Using the z-distribution, the 90% confidence interval for the difference of proportions is given by: (-0.0534, 0.0434).
<h3>What is the mean and the standard error for the distribution of differences?</h3>
For each sample, the mean and the standard error are given as follows:
.
.
For the distribution of differences, they are given by:
.
<h3>What is the confidence interval?</h3>
The interval is given by:
![\overline{p} \pm zs](https://tex.z-dn.net/?f=%5Coverline%7Bp%7D%20%5Cpm%20zs)
In this problem, we have a 90% confidence level, hence
, z is the value of Z that has a p-value of
, so the critical value is z = 1.645.
Hence the bounds of the interval are:
![\overline{p} - zs = -0.005 - 1.645(0.0294) = -0.0534](https://tex.z-dn.net/?f=%5Coverline%7Bp%7D%20-%20zs%20%3D%20-0.005%20-%201.645%280.0294%29%20%3D%20-0.0534)
![\overline{p} + zs = -0.005 + 1.645(0.0294) = 0.0434](https://tex.z-dn.net/?f=%5Coverline%7Bp%7D%20%2B%20zs%20%3D%20-0.005%20%2B%201.645%280.0294%29%20%3D%200.0434)
More can be learned about the z-distribution at brainly.com/question/25890103
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