Question

An article summarized results from a national survey of 2302 Americans age 8 to 18. The sample was selected in a way that was expected to result in a sample representative of Americans in this age group.

a) Of those surveyed, 1126 reported owning a cell phone. Use this information to construct a 90% confidence interval estimate of the proportion of all Americans age 8 to 18 who own a cell phone. (Round your answers to three decimal places.)

(___________ , ___________ )

b) Of those surveyed, 1627 reported owning an MP3 music player. Use this information to construct a 90% confidence interval estimate of the proportion of all Americans age 8 to 18 who own an MP3 music player. (Round your answers to three decimal places.)

(__________ , __________)

Answer 1

Solution :

Given that,

a)

Point estimate = sample proportion = = x / n = 1126 / 2302 = 0.489

1 - = 1 - 0.489 = 0.511

Z_/2 = 1.645

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

= 1.645 * (((0.489 * 0.511) / 2302)

= 0.017

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.489 - 0.017 < p < 0.489 + 0.017

0.472 < p < 0.506

The 90% confidence interval for the population proportion p is : (0.472 , 0.506)

b)

Point estimate = sample proportion = = x / n = 1627 / 2302 = 0.707

1 - = 1 - 0.707 = 0.293

Z_/2 = 1.645

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

= 1.645 * (((0.707 * 0.293) / 2302)

= 0.016

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.707 - 0.016 < p < 0.707 + 0.016

0.691 < p < 0.723

The 95% confidence interval for the population proportion p is : (0.691 , 0.723)