A dealer in recycled paper places empty trailers at various sites. The trailers
ID: 3223467 • Letter: A
Question
A dealer in recycled paper places empty trailers at various sites. The trailers are gradually filled by individuals who bring in old newspapers and magazines, and are picked up on several schedules. One such schedule involves pickup every second week. This schedule is desirable if the average amount of recycled paper is more than 1600 cubic feet per 2-week period. The dealer’s records for eighteen 2-week periods show the following volumes (in cubic feet) at a particular site: recycle=c(1660,1820,1590,1440,1730,1680,1750,1720,1900,1570,1700,1900,1800,1770,2010,1580,1620,1690) The mean and standard deviation are as follows: X = 1718.3 and s = 137.8
(a) In constructing confidence intervals, would we use z or t in this situation? Briefly explain why you would use one instead of the other.
(b) Estimate the true mean weight of recycled paper with 95% confidence. Interpret.
Explanation / Answer
Answer:
A dealer in recycled paper places empty trailers at various sites. The trailers are gradually filled by individuals who bring in old newspapers and magazines, and are picked up on several schedules. One such schedule involves pickup every second week. This schedule is desirable if the average amount of recycled paper is more than 1600 cubic feet per 2-week period. The dealer’s records for eighteen 2-week periods show the following volumes (in cubic feet) at a particular site: recycle=c(1660,1820,1590,1440,1730,1680,1750,1720,1900,1570,1700,1900,1800,1770,2010,1580,1620,1690) The mean and standard deviation are as follows: X = 1718.3 and s = 137.8
(a) In constructing confidence intervals, would we use z or t in this situation? Briefly explain why you would use one instead of the other.
When population standard deviation is known, we use z distribution.
When sample standard deviation is given, we use t distribution.
In this case we use t distribution.
(b) Estimate the true mean weight of recycled paper with 95% confidence. Interpret.
R output:
> # 95% CI for mean
> a <- 1718.3
> s <- 137.8
> n <- 18
> error <- qt(0.975,df=n-1)*s/sqrt(n)
> left <- a-error
> right <- a+error
> left
[1] 1649.774
> right
[1] 1786.826
95% CI = (1649.774, 1786.826)