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

**trackies** · June 10th 2008, 07:45 PM

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

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

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

**mr fantastic** · June 10th 2008, 08:10 PM

Originally Posted by trackies

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

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

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

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 ......

**trackies** · June 10th 2008, 08:18 PM

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

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

**mr fantastic** · June 10th 2008, 08:23 PM

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.

**trackies** · June 10th 2008, 08:29 PM

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

**mr fantastic** · June 10th 2008, 08:37 PM

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.

**mr fantastic** · June 11th 2008, 03:27 PM

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 .......

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