Question

In the week before and the week after a​ holiday, there were 11 comma 000 total​ deaths, and 5489 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.

____ < p < ____

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

Solution :

a) The 95% confidence interval for population proportion is given as follows :

Where, p̂ is point estimate of true proportion, q̂ = 1 - p̂, n is sample size and Z(0.05/2) is critical z-value to construct 95% confidence interval.

Sample proportion of death in the week before the holiday is,

n = 11000

Using Z-table we get, Z(0.05/2) = 1.96

Hence, 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 is,

The 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 is,

0.490 < p < 0.508

b) Since, the 95% confidence interval also contains the value of greater than 0.50, therefore we can not say that there appear to be any indication that people can temporarily postpone their death to survive the​ holiday.