Respuesta :
Using the z-distribution, as we are working with a proportion, it is found that the 99% confidence interval is (-0.049, 0.129).
What is the mean and the standard error for the distribution of differences?
For each sample, we have that:
[tex]p_N = \frac{176}{200} = 0.88, s_N = \sqrt{\frac{0.88(0.12)}{200}} = 0.023[/tex]
[tex]p_E = \frac{168}{200} = 0.84, s_E = \sqrt{\frac{0.84(0.16)}{200}} = 0.0259[/tex]
Hence, for the distribution of differences, the mean and the standard error are given as follows:
[tex]p = p_N - p_E = 0.88 - 0.84 = 0.04[/tex]
[tex]s = \sqrt{s_N^2 + s_E^2} = \sqrt{0.023^2 + 0.0259^2} = 0.0346[/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 = 2.575.
Then:
[tex]p - zs = 0.04 - 2.575(0.0346) = -0.049[/tex]
[tex]p + zs = 0.04 + 2.575(0.0346) = 0.129[/tex]
The 99% confidence interval is (-0.049, 0.129).
To learn more about the z-distribution, you can check https://brainly.com/question/25890103
