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The sample size needed is n(if a decimal number, round up to the next integer), considering the estimate of the proportion $\pi$ (if no previous estimate use 0.5) and M is the desired margin of error.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

The margin of error is of:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

Needed sample size:

The needed sample size is n. We have that:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$M = 1.96\sqrt{\frac{\pi(1-\pi)}{n}}$

$\sqrt{n}M = 1.96\sqrt{\pi(1-\pi)}$

$\sqrt{n} = \frac{1.96\sqrt{\pi(1-\pi)}}{M}$

$(\sqrt{n})^2 = (\frac{1.96\sqrt{\pi(1-\pi)}}{M})^2$

$n = (\frac{1.96\sqrt{\pi(1-\pi)}}{M})^2$

The sample size needed is n(if a decimal number, round up to the next integer), considering the estimate of the proportion $\pi$ (if no previous estimate use 0.5) and M is the desired margin of error.

4 0

3 years ago

You found the prime factorization of the number 72. Explain how you can check your answer.

ch4aika [34]

To find out whether a prime factorization is correct, verify that all the factors are indeed prime. Then multiply the factors. The product should equal 72.

5 0

3 years ago