Thread: 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.

dadon

Originally Posted by dadon

hey!

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

..and if it is possible an example.


thanks.

dadon

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

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

Originally Posted by dadon

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.

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)]

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

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

dadon

Originally Posted by dadon

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

dadon

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

do the same for positive.


thanks cheer captain

i was doing the sum in ms excel and left the brackets outs.