# For a dataset of size 58, the sample mean was calculated to be barx=$2.75 and the sample standard deviation is s=$0.86. What is the 95% confidence interval for mu?

Sep 2, 2017

The 95% confidence interval for $\mu$ is ($2.5239," "$2.9761).

#### Explanation:

The formula for a 95% confidence interval for $\mu$ is

$\overline{x} \pm \left[{t}_{\alpha / 2 , n - 1} \times \frac{s}{\sqrt{n}}\right]$

where

• $\overline{x}$ is your sample mean, the middle point of the confidence interval,
• ${t}_{\alpha / 2 , n - 1}$ is a stretching factor that tells us how many 'standard errors' wide our interval needs to be,
• $s$ is the standard deviation of the sample data points, and
• $n$ is the number of data points, also called the sample size.

$\alpha$ is just 100% minus the confidence level you wish to have (in this case 95%). So alpha = 100%-95%, which is 5%, or $0.05$.

The term $\frac{s}{\sqrt{n}}$ is known as the standard error for the estimate of $\mu$. This is different than the standard deviation of the whole sample. (More on that later.)

To find the 95% confidence interval, we just need to plug in the given values for the variables, and look up the $t$-value in a table (or use computer software).

$\textcolor{w h i t e}{=} \overline{x} \pm \left[{t}_{\alpha / 2 , n - 1} \times \frac{s}{\sqrt{n}}\right]$
$= 2.75 \pm \left[{t}_{0.05 / 2 , 58 - 1} \times \frac{0.86}{\sqrt{58}}\right]$
$= 2.75 \pm \left[{t}_{0.025 , 57} \times 0.1129\right]$
$= 2.75 \pm \left[2.002465 \times 0.1129\right]$
$= 2.75 \pm 0.2261$

$= \left(2.5239 , \text{ } 2.9761\right)$

## Bonus:

Standard deviation measures the spread of all the data as a whole. Standard error measures how close to the actual population mean $\left(\mu\right)$ our sample mean $\left(\overline{x}\right)$ is. Each additional data point affects the spread of the data as a whole less and less, allowing $\overline{x}$ to stabilize. In math terms: as $n$ increases, $s$ gets more precise, and $\frac{s}{\sqrt{n}}$ gets closer to 0.

Also: for high enough values of $n$, the $t$-distribution can be approximated by a normal distribution, allowing for easier table lookup. That is,

for sufficiently large $n$ (greater than 30, usually), it is common to approximate ${t}_{\alpha / 2 , n - 1}$ as ${z}_{\alpha / 2}$.

It is much easier to look up ${z}_{\alpha / 2}$ than it is to look up ${t}_{\alpha / 2 , n - 1}$.