Basic Statistics

ESTIMATING THE DIFFERENCE BETWEEN TWO MEANS: UNPAIRED DATA 87

It is usual to assume in this problem that the variances of the two original popu-

lations are equal; the two equal variances will be called ua. If the first population

is approximately normally distributed, we know that is approximately normally

distributed, its mean is 1-11, and its standard deviation is a/fi = ./a = a/4 g,

where n1 = 16 is the size of the first group of infants. Similarly, X2 is approximately

normally distributed with mean 1-12 and standard deviation

u/fi2 = u/& = u/3 g

where n2 = 9 is the sample size of the second group of infants.
If all possible pairs of samples were drawn from the two populations and for each
pair - F2 is calculated, what kind of distribution would this statistic have? First,
the distribution is approximately normal; this does not seem surprising. Second,
the mean of the distribution is 1-11 - p2; this seems reasonable. Third, the standard
deviation is larger than the standard deviation of the 7, distribution and is also larger
than that of the 5?2 distribution; in fact, the variance for 5?1 - Fa is equal to the sum
of the two variances. This may be somewhat surprising; evidence to support it can
be obtained in a class exercise with the samples of cholesterol levels from Problem
2.1, At any rate, it has been proven mathematically that

ua u2

• X1 2 Xz n1 n2 ($+i)

gr;.,-j?z^2 - fl- + gl- = - + - = 02

Note that when the sample sizes of the two groups are equal, nl = 122 or simply n,

then CT$~ -xz = 2a2/n. Figure 7.3 illustrates the appearance of the distributions of

XI, X2, and XI - Fa.

_-

0 Irl -L(2 1112 Irl

Figure 7.3
differences.

Two distributions of sample means compared with the distribution of their

7.5.2

It can be shown mathematically, then, that for independent samples of size n1 and
n2 from normal populations with means 1-11 and 1-12 and with equal variances u2,

Confidence Intervals for p1 - p2: Known Variance