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In the week before and the week after a​ holiday, there were 10 comma 00010,000 total​ deaths, and 49964996 of them occurred in the week before the holiday.

a. Construct a 9595​% confidence interval estimate of the proportion of deaths in the week before the holiday to the total deaths in the week before and the week after the holiday.
b. Based on the​ result, does there appear to be any indication that people can temporarily postpone their death to survive the​ holiday?

User Hbgamra
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1 Answer

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a. The proportion of deaths in the week before the holiday to the total deaths in the week before and the week after the holiday is found from:

(4996)/(10000)=0.4996
The 95% confidence interval is found from:

0.4996\pm1.96 \sqrt{(0.4996(1-0.4996))/(10000)}
CI = 0.4898, 0.5094

b. There is no strong indication that people can temporarily postpone their death to survive the​ holiday.
User Jonatanes
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