Unit 2: statistical estimation

2.5 Confidence interval for a population mean


Case I. Sampling from a normally distributed population with known population variance
As already discussed in the previous unit, the sampling distribution of the mean taken form a normal distribution is also normal with mean  and variance .
Hence the variate is distributed as a standard normal with mean O and variance 1.
i.e. ~ N(0, 1).
From the area under the standard normal curve, one can observe that the probability that the value of the sample statistic falls in the range –2.58 and 2.58 us 0.99. i.e
P{-2.58 2.58} = 0.99
Or P{ -2.58 . } = 0.99
This means that it is very likely, the probability being 0.99 that the interval ( - 2.58 , + 2.58 ) will include . In other words, if a very large number of samples, each of size n, are taken from the population and if for each such sample the above interval is determined, then in about 99% of the cases the interval will include , while in the remaining 1% it will fail to do so. One will therefore, be justified in saying, on the basis of a given sample, that lies between - 2.58 and + 2.58 . , the limits being computed form the observations in hand. These are called 99% confidence limits to , 0.99 being the confidence coefficient.
Recalling that Z_ denotes the value of Z for which the area under the standard normal curve to its right is equal to , Analogously, Z_/2 denotes the value of Z for which the area to its right is /2, and -Z_/2 denotes the value of which the area to its left is /2.

Thus P(-Z_/2 < Z_/2) = 1-

 P{-Z_/2 < < Z_/2} = 1-
 P{ - Z/2 < < + Z_/2 } = 1-
Thus the (1-) 100% confidence interval for the population mean is given by:
Z_/2 .
For example the 95% confidence interval for the population mean is given by:
P{-Z_/2 < Z < Z_/2} = 1- = 0.95
i.e 1- = 0.95   = 0.05 and /2 = 0.025
Thus the confidence interval is
Z_/2 .  Z_0.025 and from the standard normal table, Z_0.025 = 1.96 thuse the confidence interval is 1.96 . i.e. the population mean is expected in the interval ( 1.96 , 1.96 ) with probability 0.95. i.e. we are 95% confident that lies in the interval ( - 1.96 , - 1.96 )