Test about a single proportion for categorical data

Ho: Pi = Pio vs. HA: Pi NA Pio

Wald Test: Z = Pi-hat - Pio/ sqrt (Pi-hat (1 - Pi-hat)/n)

Score Test : Z = Pi-hat - Pio/ sqrt (Pio (1 - Pio)/n)

Score test is slightly more powerful in distinguising evidence against Ho.


R-code for score test : prop.test (27, 922, p=0.02, correct =FALSE)

Here is p is the population probability. and 27 is the observed instance of some phenomena in population. 27/922 will give Pi-hat - the binomial estimate in sample.
Its a Chi-square test on 1 degree of freedom.

Here the p-value is 0.0440 - where the Ho is rejected and 27/922 = 0.029 is NA to 0.02.

Obtaining Confidence Intervals
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The Wald Confidence Interval for single proportion can be obtained by inverting the Wald test statistic but it falls into problem if n is small or Pi is small.

A Score CI can be used in this case.

The R-code is

Wald

library(Hmisc)

binconf(27,922,method="asymptotic")

PointEst Lower Upper
0.02928416 0.01840125 0.04016708


Score (Wilson):

binconf(27,922,method="wilson")

PointEst Lower Upper
0.02928416 0.02020271 0.04227177

prop.test also give the score CI.

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In case observed events are small in number and normality assumption is not valid.
One can use binomial exact test to calculate the p-value and CI.

binom.test(3, 58, p = 0.02,
+ alternative = "two.sided",
+ conf.level = 0.95)
Exact binomial test
data: 3 and 58
number of successes = 3, number of trials = 58,
p-value = 0.1101
alternative hypothesis: true probability of success
is not equal to 0.02
95 percent confidence interval:
0.01079648 0.14380463
sample estimates:
probability of success
0.05172414