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

`ME = (10.08)/(2)= 5.04`

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)(s)/(√(n))` (2)

Where the margin of error is given by:

`ME=  t_(\alpha/2)(s)/(√(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 = (10.08)/(2)= 5.04`