Estimate of the population mean

• May 12th 2007, 06:17 PM
busyboy
Estimate of the population mean
Hello,everybody! I'll be glad if you help me!

Using the following data
30, 40, 50, 20, 20, 40.
construct 95% confidence interval estimate of the popolation mean.

What is the best formula for solving this problem?
E=t * S/sqrN (1)
or 1/2

E=Z * O/sqrN (2)
1/2
WHY?
• May 13th 2007, 12:36 AM
CaptainBlack
Quote:

Originally Posted by busyboy
Hello,everybody! I'll be glad if you help me!

Using the following data
30, 40, 50, 20, 20, 40.
construct 95% confidence interval estimate of the popolation mean.

The mean is m = (30+40+50+20+20+40)/6 = 33.33

Now if the distribution of the data is normal the sample mean has a
t-distribution with 5 degrees of freedom. But to use the t-distribution
we need the population standard deviation, which we will estimate with:

s = sqrt[1/(n-1) sum (x_1 - m)^2] = 12.11

Then the interval is:

(m - cv*s/sqrt(n), m + cv*s/sqrt(n))

where cv is the critical value for a 95% interval for a t-distribution with n-1
degrees of freedom.

The critical t-value for a 95% interval with 5 degrees of freedom is 2.57,
so the 95% interval would then be:

(33.33-2.57*12.11/sqrt(6), 33.33+2.57*12.11/sqrt(6))~= (20.62, 46.04)

We use the t-distribution because we have a sample size n=6, which is
a small sample so the use of the normal distribution is inappropriate here
as that requires large samples.

Note the use of the t-distribution is also inappropriate unless we have some
reason to believe the underlying distribution that the data is sampled from
is normal.

RonL