Part I: In a few days, I will be meeting with prospective students and parents a
ID: 3041346 • Letter: P
Question
Part I: In a few days, I will be meeting with prospective students and parents as part of the Presidential Scholarship day. I am struggling with what to say about this flyer posted on campus. Does itmean that 26% of students have more than five drinks per week? That seems very high. That is approximately one in four students. The estimate is from a survey of 1,284 students. - 74% , A 1. What is the margin of error of the proportion? 2. What is the 95% confidence interval? 3. I could believe the true answer is one in twenty (5%). Use HAVE O-5 ALCOHOLIC DRINKS A WEEK answer to the preceding question to consider this hypothesis. Is it possible? Is it likely? 4. How large of a sample would I need to have a margin of error of 1 percentage point or less?Explanation / Answer
Ans:
Sample proportion=0.25 (As,one in four students)
n=1284
Confidemce level=0.95
z value=1.96
a)Margin of error=1.96*sqrt(0.25*(1-0.25)/1284)=0.0237
b)95% confidence interval
=0.25+/-0.0237
=(0.2263,0.2737)
3)As,above conidence interval include 0.26,we can believe or conclude that true proportion is equal to 0.26 at the significance level of 0.05.
4)
sample size=(1.962*0.25*0.75)/0.012=7203