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A sample of 1300 people is to be used to estimate the proportion of left-handed people. Suppose the true proportion of left-handed people in the population is 12% (which is unknown to the organizers of the observational study).

A) What is the probability that as a result of the survey, the estimate will differ by more than 2% from the true value 12%? Use the normal approximation to solve this problem.

B) What should be the minimal sample size to guarantee that the obtained estimate doesn't differ by more than 2% from the true value 12% with probability at least 99%?

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23 сентября, 02:58

Статистика!

A sample of 1300 people is to be used to estimate the proportion of left-handed people. Suppose the true proportion of left-handed people in the population is 12% (which is unknown to the organizers of the observational study).

A) What is the probability that as a result of the survey, the estimate will differ by more than 2% from the true value 12%? Use the normal approximation to solve this problem.

B) What should be the minimal sample size to guarantee that the obtained estimate doesn't differ by more than 2% from the true value 12% with probability at least 99%?

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Филофей · Answer 1

Биномиальное распределение - стремится к нормальному при больших n

Дисперсия D = npq = 1300*0.12*0.88=137.28

сигма = √D = 11.72

два процента от тысячи трехсот это двадцать шесть или 26/11.72 = 2.22 сигмы. Смотрим по таблице нормального распределения - это вероятность 0.9736

По той же таблице смотрим сколько сигм вероятность 0.99

- это 2.58 сигмы.

0.02*N = 2.58√ (Npq)

N=1758 - округляем в большую сторону.