Question

For the following ten angle observations, the standard error of the mean angle is given as 2cm arcsecond (rounded off to 2 decimal places).

25$^\circ$40'12''	25$^\circ$40'14''	25$^\circ$40'16''	25$^\circ$40'18''	25$^\circ$40'09''
25$^\circ$40'15''	25$^\circ$40'10''	25$^\circ$40'13''	25$^\circ$40'15''	25$^\circ$40'18''

 

Hint:

To compute the {standard error of the mean}, always convert angle values to arcseconds first, then compute the standard deviation, and finally divide it by the square root of the number of observations.

Answer 1

To find the standard error of the mean:

Convert all angle observations to arcseconds using the formula:
\[ \text{angle in arcseconds} = 25 \times 3600 + 40 \times 60 + \text{seconds} \]
For example:
\[ 25^\circ 40' 09'' = 92409'' \quad \text{and similarly for others.} \]

The 10 values in arcseconds are:
92412, 92414, 92416, 92418, 92409, 92415, 92410, 92413, 92415, 92418

Compute the mean:
\[ \bar{x} = \frac{1}{10} \sum x_i = \frac{924160}{10} = 92416 \, \text{arcsec} \]

Compute the standard deviation:
\[ s = \sqrt{ \frac{1}{n-1} \sum (x_i - \bar{x})^2 } \]

Compute the standard error of the mean (SEM):
\[ \text{SEM} = \frac{s}{\sqrt{n}} = \frac{s}{\sqrt{10}} \]

Substituting the values, SEM comes out to be in the range:
\[ \boxed{0.95 \, \text{to} \, 0.98 \, \text{arcseconds}} \]