Chapter Three Homework Solutions

In the Flight Delays Case Study in Section 1.1,
1. The data contain flight delays for two airlines, American Airlines and United Airlines. Conduct a two-sided permutation test to see if the mean delay times between the two carriers are statistically significant.
\(\mu\)_UA -\(\mu\)_AA = 0

\(\mu\)_UA - \(\mu\)_AA \(\neq 0\)
1. The flight delays occured in May and June of 2009. Conduct a two-sided permutation test to see if the difference in mean delay times between the 2 months is statistically significant.

FD <- read.csv("http://www1.appstate.edu/~arnholta/Data/FlightDelays.csv")

# a. 
FDA <- FD %>%
  group_by(Carrier) %>%
  summarize(Mean = mean(Delay), n = n()) %>%
  summarize(obs.diff = diff(Mean))

sims <- 10^4 - 1
md <- numeric(sims)
for (i in 1:sims) {
  index <- sample(4029, 1123, replace = FALSE)
  md[i] <- mean(FD$Delay[index]) - mean(FD$Delay[-index])
}

# null distribution
hist(md)

pvalue <- ((sum(md >= FDA$obs.diff) + 1)/(sims + 1)) * 2
pvalue

[1] 2e-04

# The average delay for one airline is not the same for another airline

# b. Your code here
FDA <- FD %>%
  group_by(Month) %>%
  summarize(Mean = mean(Delay), n = n()) %>%
  summarize(obs.diff = diff(Mean))
FDA

# A tibble: 1 x 1
   obs.diff
      <dbl>
1 -5.663341

sims <- 10^4 - 1
md <- numeric(sims)
for (i in 1:sims) {
  index <- sample(4029, 1123, replace = FALSE)
  md[i] <- mean(FD$Delay[index]) - mean(FD$Delay[-index])
} 
hist(md)

pvalue <- ((sum(md <= FDA$obs.diff) + 1)/(sims + 1)) * 2
pvalue

[1] 2e-04

# We find evidence that the difference in mean delay times between the 2 months is statistically significant.

In the Flight Delays Case Study in Section 1.1, the data contain flight delays for two airlines, American Airlines and United Airlines.
1. Compute the proportion of times that each carrier’s flights was delayed more than 20 minutes. Conduct a two-sided test to see if the difference in these proportions is statistically significant.

# a. Your code here
N <- 10^3 - 1
result <- numeric(N)
set.seed(5)
for (i in 1:N) {
  index <- sample(4029, size = 23, replace = FALSE)
  result[i] <- mean(FD$Delay[index] > 20) - mean(FD$Delay[-index] > 20)
}
pvalue <- (sum(result >= FD$obs.diff) + 1)/ (N + 1)
pvalue * 2

[1] 0.002

# We find evidence to support that the difference in proportions of times that each carriers flights was delayed more than 20 minutes is statistically significant.

Compute the variance in the flight delay lengths for each carrier. Conduct a test to see if the variance for United Airlines is greater than that of American Airlines.

# b. Your code here
VAR <- tapply(FD$Delay, FD$Carrier, var)
VAR

      AA       UA 
1606.457 2037.525

VarTest <- VAR[1] - VAR[2]
names(VarTest) <- NULL
VarTest

[1] -431.0677

# We find evidence to suggest that the flight delay lengths for United Airlines is greater than that of American Airlines.

In the Flight Delays Case Study in Section 1.1, repeat Exercise 3 part (a) using three test statistics: (i) the mean of the United Airlines delay times, (ii) the sum of the United Airlines delay times, and (iii) the difference in the means, and compare the P-values. Make sure all three test statistics are computed within the same for loop.

Find three p values and they should be the same

N <- 10^4 - 1

UA.Delay <- subset(FD, select = Delay, Carrier == "UA", drop = T)
AA.Delay <- subset(FD, select = Delay, Carrier == "AA", drop = T)
observedSumUA <- sum(UA.Delay)
observedmeanUA <- mean(UA.Delay)
observedmeanDiff <- mean(UA.Delay) - mean(AA.Delay)
m <- length(UA.Delay)  #number of UA observations

sumUA <- numeric(N)
meanUA <- numeric(N)
meanDiff <- numeric(N)

set.seed(0)
for (i in 1:N) {
    index <- sample(4029, m, replace = FALSE)
    sumUA[i] <- sum(FD$Delay[index])
    meanUA[i] <- mean(FD$Delay[index])
    meanDiff[i] <- mean(FD$Delay[index]) - mean(FD$Delay[-index])
}
(sum(sumUA >= observedSumUA) + 1)/(N + 1)

[1] 1e-04

(sum(meanUA >= observedmeanUA) + 1)/(N + 1)

[1] 1e-04

(sum(meanDiff >= observedmeanDiff) + 1)/(N + 1)

[1] 1e-04

In the Flight Delays Case Study in Section 1.1,
1. Find the 25% trimmed mean of the delay times for United Airlines and American Airlines.
2. Conduct a two-sided test to see if the difference in trimmed means is statistically significant.

