Question 885374
the formula for z score is (x-m) / se
x is the sample mean
m is the population mean
se is the standard error which is equal to psd / sqrt(ss)
psd = population standard deviation
ss = sample size.
if you replace se with psd / sqrt(ss) in the formula of z = (x-m) / se, then you get:
z = (x-m) / (psd / sqrt(ss)) which can be simplified to:
z = ((x-m) * (sqrt(ss)) / psd.
if you increase the difference between x and m then the z score becomes larger because (x-m) is in the numerator of the simplified equation.
if you increase psd, then the z score becomes smaller because psd is in the denominator of the simplified equation.
if you increase the sample size, then the z score becomes larger because sqrt(ss) is in the numerator of the simplified equation.
example:
assume x = 100 and m = 80 and ss = 100 and psd = 10
formula for z score is (x-m) / se
se = psd / sqrt(ss) = 10 / sqrt(100) = 10 / 10 = 1
you have:
x = 100
m = 80
se = 1
z = (x-m) / se becomes z = (100 - 80) / 1 = 20 / 1 = 20
now increase the spread by making x = 120
you get:
z = (x-m) / se becomes z = (120 - 80) / 1 = 40 / 1 = 40
z got bigger.


now go back to the original values of:
x = 100
m = 80
psd = 10
ss = 100
increase psd from 10 to 20.
you get se = psd / sqrt(ss) = 20 / 10 = 2
you now have:
x = 100 
m = 80
se = 2
z = (x-m) / se = 20/2 = 10
z got smaller because you increased the population standard deviation.


now go back to the original values of:
x = 100
m = 80
psd = 10
ss = 100
increase the sample size from 100 to 400.
you get se = psd / sqrt(ss) = 10 / 20 = .5
you now have:
x = 100
m - 80
se = .5
z = (x-m) / se = 20/.5 = 40
z got bigger because you increased the sample size from 100 to 400.

when you increase (x-m), z gets bigger.
when you increase psd, se gets bigger which makes z get smaller.
when you increase sample size, se gets smaller which makes z get bigger.


i have to run so don't have time to edit what i just wrote.
if you understand, then you're good.
if there's anything in here that confused you then send me an email and i'll provide a correction and or a better explanation.