Hypothesis Testing > T-Score vs. Z-Score

You may want to read these articles first:

What is a t-statistic?

What is a z-score?

Watch the video or read the article below:

## T-Score vs. Z-Score: Overview

A **z-score** and a **t score **are both used in hypothesis testing. Few topics in elementary statistics cause more confusion to students than deciding when to use the z-score and when to use the t score. Generally, in elementary stats and AP stats, you’ll use a z-score in testing more often than a t score.

## T-score vs. z-score: When to use a t score

The general rule of thumb for *when* to use a t score is when your sample:

- Has a sample size below 30,
- Has an unknown population standard deviation.

You **must** know the standard deviation of the **population** *and *your sample size **should **be above 30 in order for you to be able to use the z-score. Otherwise, use the t-score.

The above chart is based on (from my experience), the “rule” you’re most likely to find in an elementary statistics class. That said, **this is one of those rules that isn’t set in stone, so you should always check with your textbook/teacher to make sure they aren’t suggesting something different.**

In real life though, it’s more common just to use the t-distribution as we usually don’t know sigma^{1}.

“When a sample has more than 30 observations, the normal distribution

canbe used in place of the t distribution. ~ Applied Statistics for Public and Nonprofit Administration (Meier et.al) p. 191.”

Note the use of the word *can *in the above quote; The use of the t-distribution is theoretically sound for all sample sizes, but you *can* choose to use the normal for sample above 30.

### T-Score vs. Z-Score: Z-score

Technically, z-scores are a conversion of individual scores into a standard form. The conversion is based on your knowledge about the **population’s standard deviation and mean. **A z-score tells you how many standard deviations from the mean your result is. You can use your knowledge of normal distributions (like the 68 95 and 99.7 rule) or the z-table to determine what percentage of the population will fall below or above your result.

The z-score is calculated using the formula:

**z=(X-μ)/σ**

**Where**:

- σ is the population standard deviation and
- μ is the population mean.

The z-score formula doesn’t say anything about sample size; The rule of thumb applies that your sample size should be above 30 to use it.

### T-Score vs. Z-Score: T-score

Like z-scores, t-scores are also a conversion of individual scores into a standard form. However, t-scores are used **when you don’t know the population standard deviation**; You make an estimate by using your sample.

**T = (X – μ) / [ s/√(n) ].**

Where:

- s is the standard deviation of the sample.

If you have a larger sample (over 30), the t-distribution and z-distribution look pretty much the same. Therefore, you can use either. That said, if you know σ, it doesn’t make much sense to use a sample estimate instead of the “real thing”, so just substitute σ into the equation in place of s:

**T = (X – μ) / [ σ/√(n) ].**

This makes the equation identical to the one for the z-score; the only difference is you’re looking up the result in the T table, not the Z-table. For sample sizes over 30, you’ll get the same result.

**References**:

1. Vermont Tech. Retrieved 11/20/2016 from https://simon.cs.vt.edu/SoSci/converted/T-Dist/.

**Next**: When to use sigma/sqrt(n)

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Isn’t the flow chart incorrect because the ability to estimate std deviation from sample std deviation allows you to use the z-test when std deviation is unknown and sample size is larger than 30?

It’s not an absolute. You should use a t test if you don’t know sigma, but you

coulduse the z-test as long as you realize it’s an estimation. The flow chart is just a general guideline.If we are already aware that the distribution of the population is normal, then can we use a z-score (still assuming we have the population standard deviation)?

It would seem to follow that the entire point of needing a sample of 30 (or more) is because of the central limit theorem, so if I am already aware that the population is normally distributed then all of my sampling distributions will be normal as well, regardless of sample size. Which presumably allows for the use of z.

Can you help clarify if this thinking is correct?

Jen,

Your thinking is correct. If you are sure your data is normally distributed, you can use a z score.

The calculation of the t-score in the video looks strange. Presumably the standard deviation should be 0.47 and not 4.7.

