SOLUTION: Good evening! I'm a bit stuck on this problem. Construct the 99% confidence interval estimate of the population proportion p if the sample size is n=1000 and the number of su

Question 1191919: Good evening!
I'm a bit stuck on this problem.
Construct the 99% confidence interval estimate of the population proportion p if the sample size is n=1000 and the number of successes in the sample is x=213. Give your answers to 4 decimal places, and use at least 3 decimal places in your critical value.
Below I typed out my answer, but it is incorrect:

I have n=1000 and p(hat)=213 and thus q(hat)=787. The z-score for a 99% C is 2.576. Using the formula I have
calculated the C.I. to be
(179.479, 246.352)
I keep calculating the same number, I even tried to round up (180, 247).
Not sure what I am doing wrong. May I please have some assistance.
Thank you, I appreciate your help.
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

You have the correct z critical value.
At 99% confidence, it is roughly z = 2.576 when rounding to three decimal places.

Where you're going wrong is when you computed the phat and qhat values.
Those should be values between 0 and 1 since they are proportions (specifically sample proportions).

phat = (number of successes)/(sample size)
phat = x/n
phat = 213/1000
phat = 0.213
Since we're dividing by 1000, it's simply a matter of moving the decimal point 3 spaces to the left (when going from 213.0 to 0.213)

The corresponding qhat value is:
qhat = 1 - phat
qhat = 1 - 0.213
qhat = 0.787
It looks like you probably did 1000-213 = 787 when you got qhat = 787 incorrectly.

The margin of error (E) is
E = z*sqrt(phat*qhat/n)
E = 2.576*sqrt(0.213*0.787/1000)
E = 0.033352
which is approximate

The lower bound (L) is
L = phat - E
L = 0.213 - 0.033352
L = 0.179648
L = 0.1796

And the upper bound (U) is
U = phat + E
U = 0.213 + 0.033352
U = 0.246352
U = 0.2464

The 99% confidence interval for the population proportion p is therefore approximately (0.1796, 0.2464) when rounding to four decimal places and when using 3 decimal places in the critical z value.

Keep in mind that the confidence interval is estimating where the population proportion (p) is likely located.
As such, the interval is most useful when the L and U values are between 0 and 1.
It's not possible for instance to have p between 179.479 and 246.352 since 0 < p < 1 should be the case.
When in doubt, try to go back to what the confidence interval is trying to measure.

side note: I appreciate you showing your work and thought process. It's a bit rare to have a student do this rather than simply dump their homework.

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