Question

Recall in our discussion of the binomial distribution the research study that examined schoolchildren developing nausea...

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 40 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 1 2 1 3 1 4 0 5 1 6 1 7 1 8 0 9 0 10 0 11 1 12 1 13 1 14 1 15 0 16 0 17 0 18 0 19 0 20 0 21 1 22 0 23 0 24 0 25 0 26 1 27 1 28 0 29 0 30 0 31 0 32 1 33 0 34 0 35 0 36 0 37 1 38 1 39 0 40 0 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. 0.248 to 0.552 b. 0.211 to 0.513 c. 0.210 to 0.472 d. 0.263 to 0.596 Question 21 A research study wishes to examine the proportion of hypertensive individuals among three different groups of exercises: marathon runners, yoga, and CrossFit. Of the 78 marathon runners, 14 are hypertensive. Of the 63 yoga practitioners, 6 are hypertensive. And there are 16 hypertensive subjects among the 54 CrossFit athletes. What is the computed test statistic based on this data? Select one: a. 6.98 b. 7.83 c. 6.23 d. 8.75

Homework Answers

Answer #1

From the given data set, we observed that the estimate of proportion is

1)

Subject Nausea and Vomiting
Yes No
Email 16 24

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

3)

Answer is a. 0.248 to 0.552

Q21

We can used chi-square test to examine the proportion of hypertensive individuals among three different groups of exercises.

Hypertensive Total
Exercises Yes No
Marathon 14 64 78
Yoga 6 57 63
CrossFit 16 38 54
Hypertensive
Exercises Yes No
Marathon 0.179487 0.820513
Yoga 0.095238 0.904762
CrossFit 0.296296 0.703704
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