A dealer in recycled paper places empty trailers at various sites. The trailers

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 performing one-sample hypothesis tests, would we use z ? or t ? in this situation? Briefly explain why you would use one instead of the other. (b) Is there sufficient evidence that the mean amount of recycled paper is more than 1600 cubic feet per 2 week period? Conduct a hypothesis test. (c) State the kind of error could have been made in context of the problem. (d) Now do part b again in R

Explanation / Answer

Part-a

As sample size n=18<30 and we do not know the population standard deviation we will use t-test

Part-b

Test statistic t=(X- - 1600)/(s/sqrt(n)) = (1718.3-1600)/(137.8/sqrt(18))= 3.6423

Degree of freedom =n-1=18-1=17

Right tailed p-value=0.00101

As p-value is less than 0.05, we reject the null hypoythesis

Part-c

As we reject the null hypothesis which may be true, so we may have committed Type-I error.

Part-d

Results using R are as follows which are same as above:

recycle=c(1660,1820,1590,1440,1730,1680,1750,1720,1900,1570,1700,1900,1800,1770,2010,1580,1620,1690)

t.test(recycle, mu=1600, alternative="greater")

> recycle=c(1660,1820,1590,1440,1730,1680,1750,1720,1900,1570,1700,1900,1800,1770,2010,1580,1620,1690)

> t.test(recycle, mu=1600, alternative="greater")

        One Sample t-test

data: recycle

t = 3.6442, df = 17, p-value = 0.001003

alternative hypothesis: true mean is greater than 1600

95 percent confidence interval:

1661.845      Inf

sample estimates:

mean of x

1718.333

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