Question

If I measure a set of times with a mean of 1.4 s, a standard deviation of 0.6 s, and an SDOM of 0.2 s, how should I report my measurement?

Answer 1

Answer:

The measurements should be reported as 1.4s ± 0.2s.

Step-by-step explanation:

The mean of a data set consisting of n independent values is:

$\bar X=\frac{1}{n}\sum (x_{1}+x_{2}+x_{3}...+x_{n})$

The standard deviation of a data set consisting of n independent values is the measure of the spread of the data. It is computed as follows:

$\sigma=\sqrt{\frac{1}{n}\sum(X_{i}-\bar X)^{2}}$

The standard deviation of mean (SDOM) is also known as the standard error.

The standard error is:

$\delta X=\frac{\sigma}{\sqrt {n}}$

If we want to report these measurements the format is: $\bar X\pm \delta X$

Given:

$\bar X=1.4s\\\delta =0.2s$

These measurements can be reported as:

$\bar X\pm \delta X=1.4s\pm 0.2s$

Thus, the measurements should be reported as 1.4s ± 0.2s.