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Lapatulllka [165]
2 years ago
8

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

ighth grade level. In an earlier study, the population proportion was estimated to be 0.21. How large a sample would be required in order to estimate the fraction of tenth graders reading at or below the eighth grade level at the 90% confidence level with an error of at most 0.03
Mathematics
1 answer:
Korolek [52]2 years ago
8 0

Answer:

A sample of 499 is needed.

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 given by:

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

In this question, we have that:

\pi = 0.21

90% confidence level

So \alpha = 0.1, z is the value of Z that has a pvalue of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.

How large a sample would be required in order to estimate the fraction of tenth graders reading at or below the eighth grade level at the 90% confidence level with an error of at most 0.03

We need a sample of n, which is found when M = 0.03. So

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

0.03 = 1.645\sqrt{\frac{0.21*0.79}{n}}

0.03\sqrt{n} = 1.645\sqrt{0.21*0.79}

\sqrt{n} = \frac{1.645\sqrt{0.21*0.79}}{0.03}

(\sqrt{n})^2 = (\frac{1.645\sqrt{0.21*0.79}}{0.03})^2

n = 498.81

Rounding up

A sample of 499 is needed.

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