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 11,000 total deaths, and 5445 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?

Solution

a)

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.495          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.004767075          
              
Now, for the critical z,              
alpha/2 =   0.05          
Thus, z(alpha/2) =    1.644853627          
Thus,              
Margin of error = z(alpha/2)*sp =    0.00784114          
lower bound = p^ - z(alpha/2) * sp =   0.48715886          
upper bound = p^ + z(alpha/2) * sp =    0.50284114          
              
Thus, the confidence interval is              
              
(   0.48715886   ,   0.50284114   ) [ANSWER]

**********************

b)

No, because the interval includes 0.50. [ANSWER]