Question

Construct a confidence interval for p1-p2 at the given level of confidence.

Answer 1

Using the z-distribution, the 90% confidence interval for the difference of proportions is given by: (-0.0534, 0.0434).

What is the mean and the standard error for the distribution of differences?

For each sample, the mean and the standard error are given as follows:

• $p_1 = \frac{30}{235} = 0.1277, s_1 = \sqrt{\frac{0.1277(0.8723)}{235}} = 0.0218$.

• $p_2 = \frac{39}{294} = 0.1327, s_1 = \sqrt{\frac{0.1327(0.8673)}{294}} = 0.0198$.

For the distribution of differences, they are given by:

• $\overline{p} = p_1 - p_2 = 0.1277 - 0.1327 = -0.005$.

• $s = \sqrt{s_1^2 + s_2^2} = \sqrt{0.0218^2 + 0.0198^2} = 0.0294$

What is the confidence interval?

The interval is given by:

$\overline{p} \pm zs$

In this problem, we have a 90% confidence level, hence$\alpha = 0.9$, z is the value of Z that has a p-value of $\frac{1+0.9}{2} = 0.95$, so the critical value is z = 1.645.

Hence the bounds of the interval are:

$\overline{p} - zs = -0.005 - 1.645(0.0294) = -0.0534$

$\overline{p} + zs = -0.005 + 1.645(0.0294) = 0.0434$

More can be learned about the z-distribution at brainly.com/question/25890103

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