You’re right. I messed up the decimal point. I made an annotation on the video. Thanks for pointing that out :)

thank’s for all

what is the difference testing your hypothesis in a research thesis with spear man’s correlation coefficient and the Z-test.

again can two of them be used in testing same hypothesis in the same study?

what is the essence of using both in a particular research study?

thanks

Tony,

You use the z-test to

testyour hypothesis.Spearman’s is just a measure of correlation. It’s not a statistical test.

So yes, you can use both :) One will test your hypothesis and the other will tell you how closely the variables are related.

Stephanie

What if you dont know the standard deviation and the same size is 125? you will need to use T table but it only goes to 100…am i missing something? Thanks

See the t-table here: t-distribution table. It goes to infinity, although the last two entries are 100 and 1000. Basically, the larger your sample size, the more the t score will look like the z score. For an exact score for any sample size, you’ll have to use a calculator.

Hi y’all – I’m in a stats class in college- anyone care to email me at blackeyedsusan at yahoo to offer me some assistance??? I don’t have anyone around me that can help- and I have two problems I need help with…… Please- anyone feel like offering a helping hand???

very nicely presented by you

but i have so many confusion

like

how Z score, t score, statine, letter grade and percentile rank used for the process of standardizing test

if you know about this than reply me on sanjay_diet@yahoo.com

Thanks

Referring to other websites and a stat major friend, I came to conclude that you should use the z-distribution even when n<30 if the population standard deviation is known. I think using t-distribution when n30 the t- and z-distributions are very similar.

It’s a judgment call. I’d disagree, but that’s statistics for you, everyone has a different opinion ;)

thanx a lot bt do we use z test is standard deviation is known and n is less than 20?

If n is very small, I’d use the z-test.

This thing about how to use z and t test is so confusing. When your sample size is less than 30 and the population standard deviation is known, which do u use for cases where your population is normally distributed and where it is not? When the sample size is greater than 30 and population standard deviation is unknown, which do you use for cases where your population is normally distributed and where it is not?When your sample size is less than 30 and the population standard deviation is unknown, which do u use for cases where your population is normally distributed and where it is not?When your sample size is greater than 30 and the population standard deviation is known, which do u use for cases where your population is normally distributed and where it is not?

“When your sample size is less than 30 and the population standard deviation is known, which do u use for cases where your population is normally distributed and where it is not?”

If you have a small sample size that you

knowis normally distributed, use the z-score.“When the sample size is greater than 30 and population standard deviation is unknown, which do you use for cases where your population is normally distributed and where it is not?”

If you don’t know the population SD, there’s no way of knowing if your data is normal. So, use the t-score.

“When your sample size is less than 30 and the population standard deviation is unknown, which do u use for cases where your population is normally distributed and where it is not?”

If you know it’s normal, use the z. Although if you don’t know the population SD, how can you know if the distribution is normal?? I’d use the t for this scenario.

“When your sample size is greater than 30 and the population standard deviation is known, which do u use for cases where your population is normally distributed and where it is not?”

If your data is normal, use the z. If not, figure out what distribution it is and use the appropriate table (i.e. t, chi square etc.). If you have no idea what the distribution is, use the t but be cautious when interpreting your results.

I studied statistics years ago but i thought that a t score was simply a z score with a mean of 66? And a standard deviation of 10.

You might be thinking of a t-score in psychometrics, which isn’t the same as the statistical t-score.

If I have multiple z-scores, can I test if they are significantly different from one another?

I’d say you could probably test the underlying data that led you to get the z-scores. Can you get more specific about what you are trying to do?

We know to use z test when n is less or greater than 30 and standard deviation is given.

And we also know to use t test when n is less than or equal to 30 and standard deviation is not given:

Q) My question is that if n is less than 30 and s.d is given as we can see that in t test formula we need to find (sp)square and one of the formulas to find sp(square) need standard deviation like :

[(n1-1)S1(square)]+[(n2-1)S2(square)] whole divided by (n1+n2-2).

Then according to this formula of Sp(square) standard deviation is given in question or research .

