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

The formula for a 95% confidence interval for #mu# is

#barx+-[t_(alpha//2, n-1)xxs/sqrtn]#

where

• #bar x# is your sample mean, the middle point of the confidence interval,
• #t_(alpha//2, n-1)# is a stretching factor that tells us how many 'standard errors' wide our interval needs to be,
• #s# is the standard deviation of the sample data points, and
• #n# is the number of data points, also called the sample size.

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

The term #s/sqrtn# is known as the standard error for the estimate of #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#-value in a table (or use computer software).

#color(white)= barx+-[t_(alpha//2, n-1)xxs/sqrtn]#
#=2.75+-[t_(0.05//2, 58-1)xx0.86/sqrt58]#
#=2.75+-[t_(0.025, 57)xx0.1129]#
#=2.75+-[2.002465xx0.1129]#
#=2.75+-0.2261#

#=(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 #(mu)# our sample mean #(barx)# is. Each additional data point affects the spread of the data as a whole less and less, allowing #barx# to stabilize. In math terms: as #n# increases, #s# gets more precise, and #s/sqrtn# gets closer to 0.

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

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

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