Question

SHOW WORKING STEPS. THANKS.

5. 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 29 schoolchildren and 47 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 0 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 0 15 1 16 0 17 0 18 0 19 1 20 0 21 0 22 0 23 0 24 0 25 1 26 0 27 1 28 0 29 1

Table 2. Adults Subject Nausea and Vomiting? 1 1 2 0 3 1 4 0 5 0 6 0 7 0 8 0 9 1 10 1 11 0 12 0 13 0 14 0 15 0 16 0 17 0 18 0 19 0 20 0 21 0 22 0 23 0 24 0 25 0 26 1 27 1 28 1 29 1 30 1 31 0 32 0 33 0 34 1 35 0 36 0 37 0 38 0 39 1 40 0 41 0 42 1 43 0 44 0 45 0 46 1 47 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. -0.2336 to 0.2278

b. -0.2077 to 0.2062

c. -0.1952 to 0.2289

d. -0.2388 to 0.1918

Answer 1

sample #1   ----->   schoolchildren
first sample size,     n1=   29          
number of successes, sample 1 =     x1=   8          
proportion success of sample 1 , p̂1=   x1/n1=   0.2759          
                  
sample #2   ----->   Adults
second sample size,     n2 =    47          
number of successes, sample 2 =     x2 =    13          
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.277          
                  
difference in sample proportions, p̂1 - p̂2 =     0.2759   -   0.2766   =   -0.0007

level of significance, α =   0.05              
Z critical value =   Z α/2 =    1.960   [excel function: =normsinv(α/2)      
                  
Std error , SE =    SQRT(p̂1 * (1 - p̂1)/n1 + p̂2 * (1-p̂2)/n2) =     0.1056          
margin of error , E = Z*SE =    1.960   *   0.1056   =   0.2069
                  
confidence interval is                   
lower limit = (p̂1 - p̂2) - E =    -0.001   -   0.2069   =   -0.2077
upper limit = (p̂1 - p̂2) + E =    -0.001   +   0.2069   =   0.2062
                  
so, confidence interval is (   -0.2077   < p1 - p2 <   0.2062   )  

b. -0.2077 to 0.2062

Thanks in advance!

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