I have the following data set:

$3.262 \\ 3.345\\3.279\\3.207\\3.366\\3.311\\3.384$

I have to find the mean and the standard error. Here is what I have done so far:

Mean: $3.307571429$

Standard Error: $0.0237886$

I am supposed to round the standard error to two significant figures. That means my standard error will be: $0.024$

In lecture we were told that we have to adjust the mean value to fit the digits of the standard error. This is what I am having trouble with. Do I have to round the last digit of the mean? In other words, should my result be:

  1. $3.308 \pm0.028$


  1. $3.307 \pm 0.028$

Thanks for your help.


Not sure if I understand exactly what your asking... In the case of rounding to the "second figure" then .0237886 rounded to the second figure is .024...Now you want to round 3.307571429 to the same digits as the standard error (.024) which is in the thousandths place...Then, 3.30757 rounded to the thousandth place is 3.308

  • $\begingroup$ Hi. I am not sure if I have to round the 3rd/4th decimal place of the mean value. I know that rounding my standard error will give me 0,024. However, do I also have to round the mean 3.308 or do I not round the last digit and leave it at 3.307? $\endgroup$ – Miran Nov 24 '16 at 18:22
  • $\begingroup$ Oh I see. So you don't want to round both the mean and standard error to 2 significant figures, but you want to round the standard error 2 significant figures and then have the mean match the same amount of digits as the standard error. So currently .024 is rounded out to the thousandth place, so I believe you do the same to the mean. Yielding 3.308 $\endgroup$ – Brandon Nov 24 '16 at 18:30
  • 1
    $\begingroup$ Sorry for the late answer. Thank you for your help. I am going to use 3.308. $\endgroup$ – Miran Nov 24 '16 at 20:10

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