Question

In a study of cell phone use and brain hemispheric dominance, an Internet survey was e-mailed to 2457 subjects randomly selected from an online group involved with ears. 935 surveys were returned. Construct a 95% confidence interval for the proportion of returned surveys.Find the best point estimate of the population proportion p.Identify the value of the margin of error E.Construct the confidence intreval.

Answer 1

Answer: a) Margin of error = 0.19, b) interval = (0.19,0.57)

Step-by-step explanation:

Since we have given that

n = 2457

x = 935

So, $p=\dfrac{935}{2457}=0.38$

At 95% confidence , z = 1.96

First we will find margin of error.

So, Margin of error is given by

$z\sqrt{\dfrac{p(1-p)}{n}}\\\\=1.96\times \sqrt{\dfrac{0.38\times 0.62}{2457}}\\\\=0.19$

95% confidence interval is given by

$p\pm \text{margin of error}\\\\=0.38\pm 0.19\\\\=(0.38-0.19,0.38+0.19)\\\\=(0.19,0.57)$

Hence, a) Margin of error = 0.19, b) interval = (0.19,0.57)