In: Statistics and Probability
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.
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