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 2346 subjects randomly selected from an online group involved with ears. 1070 surveys were returned. Construct a 90​% confidence interval for the proportion of returned surveys. Click the icon to view a table of z scores. ​a) Find the best point estimate of the population proportion p.

b. margin of error

c. construct confidence of error

Answer 1

Solution :

Given that,

n = 2346

x = 1070

= x / n = 1070 / 2346 = 0.456

a)Point estimate = 0.456

1 - = 1 - 0.456 = 0.544

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

Z/2 = Z0.05 = 1.645

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.645 * (((0.456 * 0.544) / 2346)

b) Margin of error = 0.017

A 90% confidence interval for population proportion p is ,

- E < P < + E

0.456 - 0.017 < p < 0.456 + 0.017

0.439 < p < 0.473

(0.439,0.473)

c) The 90% confidence interval is 0.439 to 0.473.