# Thread: Z-value using Z-table or t-table.

1. ## Z-value using Z-table or t-table.

Hi there. I don't really understand how the solution to the following came about:

$\displaystyle z_{a/2}=z_{0.01}=2.326$
$\displaystyle P(Z>2.326)=0.01$

The answers after dictate: "To find where this value comes from, look up 0.49 backwards on the Z table, or look up $\displaystyle t_{0.01}$ for $\displaystyle \infty$ degrees of freedom the $\displaystyle t$ table."

Firstly, I don't understand why I'm supposed to look up 0.49 on the Z table and not 0.01? I'm also confused as to how to use the table 'backwards', especially if/when interpolation comes into it.

Secondly, why does the degree of freedom = $\displaystyle \infty$?

2. Originally Posted by trackies
Hi there. I don't really understand how the solution to the following came about:

$\displaystyle z_{a/2}=z_{0.01}=2.326$
$\displaystyle P(Z>2.326)=0.01$

The answers after dictate: "To find where this value comes from, look up 0.49 backwards on the Z table, or look up $\displaystyle t_{0.01}$ for $\displaystyle \infty$ degrees of freedom the $\displaystyle t$ table."

Firstly, I don't understand why I'm supposed to look up 0.49 on the Z table and not 0.01? I'm also confused as to how to use the table 'backwards', especially if/when interpolation comes into it.

Secondly, why does the degree of freedom = $\displaystyle \infty$?
If you use the t-statistic instead of the z-statistic, n = 250 => df = 249 which is effectively df = infinite as far as the values used in the calculation are concerned.

You'd use 0.99 because the tables give you Pr(z < z*), not Pr(z > z*). So you want the value z* such that Pr(z < z*) = 0.99, which means going backwards from 0.99 in your tables.

I'm not sure where the 0.49 comes from ......

3. Ah, okay! I understand about the t-table now.

However, I don't understand where you got 0.99 from.

4. Originally Posted by trackies
Ah, okay! I understand about the t-table now.

However, I don't understand where you got 0.99 from.
You need to find the value of z* such that Pr(z > z*) = 0.01. But your tables don't give Pr(z > z*), they only give Pr(z < z*). So you need to consider Pr(z < z*) = 1 - 0.01 = 0.99.

5. I double checked and it definitely says 0.49. Perhaps a typo? But in any case, that makes perfect sense now. Thank-you!

6. Originally Posted by trackies
I double checked and it definitely says 0.49. Perhaps a typo? [snip]
I'd bet on it. Note that 0.49 is close to 0.5 and Pr(z < z*) = 0.5 => z* = 0 .....

Originally Posted by trackies
[snip]But in any case, that makes perfect sense now. Thank-you!
You're welcome.

7. Originally Posted by mr fantastic
If you use the t-statistic instead of the z-statistic, n = 250 => df = 249 which is effectively df = infinite as far as the values used in the calculation are concerned.

You'd use 0.99 because the tables give you Pr(z < z*), not Pr(z > z*). So you want the value z* such that Pr(z < z*) = 0.99, which means going backwards from 0.99 in your tables.

I'm not sure where the 0.49 comes from ......
OK, I think post #6 of this thread: http://www.mathhelpforum.com/math-he...-dont-get.html

explains why 0.49 rather than 0.99 is used .......