*This file is part of a
program based on the Bio 4835 Biostatistics class taught at Kean University
in *

**B)
Confidence interval for the difference of two population
means
**

From each of two populations an independent random sample is drawn. Sample means, and , are calculated. The difference is

- which is an unbiased estimator of the difference between the two population means, - . The variance of the estimator is

( / ) + ( / ).

Assuming the populations are normally distributed, there are three situations where we would determine the 100(1- ) percent confidence interval for - .

a) where the population variances are known (use z)

b) where the population variances are unknown but equal (use

c) where the population variances are unknown but unequal (use

The concept of t' is noted here so that readers are aware of its existence but it will not be treated further in this narrative.

Illustrative Examples

When the population variances are known, the 100(1- ) percent confidence interval for - is given by

Example 6.4.1

A research team is interested in the difference between serum uric acid levels in patients with and without Down's syndrome. In a large hospital for the treatment of the mentally retarded, a sample of 12 individuals with Down's syndrome yielded a mean of = 4.5 mg/100 ml. In a general hospital a sample of 15 normal individuals of the same age and sex were found to have a mean value of = 3.4 mg/100 ml. If it is reasonable to assume that the two populations of values are normally distributed with variances equal to 1 and 1.5, find the 95 percent confidence interval for - .

(1) Given

= 12, = 4.5, = 1

= 15, = 3.4, = 1.5

(2) Calculations

- The point estimate for - is -

- = 4.5 - 3.4 = 1.1

- The standard error is

- The 95% confidence interval is

1.1 ± 1.96 (.4282)

(.26, 1.94)

*Discussion:* As this is a z-interval, we know that the correct
value of z to use is 1.96. We interpret this interval that the difference
between the two population means is 1.1 and we are 95% confident that the true
mean lies between 0.26 and 1.94.

**Situation b) Population variances are unknown but can be assumed to be equal
( t is used)
**

If it can be assumed that the population variances are equal then each sample variance is actually a point estimate of the same quantity. Therefore, we can combine the sample variances to form a pooled estimate.

The pooled estimated of the common variance is made using weighted averages. This means that each sample variance is weighted by its degrees of freedom.

The pooled estimate of the variance comes from the formula:

Standard error of the estimate

The standard error of the estimate is

The 100(1- ) confidence interval for - is

Example

(1) Given

= 13, = 21.0, = 4.9

= 17, = 12.1, = 5.6

(2) Calculations

- The point estimate for - is -

- = 21.0 - 12.1 = 8.9

- The pooled estimate of the variance is

- The standard error is

- The 95% confidence interval is

8.9 ± 2.0484 (1.9569)

8.9 ± 4.0085

(4.9, 12.9)

Discussion: The correct value of t to use for a 95% confidence interval
with 28 degrees of freedom is 2.0484. We interpret this interval that the
difference between the two population means is estimated to be 8.9 and we are
95% confident that the true value lies between 4.9 and 12.9.