Question

15. Survey Return Rate In a study of cell phone use and brain hemispheric dominance, an Internet survey was e-mailed to 5000 subjects randomly selected from an online group involved with ears. 717 surveys were returned. Construct a 90% confidence interval for the proportion of returned surveys.

Answer 1

Answer:

0.1352 ± 0.1516

Step-by-step explanation:

The best estimate of the population is calculated as:

P = x/n

P = 717 / 500

P = 0.1434

The best estimate of the population is 0.1434

The value of margin of error is calculated as:

Z-critical value = 90%, Confidence = 0.1434

E = Z(c) * $\sqrt{P(1-P)/n}$

E = 1.645 * $\sqrt{0.1434(1-0.1434)/5000}$

E = 0.0082

A 90% confidence interval for the population proportion of returned survey is:

90%CI = P ± E

90%CI = 0.1434 ± 0.0082

90%CI = 0.1352 ± 0.1516