In a week before and the week after a holiday there were 130

In a week before and the week after a holiday, there were 13,000 total daeths, and 6435 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 apear to be any indication that people can temporarily postpone their death to survive the holiday?

a. ___< p <___ Round to 3 decimal places.

b. Based on the result does there appear to be any indication that people can temporarily postpone their death to survive the holiday? Yes or no

Solution

Here is the solution:

p-hat = 5445 / 11000 = 0.495

For a 95% confidence interval, use z = 1.96

The confidence interval is then:

p-hat - z(?(p-hat)(1 - p-hat) / n) < p < p-hat + z(?(p-hat)(1 - p-hat) / n)

0.495 - (1.96)(?(0.495)(1 - 0.495) / 11000) < p < 0.495 + (1.96)(?(0.495)(1 - 0.495) / 11000)

0.486 < p < 0.504

If people cannot temporarily postpone their death to survive the holiday, then the proportion of deaths in the week before the holiday to the total deaths in the week before and after the holiday should be less than 0.5. Since the confidence interval includes the value 0.5, there is not sufficient evidence to indicate that people can postpone their death to survive the holiday.