In the week before and the week after a holiday there were 1

In the week before and the week after a holiday. there were 10.000 total deaths, and 4990 of them occurred in the week before the holiday. a. Construct a 95% 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?

Solution

(a) Given a=1-0.95=0.05, Z(0.025) = 1.96 (from standard normal table)

p=4990/10000 = 0.499

So the lower bound is

p - Z*sqrt(p*(1-p)/n) = 0.499 - 1.96*sqrt(0.499*(1-0.499)/10000) =0.4892

So the upper bound is

p + Z*sqrt(p*(1-p)/n) = 0.499 + 1.96*sqrt(0.499*(1-0.499)/10000) =0.5088

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(b)Since the interval includes 0.5, there does not appear to be any indication that people can temporarily postpone their death to survive the holiday.