Question

In the week before and the week after a​ holiday, there were 13000 total​ deaths, and 6461 of them occurred in the week before the holiday.

a. Construct a 90​% 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

n = Sample Size = 13000

p = Sample Proportion = 6461/13000 = 0.497

So,

q = 1- p = 0.503

= 0.10

From Table, critical values of Z = 1.645

So,

Confidence Interval:

0.497 (1.645 X 0.0044)

= 0.497 0.0072

= (0.4898 , 0.5042)

Confidence interval:

0.4898 < P < 0.5042

(b)

Since the confidence interval contains both values of less than 50% and greater than 50%, there does not appear to be any indication that people can temporarily postpone their death to suvive the holiday.