In the week before and the week after a holiday, there were 10,000 total deaths,
ID: 3153581 • Letter: I
Question
In the week before and the week after a holiday, there were 10,000 total deaths, and 4970 of them occurred in the week before the holiday.
A) Construct a 90% 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?
No, because the proportion could easily equal 0.5. The interval is not less than 0.5 the week before the holiday. OR
Yes, because the proportion could not easily equal 0.5. The interval is substantially less than 0.5 the week before the holiday.
Explanation / Answer
a)
Note that
p^ = point estimate of the population proportion = x / n = 0.497
Also, we get the standard error of p, sp:
sp = sqrt[p^ (1 - p^) / n] = 0.00499991
Now, for the critical z,
alpha/2 = 0.05
Thus, z(alpha/2) = 1.644853627
Thus,
Margin of error = z(alpha/2)*sp = 0.00822412
lower bound = p^ - z(alpha/2) * sp = 0.48877588
upper bound = p^ + z(alpha/2) * sp = 0.50522412
Thus, the confidence interval is
( 0.48877588 , 0.50522412 ) [ANSWER]
*****************************
b)
As we can see, 0.5 is inside this interval. Hence,
OPTION A: No, because the proportion could easily equal 0.5. The interval is not less than 0.5 the week before the holiday. [ANSWER, A]