Question

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???

Answer 1

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

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