This is confusing me whether how to apply z-test when n is less than 30 and standard deviation is given. Plz help me

Use t test when n is less than or equal to 30 OR standard deviation is not given. I hope that clarifies it, but remember…it’s only a guideline.

Why does the chart completely go against what you state earlier on the page? got a homework question wrong following this chart >:( rediculous

Hi, Tyler,

I’m sorry that you got your homework question wrong. I took another look at the chart and I’m not seeing what’s wrong with it. That said, it is a “rule of thumb” and your professor might not be following it. Could you post the homework question? I’ll take a look and see what’s up.

Regards,

Stephanie

Your flow chart says to use the “t statistic” if there are less than 30 samples even if the population SD is known. I have two questions:

1) I think you are suggesting to use the “t distribution” for the “z statistic”. If σ is the population SD, and s is the sample SD, x is the sample mean, and mux is the hypothesized mean, then as you say

t = (x – μx) / [ s/√(n) ]

but

z = (x – μx) / [ σ/√(n) ] .

I don’t see any reason to use the estimate, s, if you actually know σ. Do you agree?

2) Assuming I’m right about #1, do you have a citation for your advice? It is quite common advice to suggest using the z distribution (Normal) for the t statistic if the sample size is above 30, basically for historical reasons (as nicely summarized here: http://stats.stackexchange.com/a/31457/13419 ), but I haven’t found anyone arguing for what you are saying.

Suppose the population distribution is not Normal, but we do know σ. If the sample size is small, then *who knows* what the distribution of the z statistic is. I could see arguing that the t distribution is better than the normal (z) distribution, because it is more conservative, but there’s nothing saying it should be closer to correct. Do you know of anyone systematically making that argument?

Here are some other places that suggest using Normal distribution for the t statistic when n is at least 30:

http://stats.stackexchange.com/questions/112162/why-not-use-the-t-distribution-to-estimate-the-mean-when-the-sample-is-large

http://ci.columbia.edu/ci/premba_test/c0331/s7/s7_4.html

http://stattrek.com/sampling/sampling-distribution.aspx

https://simon.cs.vt.edu/SoSci/converted/T-Dist/

Thank you, the flow chart really helped!

Hello, Peter,

Thanks for your thoughtful comments.

1. “I don’t see any reason to use the estimate, s, if you actually know σ. Do you agree?”

Yes, you’re right. I added a sentence clarifying that.

2. “It is quite common advice to suggest using the z distribution (Normal) for the t statistic if the sample size is above 30.”

You’re right there too. However, there’s just as much advice the other way too :). The fact is it just doesn’t matter when your sample size is over 30, as the two distributions are almost identical for large samples. In real life though, it’s more common just to use the t as we don’t know sigma. So I think it makes more sense to argue for using T as it’s more applicable to real life. There’s a ton of references out there. One reference you included (https://simon.cs.vt.edu/SoSci/converted/T-Dist/) actually states “It is usually the case that researchers do not know the population standard deviation for the variables they are studying. Therefore, researchers are more likely to use the t-distribution than a normal distribution when testing hypotheses.” OR, quoting from Applied Statistics for Public and Nonprofit Administration (Meier et.al) p. 191 “when a sample has more than 30 observations, the normal distribution

canbe used in place of the t distribution.” The use of the t-distribution is theoretically sound for all sample sizes, but you *can* choose to use the normal for sample above 30.Regards,

S

Hello, and thanks for your reply! I certainly agree that we don’t usually know σ, so the t distribution is more commonly used.

I don’t think you’ve answered my question, though. I’m asking specifically about the recommendation that your flow chart makes to use the T distribution if you know σ but the sample size is below 30. I have not found this recommendation elsewhere: can you give me a reference? Neither of the references you quote say this, by my reading.

best,

P

Here’s my interpretation of the general rules of thumb out there:

Do you know the population standard deviation?

yes -> Use the z-score.

no -> Is the sample size above 30?

yes -> Use the z-score

no -> Use the t-score.

… but I’d add that the second branch isn’t really necessary with modern computers; it can just go to z versus t.

