document.write( "Question 398352: Please help with part (a). I understand that I need to use a students t bar but my knowledge on this subject is very limited. Please disregard the text references, I am just trying to find starting points for solving this problem. \r
\n" ); document.write( "\n" ); document.write( "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. (a) Construct a 90 percent confidence interval for the proportion
\n" ); document.write( "of all kernels that would not pop. (b) Check the normality assumption. (c) Try the Very Quick Rule. Does it work well here? Why, or why not? (d) Why might this sample not be typical?\r
\n" ); document.write( "\n" ); document.write( "p = x/n = 86/773 = 0.1113
\n" ); document.write( "The z-value = 0.10 = 1.645
\n" ); document.write( "According to text on page 315, 90% = .10, 99% = .01 etc.\r
\n" ); document.write( "\n" ); document.write( "a) We use a Student's T test.\r
\n" ); document.write( "\n" ); document.write( "b) Page 374\r
\n" ); document.write( "\n" ); document.write( "c) The Very Quick Rule can be found on page 325 of the text. Test once solve (a) and (b)\r
\n" ); document.write( "\n" ); document.write( "d) This sample is not typical because the connoisseur only tested one bag of popcorn. Like in class with the M & M's, we would have to have a large sample size to make accurate predictions.
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Algebra.Com's Answer #282302 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
Please help with part (a). I understand that I need to use a students t bar but my knowledge on this subject is very limited. Please disregard the text references, I am just trying to find starting points for solving this problem.
\n" ); document.write( "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.
\n" ); document.write( "(a) Construct a 90 percent confidence interval for the proportion
\n" ); document.write( "of all kernels that would not pop.
\n" ); document.write( "---
\n" ); document.write( "p-hat = 86/773 = 0.111255
\n" ); document.write( "ME = (1.645)sqrt[0.111*0.889/773] = 0.018605
\n" ); document.write( "90% CI: 0.111-0.0186 < p < 0.111+0.0186
\n" ); document.write( "90%CI: 0.0923 < p < 0.1296
\n" ); document.write( "---
\n" ); document.write( "You could run a 1-PropZInt on x = 86, n= 773 and get this answer.
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\n" ); document.write( "\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( "\n" ); document.write( "(c) Try the Very Quick Rule. Does it work well here? Why, or why not?
\n" ); document.write( "p+-(1/sqrt(n) = 0.111+-(1/sqrt(773)) = 0.111+-0.035968
\n" ); document.write( "gives 90%CI: 0.111-0.036 < p < 0.111+0.036
\n" ); document.write( "0.075 < p < 0.147
\n" ); document.write( "Not good because p-hat = 0.111 is not near 0.5\r
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\n" ); document.write( "\n" ); document.write( "(d) Why might this sample not be typical?
\n" ); document.write( "p = x/n = 86/773 = 0.1113
\n" ); document.write( "The z-value = 0.10 = 1.645
\n" ); document.write( "According to text on page 315, 90% = .10, 99% = .01 etc.
\n" ); document.write( "a) We use a Student's T test.
\n" ); document.write( "b) Page 374
\n" ); document.write( "c) The Very Quick Rule can be found on page 325 of the text. Test once solve (a) and (b)
\n" ); document.write( "==============================================
\n" ); document.write( "d) This sample is not typical because the connoisseur only tested one bag of popcorn. Like in class with the M & M's, we would have to have a large sample size to make accurate predictions.
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H. \r
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