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Unit 29: Estimation of Parameters: Criteria for Estimates
Comparing (1) and (2), we get Notes
´
1.96 1600 æ 1.96 1600 ö 2
´
300 = or n = ç ÷ = 109.3
n è 300 ø
Since this value is greater than 109, therefore, the size of the sample should be 110.
29.3.2 Confidence Interval for Population Standard Deviation
1 2
Let S = å (X - X ) be the sample standard deviation of a random sample of size n drawn
i
n
from a normal population with standard deviation s. It can be shown that the sampling distribution
of S is approximately normal, for large values of n, with mean s and standard error . Thus,
2n
-
S
z = can be taken as a standard normal variate.
/ 2n
Example 11: A random sample of 50 observations gave a value of its standard deviation
equal to 24.5. Construct a 95% confidence interval for population standard deviation .
Solution.
( )
It is given that S = 24.5 and n = 50 (large). We know that . .S E S = . Since s is not known, we
2n
S 24.5
( )
use its estimate based on sample. Thus, we can write . .S E S = = = 2.45 .
2n 100
Hence 95% confidence interval for s is given by
24.5 – 1.96 ´ 2.45 24.5 + 1.96 ´ 2.45 or 19.7 29.3
Note More examples on confidence intervals are given later with the questions on
test of significance.
29.4 Summary
Let X be a random variable with probability density function (or probability mass function)
f(X ; , , .... ), where , , .... are k parameters of the population.
1 2 k 1 2 k
Given a random sample X , X , ...... X from this population, we may be interested in
1 2 n
estimating one or more of the k parameters , , ...... . In order to be specific, let X be a
1 2 k
normal variate so that its probability density function can be written as N(X : , ). We
may be interested in estimating m or s or both on the basis of random sample obtained
from this population.
It should be noted here that there can be several estimators of a parameter, e.g., we can
have any of the sample mean, median, mode, geometric mean, harmonic mean, etc., as an
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