Question

A recent study found that 2.56% of college graduates were unemployed and 5.35% of high school graduates (with no college degree) were unemployed. The study surveyed 586 college graduates and 467 high school graduates. Construct a 95% confidence interval for the difference of the proportion of unemployed college graduates minus the proportion of unemployed high school graduates.

Answer 1

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Answer 2

Using the z-distribution, as we are working with a proportion, it is found that the 95% confidence interval for the difference of the proportions is (-0.052, -0.0038).

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

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

$p_C = 0.0256, s_C = \sqrt{\frac{0.0256(0.9744)}{586}} = 0.0065$

$p_H = 0.0535, s_H = \sqrt{\frac{0.0535(0.9465)}{467}} = 0.0104$

Hence, for the distribution of differences, the mean and the standard error are given by:

$p = p_C - p_H = 0.0256 - 0.0535 = -0.0279$

$s = \sqrt{s_C^2 + s_H^2} = \sqrt{0.0065^2 + 0.0104^2} = 0.0123$

What is the confidence interval?

It is given by:

$p \pm zs$

In which z is the critical value.

95% confidence level, hence$\alpha = 0.95$, z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$, so $z = 1.96$.

Then, the bounds of the interval are given as follows:

$p - zs = -0.0279 - 1.96(0.0123) = -0.052$

$p + zs = -0.0279 + 1.96(0.0123) = -0.0038$

The 95% confidence interval for the difference of the proportions is (-0.052, -0.0038).

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