document.write( "Question 197259: Biting an unpopped kernel of corn 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.
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\n" ); document.write( "(a) Construct a 90 percent confidence interval for the proportion of all kernels that would not pop.
\n" ); document.write( "(b) Check the normality assumption.
\n" ); document.write( "(c) Try the Very Quick Rule. Does it work well here? Why, or why not?
\n" ); document.write( "(d) Why might this sample not be typical?
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Algebra.Com's Answer #147901 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
Biting an unpopped kernel of corn 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.
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\n" ); document.write( "sample proportion = 86/773 = 0.11\r
\n" ); document.write( "\n" ); document.write( "(a) Construct a 90 percent confidence interval for the proportion of all kernels that would not pop.
\n" ); document.write( "ME = 1.645*sqrt[0.11*0.89/773] = 0.0185
\n" ); document.write( "90% CI: 0.11 - 0.0185 < p < 0.11 + 0.0185
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\n" ); document.write( "(b) Check the normality assumption.
\n" ); document.write( "I'll leave that to you. Check your textbook.
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\n" ); document.write( "(c) Try the Very Quick Rule. Does it work well here? Why, or why not?
\n" ); document.write( "Not sure what that means.
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\n" ); document.write( "(d) Why might this sample not be typical?
\n" ); document.write( "It is only a sample of convenience sample; not a simple random sample.
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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