TESTS OF HYPOTHESIS FOR POPULATION PROPORTIONS 133The 95% confidence interval is then(.90 - .80) f. 1.96(.05) = .10 k .098
or .002 to .198.
Since the 95% confidence interval for difference in recovery rate is between .2%
and 19.8%, it may be practically 0, or it may be nearly 20%. Because both confidence
limits are positive, we decide that the difference is positive, so that the population
10-day recovery rate is higher for the first treatment than for the second. We say that
the difference between the two recovery rates is statistically significant, or that it is
a significant difference. In calling a sample difference such as .90 - .80 = .10 a
significant difference, we mean that it is large enough to enable us to conclude that
a population difference exists. Note that we are not asserting that the population
difference is large or important; we merely believe that it exists. In symbols, the 95%
confidence interval is
(PI - ~2) 1.96Jpi(l - pi)/ni + ~2(1 - ~2)/.2
The formula for the 95% confidence interval if a continuity correction factor is in-
cluded is
(PI -P2) f (1.96dP1(1 -Pi)/ni +P2(1 -p2)/n2 + 1/2(1/ni 4- l/n2))
10.6 TESTS OF HYPOTHESIS FOR POPULATION PROPORTIONSIf the sample size is large enough to use the normal curve, hypotheses concerning
population proportions may be tested in much the same way as were hypotheses
concerning population means. We will first give the tests for a single population
proportion, and then the test for the difference in two population proportions.
10.6.1 Tests of Hypothesis for a Single Population Proportion
Suppose, for example, based on considerable past experience, that it is known that the
proportion of the people with a given infectious disease who recover from it within
a specified time is .80 when treated with the standard medication. There is concern
that the medication may not always have the same success rate, due to changes in
the virus causing the disease. We want to test for possible changes in the patients
who have been treated recently and to find out whether or not T = .80. The null
hypothesis is that HO : TO = .80. We have data for 50 consecutive patients who have
been treated within the last year for the disease, and find that 35 recovered, so that
the sample proportion is p = 35/50 = .70. Under Ho, p is approximately normally
distributed with mean 30 and with standard deviationcp = Z/TO(~ - TO)/. = Jm = dZ@G = .0566
Since we wish to know whether T equals or does not equal 30, we will use a two-sided
test.