Question

In the week before and the week after a​ holiday, there were
10 comma 00010,000
total​ deaths, and
49834983
of them occurred in the week before the holiday.
a. Construct a
9090​%
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?
a.
nothingless than<pless than<nothing
​(Round to three decimal places as​ needed.)

Answer 1

Solution :

Given that,

n = 10000

x = 4983

= x / n = 4983 / 10000 = 0.498

1 - = 1 - 0.498 = 0.502

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.645 * (((0.498 * 0.502) / 10000)

= 0.008

A 90% confidence interval for population proportion p is ,

- E < P < + E

0.498 - 0.008 < p < 0.498 + 0.008

0.490 < p < 0.506

There appear to be any indication that people can temporarily postpone their death

to survive the​ holiday .