In 2015, approximately 136,943,000 people visited an emergency room in the United States. Of these, approximately 12,263,000 were subsequently admitted to the hospital. Determine an interval with 99% confidence for the number (how many) of people visiting an emergency room who were subsequently admitted to the hospital.
A) 1,097,434 to 1,098,826
B) 1,097,543 to 1,098,716
C) 1,098,129 to 1,098,129
D) 1,097,359 to 1,098,900

Question

Quizplus · Accepted Answer

We can use the formula for a confidence interval of a binomial proportion:

p ± z*sqrt((p*(1-p))/n)

where p is the sample proportion (number of admitted patients divided by total number of ER visits), n is the sample size (total number of ER visits), and z* is the critical value from the standard normal distribution for the desired level of confidence (0.99 in this case).

Plugging in the values, we get:

p = 12,263,000 / 136,943,000 = 0.0896
n = 136,943,000
z* = 2.576 (from standard normal distribution table for 99% confidence)

p ± z*sqrt((p*(1-p))/n) = 0.0896 ± 2.576*sqrt((0.0896*(1-0.0896))/136,943,000) = 0.0896 ± 0.00026

Therefore, the 99% confidence interval is [0.08934, 0.08986]. Multiplying this by the total number of ER visits, we get:

[0.08934*136,943,000, 0.08986*136,943,000] = [12,263,159, 12,276,558]

Rounding to the nearest integer, we get the interval [12,263,000, 12,276,000].

Only option D contains this interval, thus the best choice is D.

In 2015, Approximately 136,943,000 People Visited an Emergency Room in the United