Question

1. if given the choice of drinking tea or coffee, what proportion of osu students would choose coffee? a random sample of osu students is surveyed, and a 95% confidence interval is constructed to estimate this unknown population proportion. the interval ends up being from 0.570 to 0.654. based on this interval, we know the sample proportion must have been 0.612 and the margin of error must have been 0.042. note that in the space below, we’d like you to show us how you arrived at your answers!

Explanation:

Step1: Calculate sample proportion

The sample proportion \( \hat{p} \) is the mid - point of the confidence interval. The formula for the mid - point of an interval \((a,b)\) is \(\hat{p}=\frac{a + b}{2}\). Here \(a = 0.570\) and \(b=0.654\).

$$ \hat{p}=\frac{0.570 + 0.654}{2}=\frac{1.224}{2}=0.612 $$

Step2: Calculate margin of error

The margin of error \(E\) is the distance from the sample proportion to either end of the confidence interval. We can use \(E=\hat{p}-a\) (or \(E = b-\hat{p}\)). Using \(E=\hat{p}-a\), with \(\hat{p}=0.612\) and \(a = 0.570\)

$$ E=0.612-0.570 = 0.042 $$

Answer:

The sample proportion is \(0.612\) and the margin of error is \(0.042\)