Mathematics, 03.12.2020 19:50 tytybruce2
A researcher wants to construct a 90% confidence interval for the proportion of elementary school students in Seward County who receive free or reduced-price school lunches. A state-wide survey indicates that the proportion is 0.50. Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.1?
Answers: 1
is
![n = [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )](/tpl/images/0946/3247/c7923.png)
![n = [\frac{1.645 }{0.1} ]^2 * 0.50 (1 - 0.50 )](/tpl/images/0946/3247/f8b00.png)
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A researcher wants to construct a 90% confidence interval for the proportion of elementary school st...
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