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If you're randomly sampling without replacement from a finite population, you're not in a binomial sampling situation, but a hypergeometric one.

When you're in a binomial situation with population proportion $\pi$, the variance of the count, $X$ is $n\pi(1-\pi)$ and so the variance of the sample proportion $p=X/n$ is $n\pi(1-\pi)/n^2=\pi(1-\pi)/n$. This variance of the proportion is then estimated as $p(1-p)/n$.

In the case of sampling $n$ without replacement from a finite population of size $N$, the count has variance $n{K\over N}{\frac{N-K}{N}}{N-n\over N-1}$.

Since $\pi=K/N$ is the population proportion, we might write that variance of the count, $X$ as $n\pi(1-\pi){N-n\over N-1}$.

So the variance of the sample proportion can be written as $\frac{\pi(1-\pi)}{n}\cdot f$ where $f={N-n\over N-1}$.

Since $f<1$, this variance is smaller than in the binomial case (as you suggested).

$f$ is referred to as "the finite population correction" (since you can use it to 'correct' the variance you get from the binomial), but as you see, it's simply the variance from using the correct (i.e. hypergeometric) probability model.

Of course, to correct the standard error rather than the variance, you must take the square root of that factor (i.e., $\sqrt{{N-n\over N-1}}$).

I read in a blog that the formula above should be used when the population is at least 10 times bigger than the sample

I'd say 'should be used' is far too strong. The binomial formula could be used - the finite population correction factor is always right, but if you leave it out when the sample is a small fraction of the population, the correction factor will be close to 1.

what happens if a survey is close to that 1/10 sample

Let's see what happens when it's exactly $1/10.$

$f=\frac{N-n}{N-1} = \frac{0.9N}{ N-1} \approx 0.9$

Hence the correction to the standard error is about $\sqrt{0.9}$ which is about $0.95.$ If you ignore it, your standard error will be about $5.4\%$ too large.

It's up to you to figure out if that's acceptable or not.

If you're randomly sampling without replacement from a finite population, you're not in a binomial sampling situation, but a hypergeometric one.

When you're in a binomial situation with population proportion $\pi$, the variance of the count, $X$ is $n\pi(1-\pi)$ and so the variance of the sample proportion $p=X/n$ is $n\pi(1-\pi)/n^2=\pi(1-\pi)/n$. This variance of the proportion is then estimated as $p(1-p)/n$.

In the case of sampling $n$ without replacement from a finite population of size $N$, the count has variance $n{K\over N}{\frac{N-K}{N}}{N-n\over N-1}$.

Since $\pi=K/N$ is the population proportion, we might write that variance of the count, $X$ as $n\pi(1-\pi){N-n\over N-1}$.

So the variance of the sample proportion can be written as $\frac{\pi(1-\pi)}{n}\cdot f$ where $f={N-n\over N-1}$.

Since $f<1$, this variance is smaller than in the binomial case (as you suggested).

$f$ is referred to as "the finite population correction" (since you can use it to 'correct' the variance you get from the binomial), but as you see, it's simply the variance from using the correct (i.e. hypergeometric) probability model.

Of course, to correct the standard error rather than the variance, you must take the square root of that factor (i.e., $\sqrt{{N-n\over N-1}}$).

I read in a blog that the formula above should be used when the population is at least 10 times bigger than the sample

I'd say 'should be used' is far too strong. While the binomial formula could be used, the finite population correction factor is always right - but when the sample is a small fraction of the population, the correction factor will be close to 1, so if you leave it out, little harm is done.

what happens if a survey is close to that 1/10 sample

Let's see what happens when the sample is one tenth of the population.

$f=\frac{N-n}{N-1} = \frac{0.9N}{ N-1} \approx 0.9$

Hence the correction to the standard error is about $\sqrt{0.9}$ which is about $0.95.$ If you ignore it, your standard error will be about $5.4\%$ too large.

It's up to you to figure out if that amount of inaccuracy in the standard error is acceptable or not.

If you're randomly sampling without replacement from a finite population, you're not in a binomial sampling situation, but a hypergeometric one.

When you're in a binomial situation with population proportion $\pi$, the variance of the count, $X$ is $n\pi(1-\pi)$ and so the variance of the sample proportion $p=X/n$ is $n\pi(1-\pi)/n^2=\pi(1-\pi)/n$. This variance of the proportion is then estimated as $p(1-p)/n$.

In the case of sampling $n$ without replacement from a finite population of size $N$, the count has variance $n{K\over N}{(N-K)\over N}{N-n\over N-1}$.

Since $\pi=K/N$ is the population proportion, we might write that variance of the count, $X$ as $n\pi(1-\pi){N-n\over N-1}$.

So the variance of the sample proportion can be written as $\frac{\pi(1-\pi)}{n}\cdot f$ where $f={N-n\over N-1}$.

Since $f<1$, this variance is smaller than in the binomial case (as you suggested).

$f$ is referred to as "the finite population correction" (since you can use it to 'correct' the variance you get from the binomial), but as you see, it's simply the variance from using the correct (i.e. hypergeometric) probability model.

Of course, to correct the standard error rather than the variance, you must take the square root of that factor (i.e. $\sqrt{{N-n\over N-1}}$)

I read in a blog that the formula above should be used when the population is at least 10 times bigger than the sample

I'd say 'should be used' is far too strong. The binomial formula could be used - the finite population correction factor is always right, but if you leave it out when the sample is a small fraction of the population, the correction factor will be close to 1.

what happens if a survey is close to that 1/10 sample

Let's see what happens when it's exactly 1/10.

