Mathematics: G A researcher wants to construct a 95% confidence interval for the proportion of children aged 8-10 living in Atlanta who own a cell phone. Estimate the sample size needed if no estimate of p is available so that the confidence interval will have a margin of error of 0.02.

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G A researcher wants to construct a 95% confidence - id956099075, iilovejohnccena1022

Ответ:

bellad05

24.01.2022

Mathematics

The sample size needed is 2401.

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 of the interval is:

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

95% confidence level

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

Estimate the sample size needed if no estimate of p is available so that the confidence interval will have a margin of error of 0.02.

We have to find n, for which M = 0.02.

There is no estimate of p available, so we use $\pi = 0.5$

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

$0.02 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.02\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.02}$

$(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.02})^{2}$

n = 2401

The sample size needed is 2401.

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G A researcher wants to construct a 95% confidence interval for the proportion of children aged 8-10 living in Atlanta who own a cell phone. Estimate the sample size needed if no estimate of p is available so that the confidence interval will have a margin of error of 0.02.

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