Using the z-distribution, as we are working with a proportion, the 99% confidence interval for the difference in the proportions is:
[tex](0.33 - 0.22) \pm 2.576\sqrt{\left(\sqrt{\frac{0.33(0.67)}{250}}\right)^2 + \left(\sqrt{\frac{0.22(0.78)}{250}}\right)^2}[/tex]
What is the mean and the standard error of the distribution of differences?
For each sample, we have that:
[tex]p_T = 0.33, s_T = \sqrt{\frac{0.33(0.67)}{250}}[/tex]
[tex]p_E = 0.22, s_E = \sqrt{\frac{0.22(0.78)}{250}}[/tex]
For the distribution of differences, the mean is the subtraction of the means, hence:
[tex]p = p_T - p_E = 0.33 - 0.22[/tex]
The standard error is the square root of the sum of the variances, hence:
[tex]s = \sqrt{s_T^2 + s_E^2} = \sqrt{\left(\sqrt{\frac{0.33(0.67)}{250}}\right)^2 + \left(\sqrt{\frac{0.22(0.78)}{250}}\right)^2}[/tex]
What is the confidence interval?
It is given by:
[tex]p \pm zs[/tex]
99% confidence level, hence[tex]\alpha = 0.99[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.99}{2} = 0.995[/tex], so the critical value is z = 2.576.
Then, the interval is:
[tex](0.33 - 0.22) \pm 2.576\sqrt{\left(\sqrt{\frac{0.33(0.67)}{250}}\right)^2 + \left(\sqrt{\frac{0.22(0.78)}{250}}\right)^2}[/tex]
To learn more about the z-distribution, you can check https://brainly.com/question/25890103