$f=\frac{N-n}{N-1} = \frac{0.9N}{ N-1} \approx 0.9$

Hence the correction to the standard error is about $\sqrt 0.9$ which is about 0.95. If you ignore it, your standard error will be about 5.4% too large.

It's up to you to figure out if that's acceptable or not.

If you're randomly sampling without replacement from a finite population, you're not in a binomial sampling situation, but a hypergeometric one.

When you're in a binomial situation with population proportion $\pi$, the variance of the count, $X$ is $n\pi(1-\pi)$ and so the variance of the sample proportion $p=X/n$ is $n\pi(1-\pi)/n^2=\pi(1-\pi)/n$. This variance of the proportion is then estimated as $p(1-p)/n$.

In the case of sampling $n$ without replacement from a finite population of size $N$, the count has variance $n{K\over N}{\frac{N-K}{N}}{N-n\over N-1}$.

Since $\pi=K/N$ is the population proportion, we might write that variance of the count, $X$ as $n\pi(1-\pi){N-n\over N-1}$.

So the variance of the sample proportion can be written as $\frac{\pi(1-\pi)}{n}\cdot f$ where $f={N-n\over N-1}$.

Since $f<1$, this variance is smaller than in the binomial case (as you suggested).

$f$ is referred to as "the finite population correction" (since you can use it to 'correct' the variance you get from the binomial), but as you see, it's simply the variance from using the correct (i.e. hypergeometric) probability model.

Of course, to correct the standard error rather than the variance, you must take the square root of that factor (i.e., $\sqrt{{N-n\over N-1}}$).

I read in a blog that the formula above should be used when the population is at least 10 times bigger than the sample

I'd say 'should be used' is far too strong. The binomial formula could be used - the finite population correction factor is always right, but if you leave it out when the sample is a small fraction of the population, the correction factor will be close to 1.

what happens if a survey is close to that 1/10 sample

Let's see what happens when it's exactly $1/10.$

$f=\frac{N-n}{N-1} = \frac{0.9N}{ N-1} \approx 0.9$

Hence the correction to the standard error is about $\sqrt{0.9}$ which is about $0.95.$ If you ignore it, your standard error will be about $5.4\%$ too large.

It's up to you to figure out if that's acceptable or not.

If you're randomly sampling without replacement from a finite population, you're not in a binomial sampling situation, but a hypergeometric one.

When you're in a binomial situation with population proportion $\pi$, the variance of the count, $X$ is $n\pi(1-\pi)$ and so the variance of the sample proportion $p=X/n$ is $n\pi(1-\pi)/n^2=\pi(1-\pi)/n$. This variance of the proportion is then estimated as $p(1-p)/n$.

In the case of sampling $n$ without replacement from a finite population of size $N$, the count has variance $n{K\over N}{(N-K)\over N}{N-n\over N-1}$.

Since $\pi=K/N$ is the population proportion, we might write that variance of the count, $X$ as $n\pi(1-\pi){N-n\over N-1}$.

So the variance of the sample proportion can be written as $\frac{\pi(1-\pi)}{n}\cdot f$ where $f={N-n\over N-1}$.

Since $f<1$, this variance is smaller than in the binomial case (as you suggested).

$f$ is referred to as "the finite population correction" (since you can use it to 'correct' the variance you get from the binomial), but as you see, it's simply the variance from using the correct (i.e. hypergeometric) probability model.

Of course, to correct the standard error rather than the variance, you must take the square root of that factor (i.e. $\sqrt{{N-n\over N-1}}$)

If you're randomly sampling without replacement from a finite population, you're not in a binomial sampling situation, but a hypergeometric one.

When you're in a binomial situation with population proportion $\pi$, the variance of the count, $X$ is $n\pi(1-\pi)$ and so the variance of the sample proportion $p=X/n$ is $n\pi(1-\pi)/n^2=\pi(1-\pi)/n$. This variance of the proportion is then estimated as $p(1-p)/n$.

In the case of sampling $n$ without replacement from a finite population of size $N$, the count has variance $n{K\over N}{(N-K)\over N}{N-n\over N-1}$.

Since $\pi=K/N$ is the population proportion, we might write that variance of the count, $X$ as $n\pi(1-\pi){N-n\over N-1}$.

So the variance of the sample proportion can be written as $\frac{\pi(1-\pi)}{n}\cdot f$ where $f={N-n\over N-1}$.

Since $f<1$, this variance is smaller than in the binomial case (as you suggested).

$f$ is referred to as "the finite population correction" (since you can use it to 'correct' the variance you get from the binomial), but as you see, it's simply the variance from using the correct (i.e. hypergeometric) probability model.

Of course, to correct the standard error rather than the variance, you must take the square root of that factor (i.e. $\sqrt{{N-n\over N-1}}$)

I read in a blog that the formula above should be used when the population is at least 10 times bigger than the sample

I'd say 'should be used' is far too strong. The binomial formula could be used - the finite population correction factor is always right, but if you leave it out when the sample is a small fraction of the population, the correction factor will be close to 1.

what happens if a survey is close to that 1/10 sample

Let's see what happens when it's exactly 1/10.

$f=\frac{N-n}{N-1} = \frac{0.9N}{ N-1} \approx 0.9$

Hence the correction to the standard error is about $\sqrt 0.9$ which is about 0.95. If you ignore it, your standard error will be about 5.4% too large.

It's up to you to figure out if that's acceptable or not.