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 515 people from across the country, and tests them for the gene. The gene was present in 78 of the 515 people tested.

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

b) Provide the margin of error of the interval as your answer.

Round your answer to 4 decimal places.

Answer 1

Solution :

Given that,

n = 515

x = 78

Point estimate = sample proportion = = x / n = 78/ 515 = 0.151

1 - = 1-0.151 = 0.849

a) At 90% confidence level

= 1-0.90% =1-0.90 =0.10

/2 =0.10/ 2= 0.05

Z/2 = Z_{0.05 = 1.645}

Z/2 = 1.645  

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

= 1.645 * ((0.151*(0.849) /515 )

= 0.02595 = 0.0260

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

- E < p < + E

0.151- 0.0260 < p < 0.151 + 0.0260

0.125 < p < 0.177  

( 0.125 , 0.177 )