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 _____________ and the margin of error must have been _____________.

Explanation:

Step1: Calculate the 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 the margin of error

The margin of error \(E\) is half of the length of the confidence interval.
The length of the confidence interval is \(b - a\).
\(E=\frac{b - a}{2}\)
Substitute \(a = 0.570\) and \(b = 0.654\)
\(E=\frac{0.654-0.570}{2}=\frac{0.084}{2}=0.042\)

Answer:

The sample proportion must have been \(0.612\) and the margin of error must have been \(0.042\)