Question

4) 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 510 people from across the country, and tests them for the gene. The gene was present in 42 of the 510 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 :

n = 510

x = 42

= x / n = 42 / 510 = 0.082

1 - = 1 - 0.0823 = 0.9177

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 * (((0.0823 *0.9177) / 510)

= 0.0200

a ) A 90 % confidence interval for population proportion p is ,

- E < P < + E

0.0823 - 0200< p < 0.0823+ 0200

0.0623 < p < 0.1023

b) The right endpoint = 0.1023