Section 6.4.

 

We can use Minitab to find confidence interval for the mean 
of a normal distribution, knowing and not knowing the variance. 
For example,  in the car data, assuming that sigma=3,  
to find a  99 % confidence interval for mu, we do
MTB > Retrieve  'C:\ISTAT\CAR.MTW'
Retrieving worksheet from file: C:\ISTAT\CAR.MTW
Worksheet was saved on 11/25/1995
MTB > zinter .99 3 c3

THE ASSUMED SIGMA =3.00

 

N

MEAN

STDEV

SE MEAN

99.0 PERCENT C.I.

Tur_Diam

109

35.514

3.321

0.287 

( 34.773, 36.254)



The default confidence interval is the  one with 1-alpha= .95 and sigma=1.
We get that n=109, x-bar=35.514, s=3.321, s/sqrt(n)=0.287 and 
the 99 % confidence interval for mu is (  34.773,  36.254)

To find t-intervals we do 

MTB > tinter.99 c3

N

MEAN 

STDEV 

SE MEAN

99.0 PERCENT C.I

Tur_Diam

109 

35.514 

3.321 

0.318

( 34.680, 36.348)


The meaning of a 1-alpha confidence interval for a parameter is that if 
find repeatedly confidence intervals, approximately 
(1-alpha)100 percent of these confidence intervals cover the true value 
of the parameter. Using minitab, we may simulate data and
 confidence intervals. Approximately (1-alpha)100 percent of 
these confidence intervals cover the true value 
of the parameter. So, if we want to estimate the mean of distribution,
we create 500 samples of this distribution and find confidence intervals 
for the mean for each of these samples, approximately 95 % (495) of the 
confidence intervals,  contain the true value of the mean. 

Running the macro 6-4B.MTB, 500 times, we get 500 confidence intervals
for the mean of a gamma (2,3) distribution. 

MTB > Execute 'C:\ISTAT\MACROS\6-4B.MTB' 500.

************6-4B.MTB*************
random 100 c1;
gamma 2 3.

let k1=mean(c1)-stdev(c1)*1.96/sqrt(99)
let k2=mean(c1)+stdev(c1)*1.96/sqrt(99)
stack k1 c2 c2
stack k2 c3 c3
end
******************************     

The lower ends of the confidence intervals are in c2.
The upper ends of the confidence intervals are in c3.
The mean of the  gamma (2,3) distribution is 6.
We should expect roughly 500*.95=475 confidence intervals containing
the mean 6. Running the macro:

************6-4A.MTB****************** 
code (-100:6)1 (6:100)0 c2  c4.
code (-100:6)0 (6:100)1 c3  c5.
let c6=c4*c5
let k1=n(c6)*mean(c6)
print k1
end
****************************** 

we get the number of intervals containing  6 (the mean):
MTB > Execute 'C:\ISTAT\MACROS\6-4A.MTB' 1.
Executing from file: C:\ISTAT\MACROS\6-4A.MTB

K1       478.000

This is number obtained through simulations.
If you repeat the procedure, you could get something else. 


Comments to: 
Miguel A. Arcones