Confidence Intervals for large sample

[ykho22]

Hello David,

There are questions regarding the formula on confidence intervals for large samples. In reviewing your notes (Section 1. page 68/82), you have included an additional factor [sqrt( (N-n) / (N-1)) for sampling with replacement or infinite population. I also noticed that in another section "Sampling distribution of means" (Section 1 page 61/82), you included the same term for the reason of finite population. I am a bit confuse and could you confirm on the formula?

Thanks,
Peach

 

[David Harper CFA FRM]

Peach,

The second instance of the formula (p. 68) is incorrectly labeled. My apologies. You are quite correct: the formulas are the same. (i will ensure it is corrected on the summary formula sheets that will be posted soon).

The key formula (i.e., most likely to be tested) is the STANDARD ERROR (standard deviation) of the SAMPLE MEAN = sample/population standard deviation divided by the square root of the sample size (n). Note, this is exactly the same thing as: variance of the sample (sampling distribution of) mean equals sample/population variance divided by sample size (n).

So, on p. 68, the same standard error is plugged into confidence intervals for purpose of hypothesis testing. And the typical (i.e., more likely to be tested) approach is to use, for the standard deviation: the sample/population standard deviation divided by the square root of the sample size. So p. 68 is incorrectly labeled because this refers in both cases to "random sampling from infinite population or finite population with replacement." That is, without the (N-n)/(N-1) modifier, and this would typically be assumed unless stated otherwise.

Under both cases, if the random sampling is without replacement, the (N-n)/(N-1) modifier is used for the variance (and the square root of same for standard deviation). The reason is that "without replacement," as n approaches N, the sample mean must converge toward the true (population) mean such that the standard error of the sample mean can be tends toward zero with the 'increasing accuracy' of the sample mean.