The Harvard College Alcohol Study finds that 67% of college students support eff

Question

The Harvard College Alcohol Study finds that 67% of college students support efforts to "crack down on underage drinking." The study took a sample of almost 15,000 students, so the population proportion who support a crackdown is very close to p = 0.67.† The administration of your college surveys an SRS of 130 students and finds that 91 support a crackdown on underage drinking. If in fact the proportion of all students on your campus who support a crackdown is the same as the national 67%, what is the probability that the proportion in an SRS of 130 students is as large or larger than the result of the administration's sample? (Use a Normal approximation with a continuity correction to approximate the probability. Round your answer to four decimal places.)

Explanation / Answer

Given that

population proportion (p) = 67% = 0.67

q = 1-p = 1 - 0.67 = 0.33

N = population size = 15000

sample size (n) = 130

x = number of support a crackdown on underage drinking. = 91

sample proportion (p^) = x/n = 91 / 130 = 0.7

We use normal approximation when np > 10 and nq > 10

np = 130*0.67 = 87.1

nq = 130*0.33 = 42.9

Both are greator than 10 so we use normal approximation.

That is here we have to find the P(X 91).

But we cannot find exact probability in continous distribution that's why we make continuity correction.

By making continuity correction P(X > 91.5).

mean = np = 87.1

var = npq = 130*0.67*0.33 = 28.743

sd = sqrt(28.743) = 5.3612

P(X > 91.5) :

Now convert x = 91.5 into z-score.

z = (x - mean) / sd

z = (91.5 - 87.1) / 5.3612 = 0.8207

That is now we have to find P(Z > 0.8207).

This probability we can find by using EXCEL.

syntax is,

=1 - NORMSDIST(z) (EXCEL always gives left tail probability)

where z is test statistic value = 0.8207

P(Z > 0.8207) = 0.2059

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.