Respuesta :
Using the z-distribution, with a critical z-score of z = 1.645, the 90% confidence interval is (-16.4%, -9.66%).
What are the mean and the standard error for the distribution of the difference of proportions?
For each sample, the mean and the standard error are given as follows:
- [tex]p_y = \frac{461}{1214} = 0.3797, s_y = \sqrt{\frac{0.3797(0.6203)}{1214}} = 0.0139[/tex].
- [tex]p_o = \frac{560}{1098} = 0.51, s_o = \sqrt{\frac{0.51(0.49)}{1098}} = 0.0151[/tex].
Hence, for the distribution of differences, they are given as follows:
- [tex]p = p_y - p_0 = 0.3797 - 0.51 = -0.1303[/tex].
- [tex]s = \sqrt{s_y^2 + s_o^2} = \sqrt{0.0139^2 + 0.0151^2} = 0.0205[/tex]
What is the confidence interval?
The interval is:
[tex]p \pm zs[/tex]
In this problem, we have a 90% confidence level, hence[tex]\alpha = 0.9[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.9}{2} = 0.95[/tex], so the critical value is z = 1.645.
Then the bounds of the interval are:
- p - zs = -0.1303 - 1.645(0.0205) = -0.164.
- p + zs = -0.1303 + 1.645(0.0205) = -0.0966.
As a percentage, the 90% confidence interval is (-16.4%, -9.66%).
More can be learned about the z-distribution at https://brainly.com/question/25890103
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