# confidence limits

• Jan 13th 2007, 04:17 AM
confidence limits
hey!

I was wondering if anyone knew anything about 95% confidence limits from a 't' distribution?

..and if it is possible an example.

thanks.

• Jan 13th 2007, 05:26 AM
CaptainBlack
Quote:

hey!

I was wondering if anyone knew anything about 95% confidence limits from a 't' distribution?

..and if it is possible an example.

thanks.

The Wikipedia article on the t-distribution is very good and can be found here

A example is shown in the following code block showing a computer session
computing a 95% confidence interval for the mean:

Code:

```> >N=10; >x=normal(N,1)     -0.309911     -2.01066     -0.568246       -1.0841     -0.277172       -1.4568     0.658046       1.53204     -2.47296     -0.937381 >m=mean(x')     -0.692714 > >s=sqrt(sum((x'-m)^2)/(N-1))       1.19313 > >..95% region for T-distribution with 9 DF is +/-2.26 (looked up >..in a table of the T-distribution) > >..so the 95% region for the sample mean is m+/-2.26*s/sqrt(N) > >[m-2.26*s/sqrt(N),m+2.26*s/sqrt(10)]     -1.54541      0.159986 > >```
RonL
• Jan 13th 2007, 06:08 AM
thanks for that captain.

im abit confused:

s=sqrt(sum((x'-m)^2)/(N-1))

(could you please put in the values for x', m, and N and wot is s?)

also this line:

[m-2.26*s,m+2.26*s]

i cnt seem to get the same answers..

is this a general way of finding the confidence limits?

and also wots the difference in the the amount of degree of freedom you use and why is it 9 DF for this case.

thanks
• Jan 13th 2007, 07:49 AM
CaptainBlack
Quote:

thanks for that captain.

im abit confused:

s=sqrt(sum((x'-m)^2)/(N-1))

(could you please put in the values for x', m, and N and wot is s?)

s is the sample stadard deviation, x is a array containing the sample
(from a standard normal population), m is the sample mean calculated
on the previous line by mean(x'), and N is the sample size set at the top
to 10.

The ' on the x's are transposition operators because the sum and mean
functions require row data vectors withe x is a column vector.

Quote:

also this line:

[m-2.26*s,m+2.26*s]

i cnt seem to get the same answers..
There is a typo, for a 95% region for the mean that should have been

[m-2.26*s/sqrt(N),m+2.26*s/sqrt(N)]

Quote:

is this a general way of finding the confidence limits?

and also wots the difference in the the amount of degree of freedom you use and why is it 9 DF for this case.

thanks
The sample size was 10, the degrees of freedom is N-1, see the article the
I gave you a link for for futher explanation.

The DF are needed as we are doing small sample statistics, and the sample
size affects the distributions involved.

RonL
• Jan 13th 2007, 10:28 AM
hey thanks captain!

i checked the link you gave. quite good!

according to thier table 95% is 1.83 but you probably read the next coulmn for 97.5% which is 2.26 (for 9DF)

(still using 2.26) but i still can't get the answer for this line:

[m-2.26*s/sqrt(N),m+2.26*s/sqrt(10)]

-1.54541, 0.159986

i get:

-1.11406 and 0.59134

am I doing something wrong?

thanks

• Jan 13th 2007, 11:19 AM
CaptainBlack
Quote:

hey thanks captain!

i checked the link you gave. quite good!

according to thier table 95% is 1.83 but you probably read the next coulmn for 97.5% which is 2.26 (for 9DF)

(still using 2.26) but i still can't get the answer for this line:

[m-2.26*s/sqrt(N),m+2.26*s/sqrt(10)]

-1.54541, 0.159986

i get:

-1.11406 and 0.59134

am I doing something wrong?

thanks

m-(2.26*s/sqrt(N))=-0.692714-(2.26*1.19313)/sqrt(10)=-1.54541.

so what did you do for the lower limit?

RonL
• Jan 13th 2007, 02:10 PM