SOLUTION: 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

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.
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
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?
p = x/n = 86/773 = 0.1113
The z-value = 0.10 = 1.645
According to text on page 315, 90% = .10, 99% = .01 etc.
a) We use a Student's T test.
b) Page 374
c) The Very Quick Rule can be found on page 325 of the text. Test once solve (a) and (b)
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.

Answer by stanbon(75887) (Show Source): 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.
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
of all kernels that would not pop.
---
p-hat = 86/773 = 0.111255
ME = (1.645)sqrt[0.111*0.889/773] = 0.018605
90% CI: 0.111-0.0186 < p < 0.111+0.0186
90%CI: 0.0923 < p < 0.1296
---
You could run a 1-PropZInt on x = 86, n= 773 and get this answer.
==========================================================
(b) Check the normality assumption.
I'll leave that to you. Check your textbook.
==============================================
(c) Try the Very Quick Rule. Does it work well here? Why, or why not?
p+-(1/sqrt(n) = 0.111+-(1/sqrt(773)) = 0.111+-0.035968
gives 90%CI: 0.111-0.036 < p < 0.111+0.036
0.075 < p < 0.147
Not good because p-hat = 0.111 is not near 0.5
==============================================
(d) Why might this sample not be typical?
p = x/n = 86/773 = 0.1113
The z-value = 0.10 = 1.645
According to text on page 315, 90% = .10, 99% = .01 etc.
a) We use a Student's T test.
b) Page 374
c) The Very Quick Rule can be found on page 325 of the text. Test once solve (a) and (b)
==============================================
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.
==============
Cheers,
Stan H.

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