Page 400 - DMTH404_STATISTICS

Statistics

Notes 
Thus, the 99% confidence limits for m are X ± 2.58 .
n
Remarks: When s is unknown and n < 30, we use t value instead of 1.96 or 2.58 and use S in place
of s.

Confidence Interval for Population Proportion

Example 6: Obtain the 95% confidence limits for the proportion of successes in a binomial
population.
Solution.
Let the parameter p denote the proportion of successes in population. Further, p denotes the
proportion of successes in n (³ 50) trials. We know that the sampling distribution of p will be
 (1 - )
approximately normal with mean p and standard error .
n
Since p is not known, therefore, its estimator p is used in the estimation of standard error of p,

p (1 p- )
( )
i.e., . .S E p =
n
Thus, the 95% confidence interval for p is given by

æ p (1 p- ) p (1 p- ) ö
P p - 1.96    p + 1.96 ÷ = 0.95
ç
ç è n n ÷ ø

p (1 p- )
This gives the 95% fiducial limits as p ± 1.96 .
n

Example 7: In a newspaper article of 1600 words in Hindi, 64% of the words were found
to be of Sanskrit origin. Assuming that the simple sampling conditions hold good, estimate the
confidence limits of the proportion of Sanskrit words in the writer's vocabulary.
Solution.

Let p be the proportion of Sanskrit words in the writer's vocabulary. The corresponding
proportion in the sample is given as p = 0.64.

´
0.64 0.36 0.48
( )
 . .S E p = = = 0.012
1600 40
We know that almost whole of the distribution lies between 3s limits. Therefore, the confidence
interval is given by
P[p - 3S.E.(p)  p  p + 3 S.E.(p)] = 0.9973
Thus, the 99.73% confidence limits of p are 0.604 (= 0.64 - 3 ´ 0.012) and 0.676 (= 0.64 + 3 ´ 0.012)
respectively.
Hence, the proportion of Sanskrit words in the writer's vocabulary are between 60.4% to 67.6%.

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