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In the week before and the week after a​ holiday, there were 10,000 total​ deaths, and 4958 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?

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Below 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

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