Question

Severity  
  [1] 3 1 3 2 3 0 0 0 3 3 2 3 3 1 1 1 2 0 3 3 0 1 2 0 1 3 0 3 0 3 0 3 1 2 2 4 2 
 [38] 3 1 1 3 3 3 3 2 1 3 3 0 1 2 3 0 3 3 3 0 0 1 3 1 0 3 3 0 1 0 3 1 0 1 3 3 0 
 [75] 3 3 0 3 3 2 2 2 0 1 3 3 4 3 2 3 2 0 0 3 1 3 0 2 1 3 0 2 3 0 0 0 1 0 0 1 1
[112] 0 0 3 3 1 3 3 2 1 2 0 3 1 0 3 1 3 3 1 3 3 0 0 2 2 1 3 3 0 0 1 2 3 0 0 1 2
[149] 2 3 0 0 1 0 3 3 3 3 2 2 1 2 0 0 1 1 1 3 1 3 4 2 1 2 0 1 2 1 0 1 3 3 3 3 4
[186] 0 3 3 1 4 0 0 0 0 0 1 0 0 0 0
Levels: 0 1 2 3 4
> count <- summary(Severity)
> count 
 0  1  2  3  4
56 42 29 68  5
> prop.test(count[4],sum(count),p=.4,alternative='less',correct=FALSE)

        1-sample proportions test without continuity correction

data:  count[4] out of sum(count), null probability 0.4
X-squared = 3, df = 1, p-value = 0.04163
alternative hypothesis: true p is less than 0.4
95 percent confidence interval:
 0.0000000 0.3969047
sample estimates:   p 0.34 

Answer the following questions:

Why can't you use prop.test() to calculate a confidence interval using a z-distribution

Answer 1

This is because

The CI given by prop.test inverts the test to give a generally more accurate CI than the 'Wald score' CI in which SE is estimated by using p^ instead of p.

The Wald interval is asymptotically correct as promised, but can fail badly even for moderately large n.n. Difficulties are that it uses a normal approximation to binomial along with an estimated SE.