I think we may be arguing apples and apples here, after all — both viewpoints get the same results, right? If you know sigma, then the t-score equation becomes the z-score equation, and if you look up the result in either the t-dist or z-dist table you’ll get the same result (assuming sample size over 30). And I agree, with computers, much of this becomes moot.

The answer to this question was in the book I referenced; The t-distribution is usually used. However, you can

chooseto use a normal dist if you have a large sample size.The t-score formula uses “s”, but if you know sigma of course you’ll just put that in there instead. I see where the confusion lies (I think); the flow chart says to use the t-score but fails to note that you should (of course) uses sigma in the equation if you know it. That is stated in the t-score section of the article, but I have put a note under the graph to make it more obvious. Thanks for pointing that out.

Regards,

S

I’m actually asking about the branch of your diagram with *small* sample sizes and known sigma. In this case you definately do not get the same answer.

Here’s a quick test example: suppose your data are 1,2,3,4; we know that sigma=1.5, and we test for population mean = 0. The sample mean is 2.5 and the sample SD is 1.29. If we go with the z statistic:

z = 2.5 / ( 1.5 / sqrt(4) ) = 2.5

then the P-value with the Normal distribution is 0.00086; with the T(df=3) it is 0.03.

The other possibility is to compute the t statistic (i.e., using the sample SD):

t = 2.5 / ( 1.29 / sqrt(4) ) = 3.87

which gives a P-value of 0.03 with df=3.

best,

Peter

Whoops, calculation errors: with the correct numbers…

Here’s a quick test example: suppose your data are 1,2,3,4; we know that sigma=1.5, and we test for population mean = 0. The sample mean is 2.5 and the sample SD is 1.29. If we go with the z statistic:

z = 2.5 / ( 1.5 / sqrt(4) ) = 3.3333

then the P-value with the Normal distribution is 0.00086; with the T(df=3) it is 0.045.

The other possibility is to compute the t statistic (i.e., using the sample SD):

t = 2.5 / ( 1.29 / sqrt(4) ) = 3.87

which gives a P-value of 0.03 with df=3.

For small sample sizes, the branch says to use the t-distribution with the *known* standard deviation. Basically, if you plus in sigma instead of s, you get the same formula, but the distributions will give you very different results. In other words, you present two options: the z-statistic, the t statistic (i.e., using the sample SD), but you don’t give the third option (which is what I present): the t-score formula with sigma — which is essentially the z-formula. but you would look up the result in the t-distribution.

Peter, what do you think the diagram should read?

everything in the formula of z -score and t score should be shown in step by step order and derived each of the variables clearly with sample problems and other types of problems so that we know how to attack each kind of problems…

Hi, Floyd,

Check out the t-score and z-score pages for lots of examples :)

I have a question with sample siz eof 80 and I have to calculate the standard deviation ie it isnt given. I wanted to use the t table but a v value of 79 isnt available. Do i then use z ?

I would just extrapolate. For 60 degrees of freedom at alpha 10%, it’s 1.296. For 120 df, it’s 1.289. So 79 df would be around 1.292. If you want a more precise figure, try a calculator like this one: http://www.danielsoper.com/statcalc/calculator.aspx?id=10

Hi there,

If i have a small sample and the standard deviations are known but quite different would i still use the z-test? Or would i use a t-test with a transformation?

So the means are 823.3 and 522.1 and the sd’s are 566.4 and 293.0 respectively?

does a z-test for a small sample assume equal variance?

For small samples use the t-test. What do you mean by “t-test with a transformation”? i.e. why do you want to transform your data?

Can anyone PLEASE assist me with this assignment question, this was all of the information provided & I am lost at how ONE research question can fit into all 3 categories of testing??? Please respond on this webpage, I rarely check email. Thank you, T.W.

Suggest one psychological research question that could be answered by each of the following types of statistical tests:

z test

t test for independent samples, and

t test for dependent samples

The way the question is worded, I think it’s asking for one research question per test (a total of three questions).