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

4946

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

Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=4994
Sample Size(n)=10000
Sample proportion = x/n =0.499
Confidence Interval = [ 0.499 ±Z a/2 ( Sqrt ( 0.499*0.501) /10000)]
= [ 0.499 - 1.64* Sqrt(0) , 0.499 + 1.64* Sqrt(0) ]
= [ 0.491,0.507]