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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28-11-2016
Mathematics

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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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Аноним Аноним · Answer 1 · 2016-11-30T15:40:13+01:00

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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