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Unit 29: Estimation of Parameters: Criteria for Estimates
Notes
Example 8: A random sample of 500 pineapples was taken from a large consignment and
65 were found to be bad. Estimate the proportion of bad pineapples in the consignment and
obtain the standard error of the estimator. Deduce that the percentage of bad pineapples in the
consignment almost certainly lies between 8.5 and 17.5.
Solution.
Let p be the proportion of bad pine apples in the large consignment. Its estimate based on the
´
65 0.13 0.87
( ) ˆ
sample is ˆ p = = 0.13 with . .S E p = = 0.015
500 500
Thus, the 99.73% confidence limits of p are 0.13 ± 3 × 0.015, i.e., 0.085 and 0.175. Hence, the
proportion of bad pineapples in the given consignment almost certainly lies between 8.5% and
17.5%.
Remarks: The width of a confidence interval can be controlled in two ways:
(i) By adjusting the sample size: More is the sample size the narrower will be the interval.
(ii) By adjusting the level of confidence: Lower the level of confidence the narrower will be
the interval.
29.3.1 Determination of an Approximate Sample Size for a Given
Degree of Accuracy
Let us assume that we want to find the size of a sample to be taken from the population such that
the difference between sample mean and the population mean would not exceed a given value,
say Î, with a given level of confidence. In other words, we want to find n such that
P ( X - ) Î = 0.95 (say) .... (1)
Assuming that the sampling distribution of X is normal with mean m and . .S E = , we can
X
n
write
æ X ö æ X ö
-
-
P - 1.96 1.96 = 0.95 or P 1.96 = 0.95
ç ÷ ç ÷
è / n ø è / n ø
æ ö
-
or P X 1.96 × ÷ = 0.95 .... (2)
ç
è n ø
Comparing (1) and (2), we get
æ 1.96 ö 2 3.84 2
Î= 1.96× or n = ç ÷ =
n è Î ø Î 2
Remarks:
1. The sample size required with a maximum error of estimation, Î and with a given level of
2
z 2
confidence is n = , where z is the value of standard normal variate for a given level of
Î 2
2
confidence and is the variance of population.
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