Question

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)

Answer 1

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

Explanation:

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

`\hat{p}\pm z*\cdot SE`

, where
`\hat{p}` = 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.

`\hat{p}=0.24`

SE = 0.019

Significance level =
`\alpha=1-0.95=0.05`

Two-tailed value corresponds for
`\alpha=0.05` :

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 :-

`0.24\pm (1.96)\cdot (0.019)=(0.24-1.96* 0.019,\ 0.24+1.96* 0.019)`

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