Question 1046979
As to Margin of Error and Confidence Intervals
CI = mean - ME <{{{ mu }}}< mean + ME
ME = {{{z(alpha/sqrt(n))}}} {{{alpha}}} being the population standard of deviation
Or
ME = {{{t(s/sqrt(n))}}} s being the standard deviation of the sampled data,
 t-value a corrected value by sample size:  (n-1) used
Regardless which of the above are used...knowing the Population Standard Deviation or not...
Smaller the n, the Larger the Margin of Error will be.
ME = {{{t(s/sqrt(10))}}} ME will be Larger than for ex. {{{t(s/sqrt(100))}}}