Page 380 - DECO504_STATISTICAL_METHODS_IN_ECONOMICS_ENGLISH

Unit 29: Methods of Point Estimation and Interval Estimation

Notes
∑ ( − x μ ) i 2 ≤ σ ≤ ∑ ( 2 − x μ ) i 2
2
χ .025 χ 2 .975 ... (13)
Case II:

n S 2 ∑ ( − x x ) i 2
(mean unknown) — In this case = follows chi-square distribution with (n – 1) degrees
σ 2 σ 2
of freedom. Using the lower and the upper 2.5% points of chi-square distribution with (n – 1) d.f., we
have with probability 95% the following inequalities

χ 2 .975 ≤ n S 2 ≤ χ 2 .025
σ 2
from which the 95% confidence interval for σ 2 can be given as

n S 2 2 n S 2
χ 2 ≤ σ ≤ χ .025 .975 ... (14)
2
Example 13: The standard deviation of a random sample of size 12 drawn from a normal population
is 5.5 Calculate the 95% confidence limits for the standard deviation () in the population (Given
σ
χ 2 .975 = 3.82 and χ 2 .025 = 21.92 for 11 degrees of freedom).
Solution: Here n = 12 and the sample s.d. (S) = 5.5. Substituting the values in formula (14), the 95%
confidence interval for σ 2 is

12 ( × )5.5 2 ≤ σ ≤ 2 12 ( × )5.5 2
21.92 3.82

or, ≤ 16.56 σ ≤ 2 95.03

i.e., 4.1 ≤ σ ≤ 9.7
σ
The 95% confidence limits for the population s.d. () are 4.1 and 9.7.
Example 14: A sample of size 8 from a normal population yields as the ubiased estimate of population
variance the value 4.4. Obtain the 99% confidence limits for the population variance σ (Given
2
χ 2 .975 = 0.99 and χ 2 .005 = 20.3 for 7 d.f.).

2
2
Solution: Here n = 8, and the unbiased estimate s = 4.4. So, Sn 2 = ( n ) − 1 s = 7 × 4.4 = 30.8.
Hence the 99% confidence limits for σ 2 are obtained from

n S 2 ≤ σ ≤ 2 n S 2
2
2
χ .005 χ .975
30.8 2 30.8
or, ≤ σ ≤ ; i.e., ≤ 1.52 σ ≤ 2 31.1
20.3 0.99

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