A random sample of 90 observations produced a mean x̄ = 25.9 and a standard deviation s = 2.7. How do you find a 95% confidence interval for the population mean μ?

The formula for a `100(1-\alpha)%` confidence interval for `\mu` is

`bar x+-(t_(\alpha//2," "n"-1") xx s/sqrtn)`

where

• `bar x` is our sample mean,
• `t_(\alpha//2," "n"-1")` is the point on the `t`-distribution (with `n-1` degrees of freedom) with `100(\alpha/2)%` of the distribution's area to its right,
• `s` is the sample standard deviation, and
• `n` is the sample size.

Note: this formula assumes the population size `N` is unknown (or at least sufficiently large relative to `n`).

For a 95% confidence interval, `\alpha=0.05`, because

`100(1-0.05)%`
`=100(0.95)%`
`=95%`.

To compute the confidence interval desired, we simply plug in our values (and, in the case of the `t_(\alpha//2)` value, look it up) and simplify:

`color(white)=bar x+-(t_(\alpha//2," "n"-1") xx s/sqrtn)`
`=25.9+-(t_(0.025,89) xx 2.7/sqrt90)`
`=25.9+-(1.987 xx 0.2846)`
`=25.9+-(0.5655)`
`=(25.33, 26.47)`

While `s` itself represents the estimate for the population standard deviation, `s/sqrtn` represents the standard error of our estimate `bar x`—that is, it measures how far from `mu` our estimate `bar x` is likely to be. As our sample size `n` grows larger, our estimate `bar x` gets more precise, and so the standard error shrinks.

`t_(\alpha//2," "n"-1")` is like a scale factor which determines how many standard errors wide our margin of error will be. The more confident we wish to be about the interval including `mu`, the higher this `t`-value needs to be. That is, a smaller `\alpha` means a larger `t_(\alpha//2).`

To obtain different confidence intervals for `mu`, simply look up the different value necessary for `t_(\alpha//2," "n"-1")` and plug it into the formula, leaving all other values the same. I'll leave the calculation of the 90% C.I. and 99% C.I. as an exercise.