Question

Recall in our discussion of the binomial distribution the research study that examined schoolchildren developing nausea and vomiting following holiday parties. The intent of this study was to calculate probabilities corresponding to a specified number of children becoming sick out of a given sample size. Recall also that the probability, i.e. the binomial parameter "p" defined as the probability of "success" for any individual, of a randomly selected schoolchild becoming sick was given.

Suppose you are now in a different reality, in which this binomial probability parameter p is now unknown to you but you are still interested in carrying out the original study described above, though you must first estimate p with a certain level of confidence. You obtain research funding to randomly sample 13 schoolchildren with an inclusion criterion that he/she must have recently attended a holiday party, and conduct a medical evaluation by a certified pediatrician. After anxiously awaiting your pediatrician colleague to complete her medical assessments, she emails you data contained in the following table.

Subject	Nausea and Vomiting?
1	0
2	1
3	0
4	1
5	0
6	0
7	0
8	0
9	0
10	0
11	0
12	0
13	1

What is the estimated 95% confidence interval (CI) of the proportion of schoolchildren developing nausea and vomiting following holiday parties?

Please note the following: 1) 0 and 1 are defined as no and yes, respectively, which is a typical coding scheme in Public Health; 2) you might calculate a CI that is different from any of the multiple choice options listed below due to rounding differences, therefore select the closest match; and 3) you may copy and paste the data into Excel to facilitate analysis.

Select one:

a. 0.002 to 0.460

b. 0.002 to 0.520

c. 0.002 to 0.494

d. 0.002 to 0.428

Answer 1

sample success x =		3
sample size          n=		13
sample proportion p̂ =x/n=		0.2308
std error se= √(p*(1-p)/n) =		0.1169
for 95 % CI value of z=		1.960
margin of error E=z*std error   =		0.2290
lower bound=p̂ -E                       =		0.002
Upper bound=p̂ +E                     =		0.460

option A is correct