Question

a countrys tax collection agency reported that 92% of individual tax returns were filed electronically in 2017. a rand was selected. from this sample, 181 were filed electronically. complete parts a through c.
a. construct a 90% confidence interval to estimate the actual proportion of taxpayers who filed electronically in 2018. the confidence interval has a lower limit of 0.822 and an upper limit of 0.902. (round to three decimal places as needed.)
b. what is the margin of error for this sample? the margin of error is □. (round to three decimal places as needed.)

Explanation:

Step1: Recall the formula for margin of error

The margin of error \(E\) for a confidence interval of proportion is given by \(E=\text{Upper limit}-\text{Sample proportion}\) (or \(E = \text{Sample proportion}-\text{Lower limit}\)). The sample proportion \(\hat{p}=\frac{181}{n}\). But since we know the confidence interval \((L = 0.822,U=0.902)\), we can use \(E=\frac{U - L}{2}\)

Step2: Calculate the margin of error

Substitute \(L = 0.822\) and \(U = 0.902\) into the formula \(E=\frac{U - L}{2}\).

$$E=\frac{0.902-0.822}{2}=\frac{0.08}{2}=0.04$$

Answer:

The margin of error is \(0.040\)