Page 376 - DECO504_STATISTICAL_METHODS_IN_ECONOMICS_ENGLISH

Unit 29: Methods of Point Estimation and Interval Estimation

σ denoting the Standard Deviation (S.D.) of the population. Here, Notes
Sample size (n) = 100, Population size (N) = 2000

Sample mean ( ) x = 0.354, Sample S.D. (S) = .048
Since σ is not known, an approximate value of S.E. is obtained on replacing the population S.D. ()
σ
by the sample S.D. (S).

S N − n
S.E. of x = approximately
−
n N1
.048 2000 − 100
= = .0047
100 2000 − 1
The 95% confidence limits for the population mean μ are

x ± 1.96 (S.E.of x ) = 0.354 ± 1.96 × .0047
= 0.354 ± .0092 = 0.3632 and 0.3448
Thus, the 95% confidence interval is (0.3448 to 0.3632) inch.
Example 9: A random sample of 100 articles taken from an large batch of articles contains 5 defective
articles. (a) Set up 96 per cent confidence limits for the proportion of defective articles in the batch. (b)
If the batch contains 2696 articles set up 95% confidence interval for the proportion of defective
articles.
Solution: (a) The 96% confidence limits for the population proportion (P) are given by p ± 2.05
(S.E. of p), where p is the sample proportion.

PQ
S.E. of p =
n
Since the formula involves the unknown population proportion P, an approximate value of
S.E. is obtained on replacing the population proportion (P) by the sample proportion (p). Putting
n = 100 and p = 5/100 = .05, (q = 1 – p = .95)

pq .05 ×.95
S.E. of p = = = .022
n 100
Hence, the 96% confidence limits for P are
p ± 2.05 (S.E. of p) = .05 ± 2.05 × .022 = .05 ± .045
= .05 + .045 and .05 – .045
= .095 and .005
(b) The 95% confidence limits for proportion (P) are given by p ± 1.96 (S.E. of p). But, when the
population is of a finite size N,

pq N − n
S.E. of p = (approximately)
n N1
−
Here, n = 100, N = 2696, p = .05. Putting these values

05 × −.95 2696 100 2596
S.E. of p = = .022
100 2696 − 1 2696
= .022 × .963 = .022 × .98 = .0216 (approx.)
Hence, the required 95% confidence limits for P are

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