's Answer #696066 by Boreal(15235) You can put this solution on YOUR website! the margin of error should be +/-0.13, so 0.13=z*s/sqrt(n) \n" ); document.write( "1.96*s/ sqrt (n)=0.13 \n" ); document.write( "1.96*0.4=0.13* sqrt(n), after multiplying through \n" ); document.write( "divide through by 0.13 and square both sides \n" ); document.write( "n=36.37 which rounds up to 37. That is unlikely to be changed by using a t-value \n" ); document.write( " \n" ); document.write( " |