Question

Suppose you know the length of a confidence interval of a population mean is 8.4 and the sample mean (x bar) is 10. Find the ​margin of error​, and the confidence interval

Answer 1

`ME= z_(\alpha/2)(\sigma)/(√(n))`

The lenght of the interval correspond to:

`8.4 = 2ME`

`ME= (8.4)/(2)= 4.2`

And since we know the margin of error we can find the limits for the confidence interval:

`Lower = 10 -4.2=5.8`

`Upper = 10 +4.2=14.2`

Explanation:

Previous concepts

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".

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".

`\bar X=10` represent the sample mean

`\mu` population mean (variable of interest)

`\sigma` represent the population standard deviation

n represent the sample size

Solution to the problem

Assuming the X follows a normal distribution

`X \sim N(\mu, \sigma)`

The sample mean
`\bar X` is distributed on this way:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

The confidence interval on this case is given by:

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

The margin of error is given by:

`ME= z_(\alpha/2)(\sigma)/(√(n))`

The lenght of the interval correspond to:

`8.4 = 2ME`

`ME= (8.4)/(2)= 4.2`

And since we know the margin of error we can find the limits for the confidence interval:

`Lower = 10 -4.2=5.8`

`Upper = 10 +4.2=14.2`