Z test

CFA Level I Quant
#1

Hello everyone,

About statistics there is something I dont get. The Z test is usually: (X-u)/ Standard Deviation. However, page 528 in book 1 it says: Z= (X-u)/(Standard deviation/ square root of n)

where n is the sample size. Why do we devide the Standard Deviation by square root of N in the second equation? is this because it’s a sample (not the entiere population)?

Why is Z= (X-u)/(Standard deviation/ square root of n) ?

thx a lot for your help

#2

In Elan’s notes, the first formula refers to the z-score, the second to z-stat. Z-score has standard deviation as the denominator while z-stat has standard error of the sample mean as the denominator. Now, standard error is the standard deviation of the distribution of the sample mean. Are they different? When do we use one over the other? Just some follow-up questions.

#3

Thx for clarifying.

#4

You’re welcome. But I have follow up questions. Is z-score conceptually different from z-stat? Any difference in their usage/application? Anyone please? Thanks

#5

I have no idea. I’m waiting for help from someone too :slight_smile:

#6

There is a difference, but it’s very subtle.

Z Score - the one divided by just the standard deviation is used when the X that you’re converting for the z-test is the population mean.

Z Statistic - the one divided by the standard error is used when the X that you’re converting for the z-test is the sample mean.

(http://en.wikipedia.org/wiki/Standard_score#Standardizing_in_mathematical_statistics)

#7

Hello,

Thanks for your answer.

#8

Thanks for your inputs. But I need further clarifications about this because per Elan’s notes, the z-stat is used to conduct hypothesis test of the population mean when:

  1. the population is normally distributed and and its population variance is known – same numerator with the denominator being population standard deviation over the square root of n.

  2. the population variance is unknown but the sample size is large – same numerator with the denominator being sample standard deviation over the square root of n.

Both denominators are actually the formula for the standard error of the sample mean using population standard deviation when it is known or the sample standard deviation when it is not known.

Based on the above statements, z-stat is used to test population mean. When do we use z-score then?

#9

I mean z-score is calculated to standardize an observation from a normal distribution but is it in anyway related to z-stat as they have almost the same formula or as in what LePetitCaporal said, there might have been some problems where z-score formula was used over z-stat’s or vice-versa?

#10

Z score is to standardise an observation (which comes from a normal distribution).

Z statistic is to standardise the sample mean of a set of observations (which comes from a normal distribution).

An observation and the sample mean of a set of observations are standardised slightly differently.

It’s on the wiki.

#11

Which formula you use depends on the quantity for which you’re trying to create a confidence interval:

If you’re creating a confidence interval for _ a single observation _ (e.g., next month’s return), then you use standard deviation (σ) in the denominator, and you get a wider interval.

Ir you’re creating a conficence interval for _ the mean of a bunch of observations _ (e.g., the population mean), then you use standard error (σ/√n) in the denominator.

#12

Oh really. Ok thanks a lot!

#13

Really. It’s that simple.

When they give you the problem, your first step should be to ask yourself, “Is this a confidence interval for a single observation, or for the mean of a bunch of observations?”

My pleasure.

#14

Thank you all! It’s clear with me now.

#15

My pleasure.

Cool!