From central limit theorem we know that
is a standard normal variate. From the discussion of an introduction to interval estimation we know that P (–1.96 < Z< 1.96) = 0.95 has the least possible range. With this inequality we can construct a 95% confidence interval estimate of the population mean if we replace Z by
In general if the margin of error in our interval estimate is (this is also referred to as the level of significance), then the level of confidence would be or . For example, if or 5% then or 95% would be the level of confidence. To have the shortest interval the margin of error is distributed equally on the two sides of the curve, as shown in the figure below:
From the above discussion we can write the following probability statement: . The 100% confidence interval estimate of the mean can be obtained by replacing Z with , thus
The 100% lower and upper confidence limits are, therefore, .
As a specific example, let , so that or 90%. The probability statement may thus be written as
[Since , ]
Consulting the normal table, . Hence we can write P [–1.64 <Z< 1.64] = 0.90.
Replacing Z with , we have
Upon simplification, this reduces to .
Therefore, the 90% confidence limits are: .
is usually not known; therefore, for large samples it can be replaced by S (the sample standard deviation) which may be calculated from the sample by using the following formula: .
Proceeding along the same lines, the 95 and 99 percent confidence limits may be stated as
Example: A random sample of 64 MS students achieved an average score of 60 with a standard deviation score of 15. Construct a 99% confidence interval estimation for the mean score of the entire class.
Solution: We have the following data:
The formula for 99% confidence interval may be written as
The lower confidence limit is
The upper confidence limit is
Hence, the 99% confidence interval estimate for the mean score will be , i.e., the mean score is between 55 and 65.