Confidence interval and confidence level

Since interval estimators have been described as “likely” to contain the true, but unknown value of the population parameters, it is necessary to phrase such term as probability statement.

Suppose that a random sample is selected and based on the sample information, it is possible to find two random variables a and b. Then interval extending from a and b either includes the population parameter or it does not contain population parameter. However, suppose that the random samples are repeatedly selected from the population and similar intervals are found. In the long run, a certain percentage of this interval will contain the unknown value. According to the frequency concept of probability, an interpretation of such intervals follows:

If the population is repeatedly samples and intervals calculated, then in the long run 90% (or some other percentages) of the intervals would contain the true value of the unknown parameter. The interval from a and b is said to be 90% (or some other percentages) confidence interval estimator for population parameters.

Definition:

Let be unknown parameter. Suppose that based on sample information, random variables a and b are found such that

,

where -is any number between 0 and 1.

The interval from a to b is called confidence interval for .

The quantity is called the confidence level of the interval.

If the population were repeatedly sampled a very large number of times, the true value of the parameter would be contained in of intervals calculated this way.

The confidence interval calculated in this way is written as with confidence.

Let us find the confidence intervals with any required confidence level , where is any number such that .

We will use the notation for the number such that

.

A notation indicates the value in the standard normal table cuts off a right tail area of . (Fig. 6.1).

For example, if , then .

So,

and from the standard normal distribution table we obtain .

Therefore

.

Now suppose that a confidence interval is required. (Fig.6.2).

We have

By the symmetry about the mean

.

And it follows that

,

where the random variable Z follows a standard normal distribution.

Example:

Find the value of if .

Solution:

.

.

and

(Fig. 6.3).

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