Question

1. Consider the number of days absent from a random sample of six students during a semester: A= {2, 3, 2, 4, 2, 5}
   1. Compute the arithmetic mean, geometric mean, median, and mode by hand and verify the results using R
      1. Arithmetic Mean: X=i=1nXin=2+3+2+4+2+56=3

mean(data2$absent)

[1] 3

2. Geometic Mean: GMx=Πi=1nX11n=2∙3∙2∙4∙2∙516=2.79816

>gmean <- prod(data2$absent)^(1/length(data2$absent))

> gmean

[1] 2.798166

3. Median: X=12n+1th, Xi2,2,2,3,4,5, n=6=126+1th ranked value=3.5, value=2.5 days absent

>median(data2$absent)

[1] 2.5

4. Mode: Most frequent value=2

> mode <- names(table(data2$absent)) [table(data2$absent)==max(table(data2$absent))]

> mode

[1] "2"

2. Compute the variance, standard deviation, and coefficient of variation by hand and verify the results using R.
   1. Sample Variance: S2=i=1nXi-X2n-1=2-32+3-32+2-324-322-325-326-1=85=1.6

> var(data2$absent)

[1] 1.6

2. Sample Standard Deviation: s=s2=i=1nXi-X2n-1=1.6=1.26491

> sd(data2$absent)

[1] 1.264911

3. Coefficient of Variation: CV=SX=1.264912.5=.421637

> cv <- abs(sd(data2$absent)/mean(data2$absent))

> cv

[1] 0.421637

3. Compute the interquartile range of this data by hand and with R

> IQR(data2$absent, type=6)

[1] 2.25

4. Display the data above with a well‐formatted and well‐labeled pie chart created with R.  Recall that with a pie chart, the slices sum to 100 percent, so the slice labeled “2 days absent” should be 50 percent of the pie.

Pie Chart with Percentages

slices <- c(50, 16.6666667, 16.6666667, 16.6666667)

lbls <- c("2 days absent", "3 days absent", "4 days absent", "5 days absent")

pct <- round(slices/sum(slices)*100)

lbls <- paste(lbls, pct) # add percents to labels

lbls <- paste(lbls,"%",sep="") # ad % to labels

pie(slices,labels = lbls, col=rainbow(length(lbls)),

    main="Pie Chart of Absences")

5. Display the data above with a well‐labeled and well‐formatted frequency distribution (histogram) using R.

x <- data2$absent

> h<-hist(x, breaks=10, col="magenta", xlab="# of student absences",

+         main="Histogram of student absences")

> xfit<-seq(min(x),max(x),length=40)

> yfit<-dnorm(xfit,mean=mean(x),sd=sd(x))

> yfit <- yfit*diff(h$mids[1:2])*length(x)

> lines(xfit, yfit, col="blue", lwd=2)

6. Use the relationship between the mean and median to determine whether the distribution above is symmetrical, negatively skewed or positively skewed.  Explain.
   1. Mean > Median = 3 > 2.5, the distribution above is positively skewed.
7. Calculate excess kurtosis using the formula from lecture.  Is this distribution leptokurtic, mesokurtic, or platykurtic?  Explain.  Compared to the normal distribution with the same variance, does this distribution exhibit fatter tails?  Briefly explain
   1. 1ni=1nxi-X41ni=1nxi-X22=162-34+3-34+2-34+4-34+2-34+5-34162-32+3-32+2-32+4-32+2-32+5-322=103169=158≈1.875
   2. Excess kurtosis = kurtosis – 3 = 1.875 – 3 = -1.125, -1.125 < 3
   3. This distribution is platykurtic as excess kurtosis < 0, kurtosis < 3
   4. No it will exhibit thinner tails than a normal distribution.

Suppose someone claims that the population mean is 5 days absent (

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Homework Answers

Answer #1

A= {2, 3, 2, 4, 2, 5}

Null Hypothesis H0:

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Answer 1

A= {2, 3, 2, 4, 2, 5}

Null Hypothesis H0: