Respuesta :
Answer:
The standard error is [tex]s = \sqrt{\frac{\pi_1(1-\pi_1)}{n_1}+\frac{\pi_2(1-\pi_2)}{n_2}}[/tex], in which [tex]p_1,p_2[/tex] are the proportions and [tex]n_1,n_2[/tex] 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 [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
The standard error is:
[tex]s = \sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
For the difference of proportions:
For proportion 1, the standard error is:
[tex]s_1 = \sqrt{\frac{\pi_1(1-\pi_1)}{n_1}}[/tex]
For proportion 2, the standard error is:
[tex]s_2 = \sqrt{\frac{\pi_2(1-\pi_2)}{n_2}}[/tex]
For the difference:
The standard error is the square root of the sum of the squares of each separate standard error. So
[tex]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}}[/tex]
The standard error is [tex]s = \sqrt{\frac{\pi_1(1-\pi_1)}{n_1}+\frac{\pi_2(1-\pi_2)}{n_2}}[/tex], in which [tex]p_1,p_2[/tex] are the proportions and [tex]n_1,n_2[/tex] are the sample sizes.
