Question 458332
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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