Question

3) A medical researcher is interested in knowing what percentage of the U.S. population has a certain gene.

The researcher collect a random sample of 529 people from across the country, and tests them for the gene. The gene was present in 57 of the 529 people tested.

a) Find a 90% confidence interval for the true proportion of people in the U.S. with the gene.

b) Provide the right endpoint of the interval as your answer.

Round your answer to 4 decimal places.

Answer 1

Solution :

Given that,

n = 529

x = 57

Point estimate = sample proportion = = x / n = 0.1078

1 - = 0.8922

a)

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z _0.05 = 1.645

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

= 1.645 * (((0.1078 * 0.8922) / 529)

= 0.0222

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.1078 - 0.0222 < p < 0.1078 + 0.0222

0.0856 < p < 0.1300

b)

The right endpoint of the interval is 0.1300