Question

Descriptive Statistics: The Old-Fashioned Way (and with 2. In this problem, calculate the descriptive statistics below "by hand." That is, please show all of your calculations using Equation Editor (in Word, use Insert/Equation) or similar (although you must use R to verify your results). - Consider the hourly wages from a sample of six employed students duringa semester: w (10, 10, 16, 18, 18,18) A. Compute the arithmetic mean, geometric mean, median, and mode by B. Compute the variance, standard deviation, and coefficient of variation by C.Compute the interquartile range of this data by hand and with R. hand and verify the results using R. hand and verify the results using R D. Use R to create a well-labeled box plot (box and whisker plot) using the E. Display the data above with a well-formatted and well-labeled pie chart F.Display the data above with a well-labeled and well-formatted frequency G.Use the relationship between the mean and median to determine whether data above. created with R. distribution (histogram) using R: the distribution above is symmetrical, negatively skewed or positively skewed. Note you do not have to compute the skewness coefficient. Explain. H 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.

Answer 1

A. A.M. (10+ 10 + 164 184 184 18)/6-15 G.M. (1010 16 18 18 18)16- 14.50963 Media(3rd observation +4th observation)/2 - (1618)/2 17 Mode 18 (since highest frequency- 3 which is a frequency of 18)

R code:

x=c(10,10,16,18,18,18)
mean(x)#AM
(prod(x))^(1/length(x))#GM
median(x)# median
names(table(x))[table(x)==max(table(x))] # mode

Output:

> mean(x)#AM
[1] 15
> (prod(x))^(1/length(x))#GM
[1] 14.50963
> median(x)# median
[1] 17
> names(table(x))[table(x)==max(table(x))] # mode
[1] "18"

B. Sample Variance = ((102-102 +162 +182 +182 +182)-6*((10-10-16+18418418)/6)2)/5 = 15.6 Sample Standard dev 3.949684 Coefficient of variation = Stanolard deviation/ Mean ation Sample Variance (3.949684/15) 100 26.33122

R code:

x=c(10,10,16,18,18,18)
var(x)#variance
sd(x)#standard deviation
(sd(x)/mean(x))*100# coefficient of variation

Output:

> var(x)#variance
[1] 15.6
> sd(x)#standard deviation
[1] 3.949684
> (sd(x)/mean(x))*100# coefficient of variation
[1] 26.33122
C.

Q1=First quartile=2nd observation+0.25*(3rd observation-2nd observation)=10+0.5*(10-10)=11.5

Q3=3rd quartile=5th observation+0.75*(6th observation-5th observation)=18+0.75*(18-18)=18

Inter quartile range=Q3-Q1=18-11.5=6.5

R code:

x=c(10,10,16,18,18,18)
quantile(x,0.25)#Q1
quantile(x,0.75)#Q3
quantile(x,0.75)-quantile(x,0.25)#Inter quartile range

Output:

> quantile(x,0.25)#Q1
25%
11.5
> quantile(x,0.75)#Q3
75%
18
> quantile(x,0.75)-quantile(x,0.25)#Inter quartile range
75%
6.5
D.

x=c(10,10,16,18,18,18)
boxplot(x,main="boxplot of hourly wages")

boxplot of hourly wages CH

E.

R code:

x=c(10,10,16,18,18,18)
f=c(2,1,3)
pct=round(f/sum(f)*100)
lbls= paste(pct) # add percents to labels
lbls= paste(lbls,"%",sep="") # ad % to labels
pie(f,main="boxplot of hourly wages",labels = lbls)

boxplot of hourly wages 33% 17%

F.

R code:

x=c(10,10,16,18,18,18)
hist(x,main="boxplot of hourly wages")

boxplot of hourly wages CH 10 12 14 16 18