Question

If a confidence interval is given from 45.82 up to 55.90 and the mean is known to be 50.86, what is the margin of error?

Answer 1

Answer:

$ME = \frac{10.08}{2}= 5.04$

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".

$\bar X=50.86$ represent the sample mean  

$\mu$ population mean

s represent the sample standard deviation  

n represent the sample size  

Solution to the problem

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

$\bar X \pm ME$   (1)

Or equivalently:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (2)

Where the margin of error is given by:

$ME= t_{\alpha/2}\frac{s}{\sqrt{n}}$

For this case we have the confidence interval limits given (45.82, 55.90)

We can find the width of the interval like this:

$Width =55.90-45.82= 10.08$

And now the margin of error would be the half of the width since we assume that the confidence interval is symmetrical.

$ME = \frac{10.08}{2}= 5.04$