Question

According to the website www.collegedrinkingprevention.gov, “About 25 percent of college students report academic consequences of their drinking including missing class, falling behind, doing poorly on exams or papers, and receiving lower grades overall.” A statistics student is curious about drinking habits of students at his college. He wants to estimate the mean number of alcoholic drinks consumed each week by students at his college. He plans to use a 99% confidence interval. He surveys a random sample of 55 students. The sample mean is 3.69 alcoholic drinks per week. The sample standard deviation is 3.56 drinks.

Construct the 99% confidence interval to estimate the average number of alcoholic drinks consumed each week by students at this college.

( ,  )

Your answer should be rounded to 2 decimal places.

Answer 1

Given that He surveyed a random sample of 55 students. The sample mean is 3.69 alcoholic drinks per week. The sample standard deviation is 3.56 drinks.

Sample size. n = 55

Sample mean = 3.69

Sample Standard Deviation S = 3.56

Given confidence level = 99% = 0.99

= 1 - confidence level

= 1 - 0.99

= 0.01

/2 = 0.005

Z_/2 will be z-score that has an area of 0.005 to its right which is 2.5758 from online calculator

99% confidence interval to estimate the average number of alcoholic drinks consumed each week by students at this college = Z_/2 * (S / )

= 3.69 2.5758 * (3.56 / )

= 3.69 2.5758 * (3.56 / 7.416198)

= 3.69 2.5758 * 0.48003

= 3.69 1.236462

= (2.453538, 4.926462)

99% confidence interval to estimate the average number of alcoholic drinks consumed each week by students at this college = 2.45 < < 4.93 rounded to 2 decimal places