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alukav5142 [94]
2 years ago
5

A random sample of elementary school children in New York state is to be selected to estimate the proportion p who have received

a medical examination during the past year. An interval estimate of the proportion p with a margin of error of 0.06 0.06 and 99% 99 % confidence is required.
Mathematics
1 answer:
Anastaziya [24]2 years ago
3 0

Answer:

The minimum sample size required is 461.

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 zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

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

99% confidence level

So \alpha = 0.01, z is the value of Z that has a pvalue of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.

An interval estimate of the proportion p with a margin of error of 0.06. What is the minimum sample size required?

The minimum sample size required is n, which is found when M = 0.06.

We don't have an estimate for the true proportion, which means that we use \pi = 0.5. So

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

0.06 = 2.575\sqrt{\frac{0.5*0.5}{n}}

0.06\sqrt{n} = 2.575*0.5

\sqrt{n} = \frac{2.575*0.5}{0.06}

(\sqrt{n})^2 = (\frac{2.575*0.5}{0.06})^2

n = 460.5

Rounding up

The minimum sample size required is 461.

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