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# Example constructing and interpreting a confidence interval for p

Check conditions, calculate, and interpret a confidence interval to estimate a population proportion.

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• When calculating the standard error, is it better to calculate the unbiased standard deviation of the sample and divide that by the positive root of the sample size,
here sqrt( (30*(0.4)^2 + 20*(0.6)^2) / 49 / 50) = 0.0699854...,
or use the formula of the standard error of the sample distribution, using p hat as an estimate of p ?
here sqrt(0.4 * 0.6 / 50) = 0.0692820...

The two don't give the same result. Sal uses both methods in different videos without saying why, so some extra explanation would be helpful!
• I'm not an expert, but I think Sal used the incorrect formula in this video.

One thing I did notice is that 30 * 0.4^2 + 20 * 0.6^2 / 50 = 0.4 * 0.6. So while the formulas look very different, if you replace the 49 in the first equation with a 50 they will produce the same answer
(1 vote)
• How do we decide what method to use to estimate the Standard Error.

Method 1) Perform correction. So standard error = sqrt(p hat * (1- p hat) / (n-1) )

Method 2) Do not perform correction. Standard error = sqrt(p hat * (1- p hat) / n )
• It's worth noting that the 'correction' here is incorrect. The actual correction would be to first find the sample standard deviation:

S = SQRT(n * p-hat * (1- p-hat)/(n-1))

then the unbiased Standard Deviation of the Sample Distribution of p-hat is:

Std deviation of p-hat = S/SQRT(n).

As others below point out, whenever population parameters are unknown, it's probably best to use the correction method to avoid bias. I'll happily defer to any experts who can give a more valid explanation.
(1 vote)
• what about using ((std. dev. of sample parameter)/sqrt(n)) instead of standard error?
• what is the difference between s/sqrt(n) vs sqrt(p*(1-p)/n)? I believe this is the same formula but not too clear why this is the case and when to used each.
• Why Does it matter if the equation is independent or not?
• If our sampled trials are not independent then that means each successive trial will not necessarily be equivalent. Because of this, our inferences could be skewed to the right/left because our "supposed" probability will be overestimate/underestimating the real value.

Hope this helps,
- Convenient Colleague
(1 vote)
• What ment was why did Sal make it 99.5% with the 0.50% above and not below the middle 99%?
(1 vote)
• Many Z-tables show the area under the curve from -inf up to a point, so if we want to have 99% confidence, it means we want to have 0.5% area left at the left and right side. See also
• When we find z*, why can't we just find that by multiplying the standard deviation by 3? (99 percent is 3 standard deviations in a normal distribution.) In this case, couldn't we just multiply the standard error by 3? I did this, but I didn't get the right answer.
(1 vote)
• Since we have to use an estimate of the population standard deviation, rather than the actual population standard deviation, shouldn't we be using the t-statistic rather than the z-statistic?
(1 vote)
• Can somone demonstrate a problem for me?
(1 vote)
• Why didn't he use the value of .9901 from the table if he wants 99% confidence? Why did he choose what he did from the table?
(1 vote)
• Z scores give the area below a point on a curve, so if we want the critical z score for 99 percent confidence, we want the z score that gives the area under that curve, and includes that 99 percent, which, if normal distributions weren't symmetrical, would be, the z score for 0.99, but since normal distributions are symmetrical, we have the piece that is not included in the 99 percent split between both sides, which means that we would need to subtract the other side because, z scores give area, under a point.