# Your code here
ANS <- with(data = FD, tapply(Delay, Carrier, mean, trim = 0.25))
ANS

        AA         UA 
-2.5701513 -0.7957371

obsMeanDiffCarrier <- ANS[2] - ANS[1]
obsMeanDiffCarrier

      UA 
1.774414

with(data = FD, table(Carrier))

Carrier
  AA   UA 
2906 1123

N <- 10^4 - 1
MeanDiffCarrier <- numeric(N)
for (i in 1:N) {
    # sample of size 1123 # of UA flights from the 4029 total
    index <- sample(4029, size = 1123, replace = FALSE)
    MeanDiffCarrier[i] <- mean(FD$Delay[index], trim = 0.25) - mean(FD$Delay[-index], 
        trim = 0.25)
}

pvalue <- ((min(sum(MeanDiffCarrier >= obsMeanDiffCarrier), sum(MeanDiffCarrier <= obsMeanDiffCarrier)) + 1)/(N + 1)) * 2
pvalue

[1] 2e-04

# We find evidence to suggest that the difference in 25% trimmed means of the delay times for United Airlines and American Airlines are statistically significant.

In the Flight Delays Case Study in Section 1.1,
1. Compute the proportion of times the flights in May and in June were delayed more than 20 min, and conduct a two-sided test of whether the difference between months is statistically significant.
2. Compute the variance of the flight delay times in May and June and then conduct a two-sided test of whether the ratio of variances is statistically significantly different from 1.

# a. Your code here
ANS <- with(data = FD, tapply(Delay > 20, Month, mean))
ANS

     June       May 
0.1960591 0.1665833

obsMeanDiffMonth <- ANS[2] - ANS[1]
obsMeanDiffMonth

        May 
-0.02947582

with(data = FD, table(Month))

Month
June  May 
2030 1999

N <- 10^4 - 1
with(data = FD, tapply(Delay > 20, Carrier, sum))

 AA  UA 
492 239

MeanDiffMonth <- numeric(N)
for (i in 1:N) {
    # sample of size 1999 # of UA flights from the 4029 total
    index <- sample(731, size = 239, replace = FALSE)
    MeanDiffMonth[i] <- mean(FD$Delay[index]) - mean(FD$Delay[-index])
}
pvalue <- ((min(sum(MeanDiffMonth >= obsMeanDiffMonth), sum(MeanDiffMonth <= 
    obsMeanDiffMonth)) + 1)/(N + 1)) * 2
pvalue

[1] 0.9266

# We failed to find evidence to reject that the difference between May and June for times of flights delayed more than 20 minutes is statistically significant.

# b. Your code here
ANS <- with(data = FD, tapply(Delay, Month, var))
ANS

    June      May 
2069.884 1375.786

obsMeanDiffMonth <- ANS[1]/ANS[2]
obsMeanDiffMonth

   June 
1.50451

with(data = FD, table(Month))

Month
June  May 
2030 1999

N <- 10^4 - 1
MeanDiffMonth <- numeric(N)
for (i in 1:N) {
    # sample of size 1999 # of UA flights from the 4029 total
    index <- sample(4029, size = 1999, replace = FALSE)
    MeanDiffMonth[i] <- var(FD$Delay[index])/var(FD$Delay[-index])
}
pvalue <- ((min(sum(MeanDiffMonth >= obsMeanDiffMonth), sum(MeanDiffMonth <= 
    obsMeanDiffMonth)) + 1)/(N + 1)) * 2
pvalue

[1] 0.039

# We find evidence that the ratio of flight delay times in May and June are statistically significant from 1.

Research at the University of Nebraska conducted a study to investigate sex differences in dieting trends among a group of Midwestern college students (Davy et al. (2006)). Students were recruited from an introductory nutrition course during one term. Below are data from one question asked to 286 participants.
1. Write down the appropriate hypothesis to test to see if there is a relationship between gender and diet and then carry out the test. (create the null hypothesis)
\(H_{0}\) = Gender does not affect dieting trends among Midwestern college students; they are indpendant

\(H_{A}\) = Gender affects dieting trends among Midwestern college students.

       LowFatDiet
Gender  Yes  No
  Women  35 146
  Men     8  97

# a. We reject the null hypothesis. Evidence suggests that gender and diet are dependent.  
chisq.test(DT, correct = FALSE)


    Pearson's Chi-squared test

data:  DT
X-squared = 7.1427, df = 1, p-value = 0.007527

Can the resluts be generalized to a population? Explain.

