STANDARD DEVIATION OF Z test?

Dason

Ambassador to the humans
#3
Just to stay ahead of it. Under standard assumptions under the null typically anything specified as a z-test should have a standard normal distribution as the distribution of the test statistic which means... Standard deviation of 1.

The standard deviation shouldn't depend on n.

My guess is that you're treating n as the group size as opposed to the total number of samples so really it should be sqrt(2) that you're seeing values close to and not just 1.5.
 

Dason

Ambassador to the humans
#5
No. All of them. Sample sizes, parameters, assumptions made in the hypothesis.

If you want to be lazy then my last post probably is good enough to show where you probably went wrong. But if you want to understand please lay everything out.
 
#6
Just to stay ahead of it. Under standard assumptions under the null typically anything specified as a z-test should have a standard normal distribution as the distribution of the test statistic which means... Standard deviation of 1.

I defined Z test above.

The standard deviation shouldn't depend on n.

It does.

My guess is that you're treating n as the group size as opposed to the total number of samples so really it should be sqrt(2) that you're seeing values close to and not just 1.5.
n is sample size of x bar 1 and 2.
 

Dason

Ambassador to the humans
#7
My guess is that you're treating n as the group size as opposed to the total number of samples so really it should be sqrt(2) that you're seeing values close to and not just 1.5.
So yeah you need to divide by sqrt(2*n) in the denominator instead.

Then it will be a proper test where the test statistic had a standard normal distribution. So a standard deviation of 1 regardless of sample size.
 
#8
No. All of them. Sample sizes, parameters, assumptions made in the hypothesis.

If you want to be lazy then my last post probably is good enough to show where you probably went wrong. But if you want to understand lay everything out.
It's easy. No hypothesis.
ya got one Normal distribution.
sample size n
Z test = (x̄ 1 – x̄ 2) / (SIGMA / √ n)
stdev Z test?
 

Dason

Ambassador to the humans
#9
Read my last post. And apparently it isn't just that easy because you're doing it wrong. I asked you for the details because you're getting them wrong. But my last post should have clarified what you should be doing
 
#10
Lad, if you want to answer your question, do it. If you can answer my question, do it. If you don't understand my question, ask. If you want to demonstrate your vastly superior expertise, please let someone who can, answer.
 

Dason

Ambassador to the humans
#11
Lad, if you want to answer your question, do it. If you can answer my question, do it. If you don't understand my question, ask. If you want to demonstrate your vastly superior expertise, please let someone who can, answer.
And lad you're the one on thin ice. I'm trying to help. You're being obnoxious in your unwillingness to answer questions properly. I don't ask questions for the hell of it. Maybe - just maybe - the person who doesn't understand the z-test might not fully understand that the questions asked are educational and are to establish a common base for discussion because believe it or not sometimes the assumptions aren't the same or the parameters represent something else and it's good to start on the same page.

I'm sorry for the rant but dammit I'm literally trying to help and you're being a jerk in response.
 

Dason

Ambassador to the humans
#12
I already answered your question multiple times. What don't you get about needing to divide by sqrt(2*n) instead of sqrt(n).

If you're doing the typical z-test and you're using the formula you put in your original post then that only works if you're treating the 'n' in that as the total sample size. Since you're treating it as the group size and I'm assuming treating the group sizes to be equal them you need to double your n. I already gave a quick fix for that multiple times.

I don't understand how you can write giant posts about the t-test but be so unwilling to specify assumptions or generally discuss this simple problem.
 
#13
Z tests are for any Normally distributed variable, where z test = v1-v2/std error, stdev. In this case, x bars in formula means means test. the std error of means is sigma x bar = sigma x / sqrt n. The stdev of means is sigma / sqrt n, and the above formula follows. it's sigma / sqrt n; not sigma / sqrt 2 n. and z test of means requires SIGMA/sqrt n.
Z test = (x̄ 1 – x̄ 2) / (SIGMA / √ n), it has for a long time.
 

Dason

Ambassador to the humans
#14
Let's agree on some notation. X ~ N(mu, sigma^2) means that X has a normal distribution with mean mu and variance sigma^2. Now let's do some simple derivations.

Let's start simple. Let's let X1 and X2 both be normally distributed with mean mu and variance sigma^2. Then the sum (X1 + X2) is distributed normal with mean mu + mu (because the means are additive) and variance sigma^2 + sigma^2 (because variances are additive). Which gives us (X1 + X2) ~ N(2*mu, 2*sigma^2). It's also true that X1 - X2 is distributed normal but the mean in this case is (mu - mu) = 0, however the variance is still sigma^2 + sigma^2 = 2*sigma^2. Are you agreeing so far? If not gtfo.

