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. Furthermore, you would also like to collect data from adults to examine the difference between the proportion with nausea and vomiting following holiday parties of schoolchildren and adults, which will reflect any possible age differences in becoming sick. You obtain research funding to randomly sample 49 schoolchildren and 42 adults with an inclusion criterion that a given participant must have recently attended a holiday party, and conduct a medical evaluation by a certified pediatrician and general practitioner for the schoolchildren and adults, respectively. After anxiously awaiting your medical colleagues to complete their medical assessments, they email you data contained in the following tables.

Table 1. Schoolchildren Subject Nausea and Vomiting? 1 1 2 0 3 0 4 0 5 0 6 0 7 0 8 1 9 1 10 0 11 0 12 0 13 0 14 1 15 0 16 0 17 0 18 1 19 0 20 1 21 1 22 0 23 0 24 0 25 0 26 1 27 0 28 0 29 0 30 1 31 0 32 1 33 0 34 1 35 0 36 1 37 1 38 0 39 0 40 0 41 1 42 0 43 1 44 1 45 0 46 0 47 1 48 1 49 0

Table 2. Adults Subject Nausea and Vomiting? 1 1 2 0 3 0 4 0 5 0 6 1 7 0 8 1 9 0 10 1 11 0 12 0 13 0 14 1 15 0 16 1 17 0 18 0 19 0 20 0 21 1 22 0 23 1 24 1 25 0 26 0 27 0 28 0 29 0 30 1 31 0 32 0 33 1 34 0 35 0 36 0 37 0 38 1 39 1 40 0 41 0 42 0

What is the estimated 95% confidence interval (CI) of the difference in proportions between schoolchildren and adults developing nausea and vomiting following holiday parties? Assign groups 1 and 2 to be schoolchildren and adults, respectively.

Please note the following: 1) in practice, you as the analyst decide how to assign groups 1 and 2 and subsequently interpret the results appropriately in the context of the data, though for the purposes of this exercise the groups are assigned for you; 2) 0 and 1 are defined as no and yes, respectively, which is a typical coding scheme in Public Health; 3) 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 4) you may copy and paste the data into Excel to facilitate analysis.

Select one:

a. -0.1543 to 0.2244

b. -0.1365 to 0.2522

c. -0.1208 to 0.2837

d. -0.1529 to 0.2900

Answer 1

	A		B
x1                =	18	x2                =	13
p̂1=x1/n1 =	0.3673	p̂2=x2/n2 =	0.3095
n1                       =	49	n2                       =	42
estimated difference in proportion   =p̂1-p̂2   =			0.0578
std error Se =√(p̂1*(1-p̂1)/n1+p̂2*(1-p̂2)/n2) =			0.0992
for 95 % CI value of z=		1.960
margin of error E=z*std error =		0.1943
lower bound=(p̂1-p̂2)-E=		-0.1365
Upper bound=(p̂1-p̂2)+E=		0.2522

from above 95% confidence interval for difference in population proportion =(-0.1365 ,0.2522)

option B is correct