Question

Use the sample data and confidence level given below to complete parts (a) through (d).

In a study of cell phone use and brain hemispheric dominance, an Internet survey was e-mailed to 2356 subjects randomly selected from an online group involved with ears. 1161

surveys were returned. Construct a 90% confidence interval for the proportion of returned surveys.

a) Find the best point estimate of the population proportion p.

(Round to three decimal places as needed.)

b) Identify the value of the margin of error E.

(Round to three decimal places as needed.)

C. Construct a confidence intervel

Answer #1

Solution :

Given that,

(a)

Point estimate = sample proportion = = x / n = 1161 / 2356 = 0.493

Z_{/2}
= 1.645

(b)

Margin of error = E = Z_{
/ 2} *
((
* (1 -
)) / n)

= 1.645 * (((0.493 * 0.507) / 2356)

**= 0.017**

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.493 - 0.017 < p < 0.493 + 0.017

0.476 < p < 0.510

**(0.476 , 0.510)**

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