Interesting probability question that I cannot crack...help?

Question

The president of a university would like to estimate the proportion of the student population who owns a personal computer. In a sample of 500 students, 417 own a personal computer. A 99% confidence interval for the proportion of student population who own a personal computer is from ________ to ________.

Answer 1

Sample data:
p = 417/500 = 0.834
n = 500

Critical z value for 99% CI = 2.576

CI = 0.834 +/- 2.576 * sqrt[ 0.834*(1-0.834) / 500]
= 0.834 +/- 0.043
= [0.791, 0.877]

Note to Lithium:
This question is about proportion not mean. The variance of the proportion is (pq/n).
And the z value with 0.005 tailing off to the right is 2.576.
Check out the sites below.

Answer 2

The standard deviation for a binomial distribution is classically given by the formula

σ = sqrt (pq/n)

where

σ is the standard deviation of the probability that any one student has a calculator,

n is the number of trials, which in this case is the number of students in the sample, or 500,

p is the actual probability that any one student owns a personal computer, and

q = 1 - p = the actual probability that any one student does not own a personal computer.

In this sample, 417 students own a personal computer and 83 students do not. From these, we can approximate p and q to be

p = 417/500 = 0.834
q = 83/500 = 0.166

so

σ = sqrt (0.834 * 0.166 / 500) = 0.0166

for the proportion of students in this sample who own calculators.

Because the binomial distribution with this many students in the sample and both this many statistical successes (students with computers) and failures (students without them) will obey the normal distribution sufficiently closely, we approximate the binomial distribution with a normal distribution of the same mean and standard deviation.

In a normal distribution, 99 percent of the population will lie within 2.576 standard deviations of the mean. (This number can be found from a table. Certain calculators can also give this number, as can spreadsheet programs with a suitable function.) As a result, the 99% confidence interval will lie within 2.576 standard deviations of the mean, which is within the interval

417/500 - 2.576*0.0166 < p < 417/500 + 2.576*0.0166

0.791 < p < 0.877

Answer 3

I would not describe this problem as "interesting". More like "tedious" and "routine". In any case, here's how to do it:

A 99% confidence interval means the range of values that you can be 99% sure contains the TRUE mean. The true mean is almost certainly not exactly 417/500 = 0.834, but it likely to be somewhere close to there. So we have to find how far away it's likely to be.


Our mean is 0.834, as we have measured it. The standard deviation is the mean divided by the square root of the sample size, or 0.834/sqrt(500) = 0.03730. That describes how uncertain we are about our number.

Now, if 99% of a normal distribution is covered in the middle, you would expect 0.5% of it to be tailing off on either side. 99% of a normal distribution falls in-between the z-values of
-2.80705
and
+2.80705
(I looked this up in a table).

So, what numbers of students correspond to these boundaries? When we know endpoints and want to find z-values, we take the endpoints, subtract the mean, then divide by the standard deviation. Here, we have the z-values, and we want the endpoints, so we do the reverse operations:

-2.80705 * 0.03730 + 0.834
= -0.1047 + 0.834
= 0.7293

2.80705 * 0.03730 + 0.834
= 0.1047 + 0.834
= 0.9387

So, we can be 99% certain that the true mean lies in the range [0.7293,0.9387], which is our 99% confidence interval.

Answer 4

I think you need to know the number of total students, cause if there are 501 students sounds like 80% is the right answer.