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.

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 [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

The margin of error is:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

99% confidence level

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

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 [tex]\pi = 0.5[/tex]. So

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

[tex]0.06 = 2.575\sqrt{\frac{0.5*0.5}{n}}[/tex]

[tex]0.06\sqrt{n} = 2.575*0.5[/tex]

[tex]\sqrt{n} = \frac{2.575*0.5}{0.06}[/tex]

[tex](\sqrt{n})^2 = (\frac{2.575*0.5}{0.06})^2[/tex]

[tex]n = 460.5[/tex]

Rounding up

The minimum sample size required is 461.

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