document.write( "Question 978189: The percent defective for parts produced by a manufacturing process is targeted at 4%. The process is monitored daily by taking samples of sizes n = 160 units. Suppose that today’s sample contains 14 defectives. how many units would have to be sampled to be 95% confident that you can estimate the fraction of defective parts within 2% (using the information from todays sample-- that is using the results that p=.00875)?\r
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Algebra.Com's Answer #599685 by Boreal(15235) $\"\"$ $\"About$

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I am not clear about today's sample. It did not have p=0.00875 of defective parts if 14/160. I will use 0.0875, which would be consistent
\n" ); document.write( "if p=0.0875
\n" ); document.write( "1-p=0.9125
\n" ); document.write( "n is unknown
\n" ); document.write( "95% confident
\n" ); document.write( "need values of defective between 0.0675 and 0.1075\r
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\n" ); document.write( "\n" ); document.write( "z*SE=0.02 on each side, which has a probability of 0.0875
\n" ); document.write( "SE= sqrt{(0.0875)(0.9125)/n}; if the SE is 0.0102, the product will be 0.02\r
\n" ); document.write( "\n" ); document.write( "Therefore, 0.0102*=sqrt (0.0798)/sqrt(n)
\n" ); document.write( "square both sides
\n" ); document.write( "0.000104=0.0798/n
\n" ); document.write( "n=0.0798/0.000104=768 rounding up
\n" ); document.write( "if we sample 768, the SE ={ (0.0875)*(.9125)/768}^(1/2)=0.101
\n" ); document.write( "SE*1.96 (z.975)=0.02\r
\n" ); document.write( "\n" ); document.write( "The answer is 768. \n" ); document.write( "

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