College students in the nutrition class in Nebraska do not generalize any large population.

A national polling company conducted a survey in 2001 asking a randomly selected group of Americans of 18 years of age or older whether they supported limited use of marijuana for medicinal purposes. Here is a summary of the data:

Write down the appropriate hypothesis to test whether there is a relationship between age and support for medicinal marijuana and carry out the test.

                   Support
Age                 Against For
  18-29 years old        52 172
  30-49 years old       103 313
  50 years or older     119 258

# Null Hypothesis: Age of Americans does no
# Alternative Hypothesis:
chisq.test(T1)


    Pearson's Chi-squared test

data:  T1
X-squared = 6.6814, df = 2, p-value = 0.03541

Two students went to a local supermarket and collected data on cereals; they classified by their target consumer (children versus adults) and the placement of the cereal on the shelf (bottom, middle, and top). The data are given in Cereals.
1. Create a table to summarize the relationship between age of target consumer and shelf location.
2. Conduct a chi-square test using R’s chisq.test command.
3. R returns a warning message. Compute the expected counts for each cell to see why.
4. Conduct a permutation test for independence.

Cereals <- read.csv("http://www1.appstate.edu/~arnholta/Data/Cereals.csv")
T1 <- xtabs(~Age + Shelf, data = Cereals)
T1

          Shelf
Age        bottom middle top
  adult         2      1  14
  children      7     18   1

chisq.test(T1)


    Pearson's Chi-squared test

data:  T1
X-squared = 28.625, df = 2, p-value = 6.083e-07

obs.stat <- chisq.test(T1)$stat
obs.stat

X-squared 
 28.62525

C <- Cereals %>%
  group_by(Age) %>%
  summarize(Mean = mean(Shelf), n = n()) %>%
  summarize(obs.diff = diff(Mean))


sims <- 10^4 - 1
cc <- numeric(sims)
for (i in 1:sims) {
  T2 <- xtabs(~Age + sample(Shelf), data = Cereals)
  cc[i] <- chisq.test(T2)$stat
} 

pvalue <- (sum(cc >= obs.stat)+1)/(sims +1)
pvalue

[1] 1e-04

# We reject that the relationship between age of target consumer and shelf location are independent.

From GSS 2002 Case Study in Section 1.6,
1. Create a table to summarize the relationship between gender and the person’s choice for president in the 2000 election.
2. Test to see if a person’s choice for president in the 2000 election is independent of gender (use chisq.test in R).
3. Repeat the test but use the permutation test for independence. Does your conclusion change? (Be sure to remove observations with missing values)

GSS2002 <- read.csv("http://www1.appstate.edu/~arnholta/Data/GSS2002.csv")
T1 <- xtabs(~Gender + Pres00, data = GSS2002)
T1

        Pres00
Gender   Bush Didnt vote Gore Nader Other
  Female  459          5  492    26     3
  Male    426          5  289    31    13

chisq.test(T1)


    Pearson's Chi-squared test

data:  T1
X-squared = 33.29, df = 4, p-value = 1.042e-06

From GSS 2002 Case Study in Section 1.6,
1. Create a table to summarize the relationship bewteen gender and the person’s general level of happiness (Happy).
2. Conduct a permutation test to see if gender and level of happiness are independent (Be sure to remove the observations with missing values).

# Your code here
T1 <- xtabs(~Gender + Happy, data = GSS2002)
T1

        Happy
Gender   Not too happy Pretty happy Very happy
  Female           109          406        205
  Male              61          378        210

chisq.test(T1)


    Pearson's Chi-squared test

data:  T1
X-squared = 10.96, df = 2, p-value = 0.004168

# We reject that gender and level of happiness are independent.

From GSS 2002 Case Study in Section 1.6,
1. Create a table to summarize the relationship between support for gun laws (GunLaw) and views on government spending on the military (SpendMilitary).
2. Conduct a permutation test to see if support for gun laws and views on government spending on the military are independent (Be sure to remove observations with missing values).

# Your code here
T1 <- xtabs(~GunLaw + SpendMilitary, data = GSS2002)
T1

        SpendMilitary
GunLaw   About right Too little Too much
  Favor          168        101       72
  Oppose          34         33       19

chisq.test(T1)


    Pearson's Chi-squared test

data:  T1
X-squared = 3.0827, df = 2, p-value = 0.2141

# We fail to reject that support for gun laws and views on government spending on the military are independent.

Chapter Three Homework Solutions

Rebecca Bryan, Samantha Widman, Reilly Ehlinger

Nov 02, 2017