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

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

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Answer 1 · 2017-09-02T16:58:40

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

Explanation:

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

$¯ x \pm [t_{α / 2, n - 1} \times \frac{s}{\sqrt{n}}]$ $barx+-[t_(alpha//2, n-1)xxs/sqrtn]$

where

$¯ x$ $bar x$ is your sample mean, the middle point of the confidence interval,

$t_{α / 2, n - 1}$ $t_(alpha//2, n-1)$ is a stretching factor that tells us how many 'standard errors' wide our interval needs to be,

$s$ $s$ is the standard deviation of the sample data points, and

$n$ $n$ is the number of data points, also called the sample size.

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

The term $\frac{s}{\sqrt{n}}$ $s/sqrtn$ is known as the standard error for the estimate of $μ$ $mu$ . 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 $t$ $t$ -value in a table (or use computer software).

$= ¯ x \pm [t_{α / 2, n - 1} \times \frac{s}{\sqrt{n}}]$ $color(white)= barx+-[t_(alpha//2, n-1)xxs/sqrtn]$
$= 2.75 \pm [t_{0.05 / 2, 58 - 1} \times \frac{0.86}{\sqrt{58}}]$ $=2.75+-[t_(0.05//2, 58-1)xx0.86/sqrt58]$
$= 2.75 \pm [t_{0.025, 57} \times 0.1129]$ $=2.75+-[t_(0.025, 57)xx0.1129]$
$= 2.75 \pm [2.002465 \times 0.1129]$ $=2.75+-[2.002465xx0.1129]$
$= 2.75 \pm 0.2261$ $=2.75+-0.2261$

$= (2.5239, 2.9761)$ $=(2.5239," "2.9761)$

Bonus:

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

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

for sufficiently large $n$ $n$ (greater than 30, usually), it is common to approximate $t_{α / 2, n - 1}$ $t_(alpha//2, n-1)$ as $z_{α / 2}$ $z_(alpha//2)$ .

It is much easier to look up $z_{α / 2}$ $z_(alpha//2)$ than it is to look up $t_{α / 2, n - 1}$ $t_(alpha//2, n-1)$ .

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

1 Answer

Explanation:

Bonus:

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Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Bonus:

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Impact of this question

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