in the week before and the In the week before and the week after a holiday, ther

Question

in the week before and the In the week before and the week after a holiday, there were 13.000 total deaths, and 6409 of them occurred in the 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? b. Based on the result does there appear to be any indication that people can temporarily postpone their death to the holiday? No. because the proportion could easily equal 0 5 The interval is not less than 0 5 the week before the holiday 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)

n = 13000    

p = 6409/13000 = 0.493    

% = 95    

Standard Error, SE = {p(1 - p)/n} =    (0.493(1 - 0.493))/13000 = 0.00438486

z- score = 1.959963985    

Width of the confidence interval = z * SE =     1.95996398454005 * 0.00438486031704546 = 0.00859417

Lower Limit of the confidence interval = P - width =     0.493 - 0.00859416829864798 = 0.48440583

Upper Limit of the confidence interval = P + width =     0.493 + 0.00859416829864798 = 0.50159417

The 95% confidence interval is [0.484, 0.502]

(b) Option A

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