Question

According to the National Institute on Drug Abuse (www.drugabuse.gov), 77% of the emergency room patients who mentioned MDMA (Ecstasy) as a factor in their admission were age 25 and under. Assume that the sample size is 200. Calculate the margin of error for the confidence level 95%. (Use decimal notation. Give your answer to four decimal places.)

E=

Construct a 95% confidence interval for the population proportion of all emergency room patients mentioning MDMA (Ecstasy) as a factor in their admission who are age 25 and under. (Use decimal notation. Give your answers to four decimal places.)

lower bound:

upper bound:

Answer 1

Solution :

Given that,

n = 200

Point estimate = sample proportion = = 0.77

1 - = 0.23

Z_/2 = Z _0.025 = 1.96

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

= 1.96 * (((0.77 * 0.23) / 200)

E = 0.0583

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.77 - 0.0583 < p < 0.77 + 0.0583

0.7117 < p < 0.8283

Lower bound = 0.7117

Upper bound = 0.8283