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)

Step-by-step 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 :-

$0.24\pm (1.96)\cdot (0.019)=(0.24-1.96\times 0.019,\ 0.24+1.96\times 0.019)$

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