Question

Suppose the number of inches of rainfall each year in a city is normally distributed. For a random sample of years, the confidence interval (3.9,7.7) is generated. Find the margin of error

Answer 1

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

`\bar X \pm z_(\alpha/2)(\sigma)/(√(n))` (1)

Or equivalently:

`\bar X \pm ME`

For this case we have the interval given (3.9, 7.7) and we want to find the margin of error. Using the property of symmetry for a confidence interval we can estimate the margin of error with this formula:

`ME= (Upper -Lower)/(2)= (7.7-3.9)/(2)= 1.9`

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` represent the sample mean for the sample

`\mu` population mean (variable of interest)

Solution to the problem

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

`\bar X \pm z_(\alpha/2)(\sigma)/(√(n))` (1)

Or equivalently:

`\bar X \pm ME`

For this case we have the interval given (3.9, 7.7) and we want to find the margin of error. Using the property of symmetry for a confidence interval we can estimate the margin of error with this formula:

`ME= (Upper -Lower)/(2)= (7.7-3.9)/(2)= 1.9`