Question #350956

A nutritionist wants to know the population proportion of grade 11 students who eat vegetables . Pegged at a confidence of 95% survey among 1200 respondents was conducted and 200 said that they eat vegetables. What is the margin of error???

Expert's answer

The critical value for "\\alpha = 0.05" is "z_c = z_{1-\\alpha\/2} = 1.96."

"ME=z_c\\times\\sqrt{\\dfrac{\\hat{p}(1-\\hat{p}}{n}}""\\hat{p}=\\dfrac{x}{n}=\\dfrac{200}{1200}=\\dfrac{1}{6}""ME=1.96\\times\\sqrt{\\dfrac{\\dfrac{1}{6}(1-\\dfrac{1}{6})}{1200}}\\approx0.021"
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