Question

The proportion of college students who use electronic cigarettes is analyzed by a student organization. A sample of 89 students showed that 51 of them smoked electronic cigarettes. Using a α = 0.06 level of significance, find a confidence interval for the proportion of college students who use electronic cigarettes.

Identify the true statement.
I. If 578 students are enrolled in a college, at most 388 of students are likely to use electronic

cigarettes.
II. If 578 students are enrolled in a college, at most 408 of students are likely to use electronic

cigarettes.
III. If 578 students are enrolled in a college, at least 244 of students are likely to use electronic

cigarettes.
IV. If 578 students are enrolled in a college, at least 274 of students are likely to use electronic

cigarettes.

1. I only

2. IV only

3. I and IV only

4. I, II, and III only

5. I and II only

Answer 1

Answer)

P = 51/89

N = 89

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 51

N*(1-p) = 38

Both the conditions are met so we can use standard normal z table to estimate the interval

Margin of error (MOE) = z*standard error

Standard error = √P*(1-p)/√n

Critical value z from z table for 94% confidence level is = 1.88

Z = 1.88

(Confidence level = 1 - alpha = 1 - 0.06 = 0.94)

After substitution

MOE = 0.09857112782

Lower limit = P - MOE = 0.47446258004

Upper limit = P + MOE = 0.67160483568

So, lower limit = 0.47446258004*578 = 274

Upper limit = 0.67160483568*800 = 388

So i and iv are correct only