Statistics problem regarding confidence intervals

Question

There were 10, 000 total deaths and 4,981 occured the week before the holiday.  Construct a 95% confidence interval estimate of the proporation of details in the week efore and the week after the holdiay. Based on the results does ther appear to be any indication that people can temporarely postpone thier death to survive the holiday.

Answer 1

Proportion of deaths, p, before holiday=4981/10000=0.4981. So 1-p=0.5019. The standard error is sqrt(0.4991*0.5019/10000)=0.005. a=1-(confidence level)=1-0.95=0.05 and the critical probability is 1-1/2(0.05)=0.975. This corresponds to z-score of 1.96 in the normal distribution table. To get the margin of error we multiply it by the standard error=1.96*0.005=0.0098. So the confidence interval is 0.4981+0.0098 or the interval 0.4883 to 0.5079 that is the number of deaths is 95% likely to fall between these limits: 4883 and 5079. 0.5019 equivalent to 5019 deaths in 10,000 in or after the holiday is within the interval, so there is only a 5% chance that "postponing" death is indicated. Therefore we can be confident to the degree of 95% that people will survive their holiday.