# 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?

## A 90% confidence interval for the population mean μ? A 99% confidence interval for the population mean μ?

Jan 29, 2017

The 95% confidence interval for $\mu$ is $\left(25.33 , 26.47\right)$.
The 90% confidence interval is $\left(25.48 , 26.37\right)$.
The 99% confidence interval is $\left(25.15 , 26.65\right)$.

#### Explanation:

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

$\overline{x} \pm \left({t}_{\setminus \alpha / 2 , \text{ "n"-1}} \times \frac{s}{\sqrt{n}}\right)$

where

• $\overline{x}$ is our sample mean,
• ${t}_{\setminus \alpha / 2 , \text{ "n"-1}}$ 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, $\setminus \alpha = 0.05$, 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 ${t}_{\setminus \alpha / 2}$ value, look it up) and simplify:

$\textcolor{w h i t e}{=} \overline{x} \pm \left({t}_{\setminus \alpha / 2 , \text{ "n"-1}} \times \frac{s}{\sqrt{n}}\right)$
$= 25.9 \pm \left({t}_{0.025 , 89} \times \frac{2.7}{\sqrt{90}}\right)$
$= 25.9 \pm \left(1.987 \times 0.2846\right)$
$= 25.9 \pm \left(0.5655\right)$
$= \left(25.33 , 26.47\right)$

While $s$ itself represents the estimate for the population standard deviation, $\frac{s}{\sqrt{n}}$ represents the standard error of our estimate $\overline{x}$—that is, it measures how far from $\mu$ our estimate $\overline{x}$ is likely to be. As our sample size $n$ grows larger, our estimate $\overline{x}$ gets more precise, and so the standard error shrinks.

${t}_{\setminus \alpha / 2 , \text{ "n"-1}}$ 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 $\setminus \alpha$ means a larger ${t}_{\setminus \alpha / 2} .$

To obtain different confidence intervals for $\mu$, simply look up the different value necessary for ${t}_{\setminus \alpha / 2 , \text{ "n"-1}}$ 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.