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The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the eighth grade level. In an earlier study, the population proportion was estimated to be 0.23.

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 98% confidence level with an error of at most 0.03? Round your answer up to the next integer.


Sagot :

Using the z-distribution, it is found that a sample size of 1066 is required.

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]\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 z-score that has a p-value of .

The margin of error is:

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

From the previous study, the estimate is of 0.23, hence [tex]\pi = 0.23[/tex].

98% confidence level, hence[tex]\alpha = 0.98[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.98}{2} = 0.99[/tex], so [tex]z = 2.327[/tex].

The sample size is n for which M = 0.03, hence:

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

[tex]0.03 = 2.327\sqrt{\frac{0.23(0.77)}{n}}[/tex]

[tex]0.03\sqrt{n} = 2.327\sqrt{0.23(0.77)}[/tex]

[tex]\sqrt{n} = \left(\frac{2.327\sqrt{0.23(0.77)}}{0.03}\right)[/tex]

[tex](\sqrt{n})^2 = \left(\frac{2.327\sqrt{0.23(0.77)}}{0.03}\right)^2[/tex]

[tex]n = 1065.5[/tex]

Rounding up, a sample size of 1066 is required.

A similar problem is given at https://brainly.com/question/12517818