Question 458332: 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 of 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?
Answer by edjones(8007)   (Show Source):
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We know N, the sample size, from the problem: N = 773
From your problem's givens, we can get p and q:
p = 86/773 = 0.1113
q = 1 - p = 0.8887
From a z table, the value for the 90% interval is:
1.6449
Use the formula for the interval around a proportion:
p - z*sqrt(pq/N) to p + z*sqrt(pq/N)
0.1113 - 1.6449*sqrt(0.1113*0.8887/773) to 0.1113 + 1.6449*sqrt(0.1113*0.8887/773)
0.09269 to 0.12991
No, the very quick rule won't work here, since the p value is very small.
Yes, normality will hold, since Np and Nq are both large (86 and 687).
This sample might not be typical, for example, since the person is a "self-confessed connoisseur of cheap popcorn", he might be better at making the popcorn than other people, so he might have a lower number of unpopped kernals than a normal eater.
3 years ago
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