Question

A college wishes to know the proportion of American adults who are bilingual. The survey includes asking 280 randomly selected American adults and finds that 123 are bilingual.

1. Find the point estimate of the percentage of adults who speak two or more languages.

1. Find a 99% confidence interval estimate of the percentage of Americans who are bilingual? Are the majority of Americans bilingual? Explain.

Answer 1

Solution :

Given that,

n = 280

x = 123

Point estimate = sample proportion = = x / n = 123/280=0.439

1 -   = 1- 0.439 =0.561

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

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

= 2.576* (((0.439*0.561) / 280)

E = 0.0764

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.439-0.0764 < p <0.439+ 0.0764

0.3626< p < 0.5154

The 99% confidence interval for the population proportion p is : 0.3626, 0.5154