document.write( "Question 217066: 8.64 Biting an unpopped kernel of popcorn hurts! As an experiment, a self-confessed connoisseur of cheap popcorn carefully counted 773 kernels and put them in a popper. After popping, the unpopped kernels were counted. There were 86. \r
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document.write( "(a) Construct a 90 percent confidence interval for the proportion of all kernels that would not pop.
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Algebra.Com's Answer #163731 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! As an experiment, a self-confessed connoisseur of cheap popcorn carefully counted 773 kernels and put them in a popper. After popping, the unpopped kernels were counted. There were 86. \n" ); document.write( "(a) Construct a 90 percent confidence interval for the proportion of all kernels that would not pop. \n" ); document.write( "-------------------- \n" ); document.write( "sample proportion: 86/773 = 0.11 \n" ); document.write( "standard error: E= 1.645*sqrt[0.11*0.89/773]=0.021.. \n" ); document.write( "-------------------- \n" ); document.write( "90 CI: 0.11-0.021 < p < 0.11+0.021 \n" ); document.write( "90 CI: 0.089 < p < 0.131 \n" ); document.write( "-------------------- \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " \n" ); document.write( " |