Question

g if we want to calculate a confidence interval of the difference of two proportions what is the standard error (do not pool for this answer)

Answer 1

Answer:

The standard error is $s = \sqrt{\frac{\pi_1(1-\pi_1)}{n_1}+\frac{\pi_2(1-\pi_2)}{n_2}}$, in which $p_1,p_2$ are the proportions and $n_1,n_2$ are the sample sizes.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

The standard error is:

$s = \sqrt{\frac{\pi(1-\pi)}{n}}$

For the difference of proportions:

For proportion 1, the standard error is:

$s_1 = \sqrt{\frac{\pi_1(1-\pi_1)}{n_1}}$

For proportion 2, the standard error is:

$s_2 = \sqrt{\frac{\pi_2(1-\pi_2)}{n_2}}$

For the difference:

The standard error is the square root of the sum of the squares of each separate standard error. So

$s = \sqrt{(\sqrt{\frac{\pi_1(1-\pi_1)}{n_1}})+(\sqrt{\frac{\pi_2(1-\pi_2)}{n_2}})^2} = \sqrt{\frac{\pi_1(1-\pi_1)}{n_1}+\frac{\pi_2(1-\pi_2)}{n_2}}$

The standard error is $s = \sqrt{\frac{\pi_1(1-\pi_1)}{n_1}+\frac{\pi_2(1-\pi_2)}{n_2}}$, in which $p_1,p_2$ are the proportions and $n_1,n_2$ are the sample sizes.