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 90% confidence interval. He surveys a random sample of 62 students. The sample mean is 3.87 alcoholic drinks per week. The sample standard deviation is 3.55 drinks. Construct the 90% 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

Solution :

Given that,

= 3.87

s = 3.55

n = 62

Degrees of freedom = df = n - 1 = 62 - 1 = 61

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,61 = 1.670

Margin of error = E = t_/2,df * (s /n)

= 1.670 * (3.55 / 62)

= 0.75

The 90% confidence interval estimate of the population mean is,

- E < < + E

3.87 - 0.75 < < 3.87 + 0.75

3.12 < < 4.62

(3.12 , 4.62)