6. 10. Confidence interval for the variance of a normal

distribution

We may often need to estimate confidence interval for the population variance (or standard deviation).

Like every sample statistic, the sample variance is a random variable and it possesses a sampling distribution. If all the possible samples of a given size are taken from a population and their variances are calculated, the probability distribution of these variances is called the sampling distribution of the sample variance.

The random variable

follows a Chi–square distribution with degrees of freedom.

To find the formula for calculating Confidence intervals for the variance, it is necessary to introduce new notations.

We will denote the number for which (Fig.6.13)

Similarly, it follows that is defined as

And is defined as .

Then it follows that

In the end, as it shown in Fig.6.14

Using usual procedure we obtain confidence interval for population variance as

where follows a Chi – square distribution with degrees of freedom.

Example:

The variance in drug weights is very critical in the pharmaceutical industry. For a specific drug, with weights measured in grams, a sample of 18 units provided a sample variance of

a) Construct a 90 % confidence interval estimate for the population variance for the weights of this drug.

b) Construct a 90 % confidence interval estimate for the population standard deviation for the weights of this drug.

Solution:

From the given information

; ;

and for a 90 % confidence interval, ; .

confidence interval for is given by

a) From the Chi- square table we obtain that

;

After substitution we obtain

b) We can obtain the confidence interval for the population standard deviation by taking the positive square root of the two limits of the above confidence interval for the population variance. Thus, 90 % confidence interval estimate for the population standard deviation is

Hence, the standard deviation of all investigated drug is between 0.47 and 0.84 grams at a 90 % confidence level.

Exercises

1. The sample of 15 bus arrivals showed a sample variance at min.

a) Construct a 95 % confidence interval of the variance for the population of arrival times.

b) Suppose that the sample variance of had been obtained from a sample of 26 bus arrivals. Determine a 95 % confidence interval of the variance for the population of arrival times.

2. From production process a random sample of 25 a certain brand of light bulbs was taken. The variance of the lives of these bulbs was found to be 4710 hours. Assume that the lives of all such bulbs are approximately normally distributed.

a) Make a 99 % confidence interval for the variance and standard deviation of the lives of all such bulbs.

3. The time required to complete a certain operation by the sample of 25 employees of auditing company has a standard deviation of 3.1 min. Construct a 98 % confidence interval for .

4. From a data set of n=10 observation, one has calculated the 95 % confidence interval for and obtained the result (0.81; 2.15).

Calculate a 90 % confidence interval for .

5. Given the sample data

12, 18, 9, 15, 14

Construct a 95 % confidence interval for .

6. A sample of 7 observations taken from a population produced the following data

10, 8, 13, 15, 6, 8, 13

Make the 98 % confidence intervals for the population variance and standard deviation.

7. A random sample of 25 customers taken from the bank gave the variance of the waiting times equal to 7.9 min. Construct 99 % confidence intervals for the population variance.

8. Suppose that based on a random sample of size 10 from a normal distribution, one has found the 95 % confidence interval for the population mean to be (36.2; 45.8). Using this result determine a 95 % confidence interval for the population standard deviation.

Answers

1. a) (2.25; 10.44); b) (2.15; 6.67); 2. a) (2481.1; 11429.7) and (49.8; 106.9); 3. (2.32; 4.61) 4. (0.86; 1.94); 5. (2.25; 9.66); 6. (3.91; 75.36) and

(1.98; 8.68); 7. (4.16; 19.18); 8. (4.62; 12.25).