Unit 2: statistical estimation

2.5 Confidence interval estimation
As defined earlier, interval estimate states the range within which a population parameter probably lies. The interval within which a population parameter is expected to occur is called a confidence interval. Suppose we want to estimate a population parameter  on the basis of the sample observations we compute two numbers L and R, corresponding to a pre assigned probability 1-  so that,
P(L  R) = 1 - .
This procedure of estimating  is known as interval estimation. The pre assigned probability 1 -  indicates the degree of confidence we like to place in the estimation procedure and is known as confidence co-efficient. The quantities L and R are known as lower and upper confidence limits respectively. The quantity W = R – L is known as the length of the confidence interval. For example, the confidence interval for the population mean is the interval that has a high probability of containing the population mean, . Two confidence intervals are used extensively: the 95 percent confidence interval and the 99 percent confidence interval. Other confidence intervals may also be used.
The 95 percent confidence interval means that about 95 percent of the similarly constructed intervals will contain the parameter being estimated. If we use the 99 percent level of confidence, then we expect about 99 percent of the intervals to contain the parameter being estimated.
Another interpretation of the 95 percent confidence interval is that 95 percent of the sample means for a specified sample size will be within 1.96 standard deviations of the hypothesized population mean. Similarly, for a 99 percent confidence interval, 99 percent of the sample means will be within 2.58 standard deviations of the hypothesized population mean.









-2.58 -1.96 0 1.96 2.58

Z value


95%
99%

Where the values 1.96 and 2.58 are the Z values corresponding to the 95 percent and 99% level of confidence. i.e. p{-1.96Z1.96} = 0.95 and P{-2.58  Z  2.58} = 0.99 or the area under the standard normal between –1.96 and 1.96 equals 0.95 and the area between –2.58 and 2.58 is 0.99. Since normal distribution is symmetric, the area to the right and left of the mean is equal. For a 95 percent confidence level, the area to the right and left of the mean is = 0.4750.

As mentioned earlier, in interval estimation we want to find an interval with lower end point L and upper end point R so that the probability or confidence level is (1 - ) that it will contain the unknown population parameter  i.e. P.(L    R) = 1 -
Where L and R are the lower and upper confidence limits and 1 -  is the associated probability or confidence level of estimation. For a 95 percent confidence level, the probability is 0.95 that the population parameter is between L and R. i.e. P(L    R) = 1 - = 0.95.
If we solve for , 1 -  = 0.95   = 0.05. Which is the luck of confidence. i.e. if we are 95 percent confident then  = 0.05 = 5 percent the luck of confidence which is also known as level of significance. For a 99 percent confidence interval, equating 1- = 0.99  = 0.01. Thus for a 99 percent confidence interval,  (which is level of significance) = 0.01.
Let Z be a standard normal variate. Let Z be a number so that for a given probability , P(Z>Z) = . Then Z will be referred to as the upper  percent point.
For example, if  = 0.05 we have,
P(Z>Z_0.05 ) = 0.05

area =  = 0.05

Z_0.05

0

5
Knowing that the area under the normal curve equals 1 and the area to right of Z = 0 equals 0.5, then the area between Z = 0 and Z_0.05 = 0.5 – 0.05 = 0.45

thus P(Z>Z_0.05) = 0.05
 0.5 – A (Z_0.05) = 0.05
 A (Z_0.05) = 0.45
Thus from the table of standard normal variate Z_0.05 = 1.64.
A (Z_0.05) is the area under the curve form 0 to Z_0.05.
: . If  = 0.05 then Z = 1.64.
Similarly, if  = 0.05, then /2 = 0.025
And Z_/2 = Z_0.025 = P(Z>Z_0.025) = 0.025
 - A(Z_0.025) = 0.025
 A (Z_0.025) = 0.4750.
i.e. the area between 0 and Z_0.025 = 0.4750 and reading form the table, if the area under the curve = 0.4750, then the Z value = 1.96. Thus Z_/2 = Z_0.025 = 1.96 for  = 0.05 and if  = 0.01, then Z_/2 = Z_0.005 is the P(Z > Z_/2) = /2
 P(Z > Z_0.005) = 0.005
- A(Z_0.005) = 0.005
 A(Z_0.005) = 0.4950
i.e the area between Z = 0 and Z0.005 = 0.4950 thus reading from the table of standard normal, Z_0.005 = 2.58

As a summary;
If  = 5% = 0.05 then confidence level is 95% and
/2 = 0.025 and Z/2 = Z_0.025 = 1.96
If  = 1% = 0.01 then confidence level is 99% and
/2 = 0.005 and Z_/2 = Z_0.005 = 2.58