Question

How to find 95% margin of error from 95% confidence interval?

When people smoke, the nicotine they absorb is converted to cotinine, which can be measured. The cotinine levels in the random sample of 100 smokers were measured and a 95% confidence interval for the population mean and cottoning level was estimated to be (130.47, 210.53)
(a) Find the 95% margin of error
(b)Determine the sample mean
(c) Determine the sample standard deviation.

Answer 1

(a) `ME = 40.03`
(b) `barx = 170.5`
(c) `s = 204.23`

Explanation:

(a)

We typically associate the percentage 95% to just the confidence interval (and not the margin of error).

A confidence interval (C.I.) for a mean is mathematically defined to be

`barx +- ME`

where `barx` is the sample mean, and `ME` is the margin of error. Since each bound on the C.I. is one `ME` away from the sample mean, the whole C.I. is twice the size of the `ME`.

In other words,

`"upper CI bound " -  "lower CI bound"=2 * ME`

Solving for `ME` gives

`ME = ("upper CI bound " -  "lower CI bound")/2`

`color(white)(ME) = (210.53-130.47)/2`

`color(white)(ME) = 40.03`

(b)

The sample mean is halfway between the endpoints of the C.I. That is, it is the average of these endpoints:

`barx = ("upper CI bound " +  "lower CI bound")/2`

`color(white)(barx) = (201.53+130.47)/2`

`color(white)(barx) = 170.5`

(c)

To solve for sample standard deviation `s`, we use the mathematical definition for a margin of error for `hatmu`, which is

`ME = z_(alpha//2) * s/sqrtn`

provided the sample size is large enough (usually at least 30 or 40).

Solving this for `s` gives

`s = (ME * sqrt n)/z_(alpha//2)`

For a 95% C.I., we have `alpha = 0.05`, and `z_(alpha//2)=1.96` (from `z`-table lookup). We've already found `ME = 40.03` and we know `n=100`, so we get

`s = (40.03 * sqrt 100)/1.96`

`color(white)s = 204.23`