Question

The following confidence interval is obtained for a population proportion, p: 0.408 < p < 0.432

Use these confidence interval limits to find the margin of error, E.

A. 0.012
B. 0.013
C. 0.024
D. 0.420

Answer 1

Answer:

For this case the wisth of the interval represent 2ME and we have this:

$0.432-0.408 = 2ME$

And if we solve for ME we got:

$ME = \frac{0.432-0.408}{2}=0.012$

So then the correct answer would be:

A. 0.012

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

The confidence interval for the mean is given by the following formula:  

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

The margin of error is given by:

$ME= z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

For this case the wisth of the interval represent 2ME and we have this:

$0.432-0.408 = 2ME$

And if we solve for ME we got:

$ME = \frac{0.432-0.408}{2}=0.012$

So then the correct answer would be:

A. 0.012