In 2012, about 24% of high-school seniors reported binge drinking (defined as five or more drinks in a row in the past two weeks), a substantial drop since the late 1990s. a simple random sample of 500 high-school seniors is to be taken. the standard error is 0.019. what is the 95% confidence interval for the proportion of high-school seniors in the sample who would report binge drinking? (0.24 - 1.96*0.019, 0.24 + 1.96*0.019) o (0.24 - 1.645*0.019,0.24 + 1.645*0.019) (0.019 - 1.96*0.24, 0.019 +1.96*0.24) (0.019 - 1.645*0.24, 0.019 +1.645*0.24)

(0.24 - 1.96*0.019, 0.24 + 1.96*0.019)

Step-by-step explanation:

We know that the confidence interval population proportion (p) is given by :-

, where = sample proportion.

z* = Critical value (Two -tailed)

SE = standard error

Given : In 2012, about 24% of high-school seniors reported binge drinking (defined as five or more drinks in a row in the past two weeks), a substantial drop since the late 1990s.

SE = 0.019

Significance level =

Two-tailed value corresponds for :

z*=1.96   [Using z-table]

Now, the 95% confidence interval for the proportion of high-school seniors in the sample who would report binge drinking will be :-

Hence, the correct answer = (0.24 - 1.96*0.019, 0.24 + 1.96*0.019)

50-38=12

step-by-step explanation:

here is your mark brainliest