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 12,000 total deaths, and 5970 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?

PLEASE SOLVE ALL PARTS ASAP

Solution

In the week before and the week after a holiday, there were 12,000 total deaths, and 5970 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.

The 95% confidence interval for the proportion of deaths is,

p^ - E < P < p^ + E

We can find this confidence interval using TI-83 calculator.

Steps :

STAT --> TESTS --> A : 1-PropZInt --> ENTER --> x = 5970 --> n = 12000 --> C-level = 0.95 --> Calculate

Output is :

(0.48855, 0.50645)

p^ = 0.4975

n = 12000

The 95% confidence interval for proportion of deaths in the week before the holiday to the total deaths in the week before and the week after the holiday is (0.48855, 0.50645)

We see that sample proportion (p^) is lie in the confidence interval.

–This would be no because the interval is not less than 0.5 (might close to it) and thus there is no proof that the number of deaths is different the week before and the week after a holiday

–If the confidence interval only contains proportions below 0.5 then there is evidence that the number of deaths the week before the holiday is different than the week after the holiday