[1] So when we have X1 - X2 where they are X1 and X2 are both normal distributions with a mean of mu and variance of sigma^2 then X1-X2 ~ N(0, 2*sigma^2). Can we agree so far?

[2] Can we also agree that if X1bar is a mean taken from n samples from a Normal(mu, sigma^2) distribution that the distribution of X1bar ~ Normal(mu, sigma^2/n)? If not gtfo.

So in your problem we have n samples from a N(mu, sigma) distribution and we take the mean. We call that X1bar. We also have n samples from a N(mu, sigma) distribution and we take the mean. We call that X2bar.

So in your problem we're looking at X1bar - X2bar. Hey we know the distribution of X1bar to be Normal(mu, sigma^2/n) by using [2] and we know that X2bar is distributed Normal(mu, sigma^2/n) by using [2]. So we know that both are normally distributed, we know the means, we know the variances - hell there is nothing from stopping us from using [1] here.

X1bar - X1bar ~ Normal( [mean of X1bar] - [mean of X2bar], [variance of X1bar] + [variance of X2bar]) which is Normal(0, sigma^2/n + sigma^2/n) which is Normal(0, (2*sigma^2)/n)). Can we agree so far?

Literally I'm going to stop right here. We are like one step away from deriving the final answer but I've put in too much work typing this up and I need to at least make sure we're on the same page up to here. If you disagree with anything I've written let me know.
 

Dason

Ambassador to the humans
#15
So please take the time to do what I would ask of a stats 101 student and lay down your assumptions about the variables and derive the distribution of the test statistic provided. Literally the only reason we teach the z-test directly is because it's easy enough to derive the distribution of the test statistic directly.

So lay down your assumptions. Go through the math. Consider the possibility that maybe you interpreted something wrong because if you're doing a z-test then dammit the test distribution better have a standard normal distribution or you messed up.

Edit: The fact that you're having trouble with this is why I was asking about your mathematical background in the t-test thread. I don't want to be harsh but I just honestly don't have faith that you have a strong enough mathematical background to deal with this stuff. If you're teaching courses and attempting to write a book about stats then this question you're asking should be the absolutely bare minimum simplest thing to do. You should be able to do it in your sleep.
 

hlsmith

Less is more. Stay pure. Stay poor.
#16
@joeb33050

The standard deviation in the z-test is just the known population standard deviation. If you don't know it, use a t-test. BOOM! If you are looking for something else you have to write out your question better.

Side note, you may not realize it, but the way you write these post is very annoying. There is a latent layer of arrogance conveyed. You may want to work on this.
 
#17
I am looking for the standard deviation OF Z test, a statistic that I define here as (x bar 1 - x bar 2) / (sigma / sqrt n). This is how we find the Z value associated with the probability that Z <= Z test.
A Monte Carlo estimate shows stdev = 1.605, n=10; to stdev = 1.430, n = 100. For samples from populations with different mu's.
These numbers are so alarming that they cannot be true. I must have error/s in the estimate.
I would like to know the Standard Error or the standard deviation of Z test, as shown, to know that my MC estimate is wrong.
 

hlsmith

Less is more. Stay pure. Stay poor.
#18
You hadn't mentioned the MC part - this helps us understand where you are coming from. And you kept saying ZTEST. But you mean the standard deviation of the test statistic given sampling variability. As you increase your sample size any test statistic is going to get larger right? Since sample size is in the denominator. So what does this mean? Also, as @Dason mentioned your underlying parameters matter here right??
 

Dason

Ambassador to the humans
#19
You can't just say z-test. That's a test statistic appropriate for a different z-test. But you're doing the test for your data wrong.

And you're being too much of an ass to realize that what you're doing isn't appropriate but still have the gall to ask "why is this different than expected"? I've explained it multiple times. Don't be so **** thick.

Seriously have you read any of my posts? Do you have a direct response to anything I've written so far.

You have one post to respond appropriately or I'll ban you. I'm sick of you thinking you're better than everybody while still not having the most basic understanding of probability.

So please prove me wrong and actually act like you want to learn and have read what anybody else has written. If you don't you'll have shown that you don't respect anybody else or anybody else's time.
 
#20
You hadn't mentioned the MC part - this helps us understand where you are coming from. And you kept saying ZTEST. But you mean the standard deviation of the test statistic given sampling variability.

What is the standard deviation of Z test = (x̄ 1 – x̄ 2) / (SIGMA / √ n)? Z test, in italics, = (x̄ 1 – x̄ 2) / (SIGMA / √ n). That is my definition of the term.


As you increase your sample size any test statistic is going to get larger right? Since sample size is in the denominator. So what does this mean?

I attempted to show what it means, here.

Also, as @Dason mentioned your underlying parameters matter here right??

I don't think so, mu is mu, sigma is sigma. What else?
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