Suppose you know σ and you want an 85 percent confidence level. What value would you use to multiply the standard error of the mean by?

1 Answer
Apr 29, 2018

If the sample size is #<= 40#, use #t_(alpha//2," " n–1)#.
If the sample size is #>40#, use #z_(alpha//2)" "~~1.44#.

(#alpha = 100% - 85%=0.15#)

Explanation:

#t# and #z# as mentioned above are multipliers that determine how many standard errors wide our margin of error will be for a desired confidence level.

When estimating the population mean #mu# with a small sample size #(n<=40),# the statistic #(barX-mu)/sigma# is assumed to come from a #t# distribution with #n-1# degrees of freedom (d.f.). This is when we use #t_(alpha//2," " n–1)#.

The family of #t# distributions looks similar to the standard normal #z# distribution, but with heavier tails. As the sample size #n# increases, the #t_(n–1)# distribution approaches the #z# distribution.

When the sample size is sufficiently large (i.e. #>40#), the #t# distribution (with #n-1# d.f.) and the #z# distribution are approximately equivalent, so it is widely accepted to use the #z# distribution when #n>40#.

For an 85% C.I., we have #alpha = 0.15#. Assuming your sample size is large enough, you would multiply the standard error #sigma/sqrtn# for the mean #mu# by

#z_(alpha//2) = z_(0.15//2)=z_0.075~~1.44#

This value of 1.44 can be found by reverse-lookup in a #z#-table, or with Statistics software.