Question

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%?

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.

Answer 2

Sure, let's break this down:

A) To find the probability that the estimate will differ by more than 2% from the true value of 12%, we can use the normal approximation to the binomial distribution since the sample size is large enough (1300 people) and the success-failure condition is met.

The standard deviation of the sampling distribution for proportions can be calculated using the formula:

\[ \sqrt{\frac{p(1-p)}{n}} \]

Where: - \( p \) is the true proportion of left-handed people in the population (0.12 in this case). - \( n \) is the sample size (1300).

Substitute the values:

\[ \sqrt{\frac{0.12 \times (1-0.12)}{1300}} \approx \sqrt{\frac{0.1056}{1300}} \approx \sqrt{0.00008123076923076923} \approx 0.009014 \]

Now, to find the probability that the estimate will differ by more than 2% from the true value (12%), you can use the standard normal distribution.

First, find the z-scores for the boundaries (2% above and below 12%):

Upper boundary: \( 0.12 + 0.02 = 0.14 \) Lower boundary: \( 0.12 - 0.02 = 0.10 \)

Now find the z-scores for these boundaries:

For the upper boundary: \[ Z = \frac{0.14 - 0.12}{0.009014} \approx \frac{0.02}{0.009014} \approx 2.22 \]

For the lower boundary: \[ Z = \frac{0.10 - 0.12}{0.009014} \approx \frac{-0.02}{0.009014} \approx -2.22 \]

Then, using a standard normal distribution table or calculator, find the probabilities associated with these z-scores.

The probability that the estimate will differ by more than 2% from the true value of 12% is the sum of the probabilities above and below these z-scores (i.e., the area beyond 2.22 standard deviations from the mean in both tails of the distribution).

B) To determine the minimal sample size to guarantee that the obtained estimate doesn't differ by more than 2% from the true value of 12% with a probability of at least 99%, you'll need to use the margin of error formula for proportions:

\[ \text{Margin of Error} = Z \times \sqrt{\frac{p(1-p)}{n}} \]

Rearrange the formula to solve for the sample size (\( n \)):

\[ n = \frac{Z^2 \times p(1-p)}{\text{Margin of Error}^2} \]

You'll need to find the appropriate Z-score corresponding to the desired confidence level (99%) and then substitute the other values to solve for the sample size (\( n \)).