[lisakki]

Can someone good with TI calculators and stat help me out with this? This is driving me nuts, and I know I'm just missing something obvious.

So with a data set 1,2,3,4,5

Mean=3
Sample Standard deviation=1.41
n=5

To calculate the z-score of the value 4, I simply go: (4-3)/1.41=.709

Simple, right?

So I try to do the same exact thing on my TI 89 using Z Test:

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List4={1,2,3,4,5}

I get z=-1.58586, which is obviously not right.

So, what am I entering wrong?

[Rudy]

Hint: You're not considering the effects of the sample size, n=5, on your answer.


Further hint. z is not the difference divided by the standard deviation, it is the difference divided by the standard error. Your calculator is correct.

[lisakki]

	Originally Posted by RudyHenkel To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	Hint: You're not considering the effects of the sample size, n=5, on your answer. Further hint. z is not the difference divided by the standard deviation, it is the difference divided by the standard error. Your calculator is correct.

The calculator does take the sample size into consideration. From that screen, after I press enter, the list of data that appears has an entry with n=5.

How can z not be the difference divided by standard deviation?

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Isn't this the definition of a z-score?

4=Raw number to be standardized
3=Mean of the set
sigma=standard deviation of the sample/population whatever the case is

Edit: Sorry I completely misunderstood what you said. I calculated the thing using standard error, and then got the same number as my calculator.

I still don't understand why this works though. In my book, sample standard deviation doesn't equal standard error, and the book's definition says exactly the definition I pasted above.

[Rudy]

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I don't know if your book is mistaken, or if you are misinterpreting it, but for a z-test, you divide by standard error, not standard deviation.

You may be confusing it with a different context in which the z-table is used. If you have a very, very large sample size, then you can use the z-value that you describe (divide the difference by the standard deviation) to determine what portion of the population is above or below a certain value, assuming that population follows a standard normal distribution. A good example of this is with IQ scores.

[lisakki]

Does the Z-Test calculate the z-score? Or are they two different things? I might just be getting them confused.

This is the article I pulled that formula from:
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Btw thanks alot for helping me.

[Rudy]

In that link (the one you gave,) look at the third formula under "Standardizing in mathematical statistics". That's the one you should be using.

[nacht]

What you are doing with your "Z-score" is called normalizing: you are transforming a N(µ, sigma) distribution to a N(0,1) distribution.

A Z-test is based on this calculation:

Z = (x-µ) / (s/sqrt(n))

Which is the Z-score you are looking for.

This assumes you know the µ of the population, of course, otherwise you should be using a student-t test.

[CliffClaven]

Your standard deviation is 1.21, not 1.41