Question

In the week before and the week after a​ holiday, there were 10 comma 000 total​ deaths, and 4950 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?

Answer 1

a)

p̂ = X / n = 4950/10000 = 0.495
p̂ ± Z(α/2) √( (p * q) / n)
0.495 ± Z(0.05/2) √( (0.495 * 0.505) / 10000)
From Z table, Z(α/2) = Z(0.05/2) = 1.96
Lower Limit = 0.495 - 1.96 * √( (0.495 * 0.505) / 10000) = 0.485
upper Limit = 0.495 + 1.96 * √( (0.495 * 0.505) / 10000) = 0.505

95% Confidence interval is ( 0.485 , 0.505 )

b)

Since confidence interval contains 0.50, it does not appear to be any indication that

people can temporarily postpone their death to survive the​ holiday