document.write( "Question 1086483: A study of nickels showed that the standard deviation of the weight of nickels is 150 milligrams. A coin counter manufacturer wishes to find the 80% confidence interval for the average weight of a nickel. How many nickels does he need to weigh to obtain an average accurate to within 10 milligrams?
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document.write( "a. 225
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document.write( "b. 369
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document.write( "c. 613
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document.write( "d. 159
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Algebra.Com's Answer #700686 by jim_thompson5910(35256)![]() ![]() ![]() You can put this solution on YOUR website! We will use this formula \n" ); document.write( "n = ( (z*sigma)/E )^2\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The variables are \n" ); document.write( "n = sample size \n" ); document.write( "z = critical value \n" ); document.write( "sigma = population standard deviation \n" ); document.write( "E = margin of error \n" ); document.write( "---------------------------------------- \n" ); document.write( "Use a table or calculator to find that \n" ); document.write( "Critical value = z = 1.28 (at 80% confidence)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "In this case, we're given \n" ); document.write( "sigma = 150 \n" ); document.write( "E = 10\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Plug those values into the formula to get: \n" ); document.write( "n = ( (z*sigma)/E )^2 \n" ); document.write( "n = ( (1.28*150)/10 )^2 \n" ); document.write( "n = ( 192/10 )^2 \n" ); document.write( "n = ( 19.2 )^2 \n" ); document.write( "n = 368.64 \n" ); document.write( "n = 369 ... round up to the nearest whole number\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Final answer is choice B) 369 \n" ); document.write( " |