Question
In: Statistics and Probability
A college wishes to know the proportion of American adults who are bilingual. The survey includes...
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.
- Find the point estimate of the percentage of adults who speak two or more languages.
- Find a 99% confidence interval estimate of the percentage of Americans who are bilingual? Are the majority of Americans bilingual? Explain.
Solutions
Expert Solution
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* (
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
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