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Troyanec [42]
3 years ago
9

Reading proficiency: An educator wants to construct a 99.5% confidence interval for the proportion of elementary school children

in Colorado who are proficient in reading.
The results of a recent statewide test suggested that the proportion is 0.69.

Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.07?
Mathematics
1 answer:
Shkiper50 [21]3 years ago
4 0

Answer:

A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07

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 given by:

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

In this problem, we have that:

p = 0.69

99.5% confidence level

So \alpha = 0.005, z is the value of Z that has a pvalue of 1 - \frac{0.005}{2} = 0.9975, so Z = 2.81.

Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.07?

This is n when M = 0.07. So

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

0.07 = 2.81\sqrt{\frac{0.69*0.31}{n}}

0.07\sqrt{n} = 1.2996

\sqrt{n} = \frac{1.2996}{0.07}

\sqrt{n} = 18.5658

(\sqrt{n})^{2} = (18.5658)^{2}

n = 345

A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07

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