converting T-scores to percentiles, WITHOUT a table?

[Matt915]

Hey,

I have a stats exam coming up, and this practice problem is kicking the shit out of me, can someone help? I can check on a chart and see that in this example the answer for a t-score of 62.5 is the 89.4th percentile...but I keep coming up with 87.5.....here's the problem:

person's raw score = 600
sample mean= 500
sample sd= 80
then z= 1.25

Since a T-score is calculated in this example as:
T= 50+10z
T= 50 + 12.5= 62.5

Based on the normal curve, a t-score of 60 is exactly 1 standard deviation above the mean, which means it is 34.13% above the mean for a percentile of 84.13%....but I still have the 2.5 of the 62.5 left over. Since that's a quarter of the way between 60 and 70 for t-scores I I tried multiplying .25*13.59 (13.59 being the % of the normal curve between 1 and 2 SDs above the mean). Which gives me 3.4% and add that to 34.13% which is the 1 sd above the mean part I already had for a grand total of 87.53 (50+3.4+34.13).

And that answer is completely wrong. WTF am I doing wrong here?

 

[TecHNooB]

maybe you're supposed to estimate the cdf linearly. its (x-mean)/sigma iirc.

 

[Fenixgoon]

i haven't used stats in a while, but don't you need the # of samples as well? t is used when the # of samples < 30.

also, IIRC, you need the number of factors or whatever..its just n-1.

like i said though..i haven't used stats in a while

 

[Matt915]

It's assuming scores on a normal curve so most of that stuff doesn't matter in this particular example

 

[Whisper]

My guess is that the error is coming in to play because your method of estimating the percentage attributed to 2.5 T-score points essentially breaks the area in the normal curve between T-scores of 60 and 70 into four equal parts. However, the normal curve in that region is curvilinear; thus, the first quartile in that area is going to have a larger change in percentiles per T-score point than the last quartile.

As for how to actually calculate it...I could be wrong, but I believe that would require you to figure out the area under the curve...?

 

[DrPizza]

I'm too rusty - to do it without a table though, don't you integrate the normal distribution curve from 0 to your number of standard deviations above or below? (And, I think you can do pretty much the same process using the z-scores. It's been 20 years or longer since I did that stuff though.)

Wait a second -
http://www.stat.yale.edu/Courses/1997-98/101/normal.htm

I don't see how you can do it without a table (in a reasonable amount of time.)

 

[Matt915]

yah you guys are right, Whisper that was the error I was making. This is more of a basic stats class. I've had basic stats before, and actually tutored people. I have to write to the prof, I think she made a mistake somewhere. She's it repeatedly on examples for other exams (she had an example where there was a correlation over 1.00 at one point....)

 

[RbSX]

Hey,

I have a stats exam coming up, and this practice problem is kicking the shit out of me, can someone help? I can check on a chart and see that in this example the answer for a t-score of 62.5 is the 89.4th percentile...but I keep coming up with 87.5.....here's the problem:

person's raw score = 600
sample mean= 500
sample sd= 80
then z= 1.25

Since a T-score is calculated in this example as:
T= 50+10z
T= 50 + 12.5= 62.5

Based on the normal curve, a t-score of 60 is exactly 1 standard deviation above the mean, which means it is 34.13% above the mean for a percentile of 84.13%....but I still have the 2.5 of the 62.5 left over. Since that's a quarter of the way between 60 and 70 for t-scores I I tried multiplying .25*13.59 (13.59 being the % of the normal curve between 1 and 2 SDs above the mean). Which gives me 3.4% and add that to 34.13% which is the 1 sd above the mean part I already had for a grand total of 87.53 (50+3.4+34.13).

And that answer is completely wrong. WTF am I doing wrong here?

The answer is without the table, you can't answer this question without using excel.

To solve this question you need to use T in excel, or the degrees of freedom and the T table.

What you're thinking of kind of works if you're talking about z scores and standard deviations with a normal distribution, but since you're talking about under the long tail.. it doesn't.

 

[Whisper]

I'm too rusty - to do it without a table though, don't you integrate the normal distribution curve from 0 to your number of standard deviations above or below? (And, I think you can do pretty much the same process using the z-scores. It's been 20 years or longer since I did that stuff though.)

Wait a second -
http://www.stat.yale.edu/Courses/1997-98/101/normal.htm

I don't see how you can do it without a table (in a reasonable amount of time.)

Yeah, it's a pain in the a$$. T-scores can easily be converted to z-scores, but after that, unless you're in a master's or doctoral-level stats program, you likely aren't going to be computing percentiles without a table.

 

[gorcorps]

yah you guys are right, Whisper that was the error I was making. This is more of a basic stats class. I've had basic stats before, and actually tutored people. I have to write to the prof, I think she made a mistake somewhere. She's it repeatedly on examples for other exams (she had an example where there was a correlation over 1.00 at one point....)

She may not be wrong with the correlation. I remember utilizing some sort of non standard correlation where the results could be >1 and even negative. It related to the slope of the trendline in some way. I'm not sure what it was, and "correlation" probably isn't the right term, but I do remember seeing something similar during our correlation stuff.