Homework #5

You can also get the PDF format of Wenqiang’s Homework.

Exercise 6.44

• (a). What are the populations of interest?
  • The plants of tobacco treated with two different fumigants
  • The variable being studies is the number of parasites from the plants
• (b). Yes. The p-value is 2.089e-05, which is a strong evidence to indicate a difference in the mean level of parasites for the two fumigants.

F1 = c(77,40,11,31,28,50,53,26,33)
F2 = c(76,38,10,29,27,48,51,24,32)

t.test(F1, F2, paired = T, conf.level = 0.90)

## 
##  Paired t-test
## 
## data:  F1 and F2
## t = 8.8544, df = 8, p-value = 2.089e-05
## alternative hypothesis: true difference in means is not equal to 0
## 90 percent confidence interval:
##  1.228866 1.882245
## sample estimates:
## mean of the differences 
##                1.555556

• (c). the size of diference = 1.555556, 90% CI = [1.228866, 1.882245]

Exercise 6.46

abundance = c(5124,2904,3600,2880,2578,4146,1048,1336,394,7370,6762,744,1874,
              3228,2032,3256,3816,2438,4897,1346,1676,2008,2224,1234,1598,2182)
depth = c(rep(40, 13), rep(100, 13))
oil_traj = c(rep(c("within", "outside"), c(7,6)), 
             rep(c("within", "outside"), c(7,6)))
pop_abundance = data.frame(abundance, as.factor(depth), oil_traj)

library(dplyr)

## 
## Attaching package: 'dplyr'

## The following object is masked from 'package:gridExtra':
## 
##     combine

## The following object is masked from 'package:MASS':
## 
##     select

## The following objects are masked from 'package:plyr':
## 
##     arrange, count, desc, failwith, id, mutate, rename, summarise,
##     summarize

## The following objects are masked from 'package:stats':
## 
##     filter, lag

## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union

within = filter(pop_abundance, oil_traj=='within')$abundance
outside = filter(pop_abundance, oil_traj=='outside')$abundance

t.test(within, outside, var.equal = FALSE)

## 
##  Welch Two Sample t-test
## 
## data:  within and outside
## t = 0.89466, df = 16.224, p-value = 0.384
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -877.7749 2162.1558
## sample estimates:
## mean of x mean of y 
##  3092.357  2450.167

par(mfcol=c(1,2))
qqnorm(within, main="Within oil trajectory")
qqline(within)

qqnorm(outside, main="Outside oil trajectory")
qqline(outside)

par(mfcol=c(1,1))
boxplot(within, outside, names=c("within", "outside"))

Exercise 7.25(a)

• Data input

portfolio1 = c(130,135,135,131,129,135,126,136,127,132)
portfolio2 = c(154,144,147,150,155,153,149,139,140,141)
sd(portfolio1)^2

## [1] 12.93333

• Levene test
  • The p-value = 0.07724 > 0.05 from the levene test below indicates that we failed to reject the \(H_{0}\) hypothesis. There is no strong evidence to support that portfolio 2 has higher risk than portfolio 1.

library(car)

## 
## Attaching package: 'car'

## The following object is masked from 'package:dplyr':
## 
##     recode

## The following objects are masked from 'package:HH':
## 
##     logit, vif

dt = data.frame(y=c(portfolio1, portfolio2), 
                group=rep(c("1", "2"), c(length(portfolio1), length(portfolio2))))
leveneTest(y~group, dt)

## Levene's Test for Homogeneity of Variance (center = median)
##       Df F value  Pr(>F)  
## group  1  3.5122 0.07724 .
##       18                  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Exercise 7.26

• Tests for two population variances
  • $H_{0}: two variances are eqaul vs. H_{a}: two varainces are not equal
  • P-value = 0.1485 > 0.05, fail to rejct \(H_{O}\).
  • Conclusion: the two variances are equal

var.test(portfolio1, portfolio2)

## 
##  F test to compare two variances
## 
## data:  portfolio1 and portfolio2
## F = 0.36421, num df = 9, denom df = 9, p-value = 0.1485
## alternative hypothesis: true ratio of variances is not equal to 1
## 95 percent confidence interval:
##  0.09046343 1.46628824
## sample estimates:
## ratio of variances 
##          0.3642053

• Tests for normality
  • QQ plots show that the two samples are approximately normally distributed.

par(mfcol=c(1,2))
qqnorm(portfolio1, main = "Portfolio 1")
qqline(portfolio1)

qqnorm(portfolio2, main = "Portfolio 2")
qqline(portfolio2)

par(mfcol=c(1,1))

• Since the two samples are randomly selected, normally distributed and have equal variance, I choose the pooled variance t-test to detect if there is any differences in the average returns for the two portfolios.

  • \(H_{0}: \mu_{1} = \mu_{2}\) vs. \(H_{a}: \mu_{1} \neq \mu_{2}\)
  • P-value = 1.314e-06 < 0.0001, reject \(H_{0}\)
  • Conclusion: there is strong evidence that a difference in the mean returns of the two portfolios exits.

t.test(portfolio1, portfolio2, var.equal = T)

## 
##  Two Sample t-test
## 
## data:  portfolio1 and portfolio2
## t = -7.0877, df = 18, p-value = 1.314e-06
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -20.22415 -10.97585
## sample estimates:
## mean of x mean of y 
##     131.6     147.2