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.

FD <- read.csv("http://www1.appstate.edu/~arnholta/Data/FlightDelays.csv")
# a.
library(tidyverse)
m_delays <- FD %>%
  group_by(Carrier) %>%
  summarize(Mean = mean(Delay), n()) %>%
  summarize(obs_diff = diff(Mean))
m_delays

# A tibble: 1 x 1
  obs_diff
     <dbl>
1 5.885696

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

pvalue <- ((sum(ts >= m_delays$obs_diff) + 1)/ (sims + 1)) * 2
pvalue

[1] 4e-04

\(H_0: \mu_U - \mu_A = 0\)

\(H_A: \mu_U - \mu_A\neq 0\)

There is evidence to support the claim that the average delay times between American Airlines and United Airlines are different in some way.

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.

# b. 
m_delays2 <- FD %>%
  group_by(Month) %>%
  summarize(Mean = mean(Delay), n()) %>%
  summarize(obs_diff = diff(Mean))
m_delays2

# A tibble: 1 x 1
   obs_diff
      <dbl>
1 -5.663341

sims <- 10^4 - 1
ts <- numeric(sims)
for(i in 1:sims) {
  index <- sample(4029, 1999, replace = FALSE)
  ts[i] <- mean(FD$Delay[index]) - mean(FD$Delay[-index])
}
hist(ts)

pvalue <- ((sum(ts <= m_delays2$obs_diff) + 1)/ (sims + 1)) * 2
pvalue

[1] 2e-04

\(H_0: \mu_M - \mu_J = 0\)

\(H_A: \mu_M - \mu_J\neq 0\)

There is evidence to suggest that average delay times in May and June are different or are significant in some way.

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.
Prop <- FD %>%
  group_by(Carrier) %>%
  summarize(Proportion = mean(Delay > 20), n()) %>%
  summarize(obs_diff = diff(Proportion))
Prop

# A tibble: 1 x 1
    obs_diff
       <dbl>
1 0.04351791

sims <- 10^4 - 1
ts <- numeric(sims)
for(i in 1:sims){
  index <- sample(4029, 1123, replace = FALSE)
  ts[i] <- mean(FD$Delay[index] > 20) - mean(FD$Delay[-index] > 20)
}
hist(ts)

pvalue <- ((sum(ts >= Prop$obs_diff) + 1)/ (sims + 1)) * 2
pvalue

[1] 0.0026

\(H_0: P_U - P_A = 0\)

\(H_A: P_U - P_A\neq 0\)

There is evidence to suggest that the proportions of flights being delayed for more than 20 minutes are different based on carrier.

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.
Var <- FD %>%
  group_by(Carrier) %>%
  summarize(Variance = var(Delay), n()) %>%
  summarize(obs_diff = diff(Variance))
Var

# A tibble: 1 x 1
  obs_diff
     <dbl>
1 431.0677

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

pvalue <- (sum(ts >= Var$obs_diff) + 1)/ (sims + 1)
pvalue

[1] 0.1479

\(H_0: V_U - V_A = 0\)

\(H_A: V_U - V_A > 0\)

There is no evidence to suggest that the variance in the flight delay lengths for United Airlines is greater than the variance for 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.

FDUA <- FD %>%
  group_by(Carrier == "UA") %>%
  summarize(Mean = mean(Delay), Sum = sum(Delay), n())
FDUA

# A tibble: 2 x 4
  `Carrier == "UA"`     Mean   Sum `n()`
              <lgl>    <dbl> <int> <int>
1             FALSE 10.09738 29343  2906
2              TRUE 15.98308 17949  1123

obs <- FD %>%
  group_by(Carrier) %>%
  summarize(Mean = mean(Delay), n()) %>%
  summarize(obs_diff = diff(Mean))
obs

# A tibble: 1 x 1
  obs_diff
     <dbl>
1 5.885696

sims <- 10^4 - 1
ts1 <- numeric(sims)
ts2 <- numeric(sims)
ts3 <- numeric(sims)
for(i in 1:sims) {
  index <- sample(4029, 1123, replace = FALSE)
  ts1[i] <- mean(FD$Delay[index])
  ts2[i] <- sum(FD$Delay[index])
  ts3[i] <- mean(FD$Delay[index]) - mean(FD$Delay[-index])
}
pvalue1 <- (sum(ts1 >= FDUA$Mean) + 1) / (sims + 1)
pvalue1

[1] 0.4737

pvalue2 <- (sum(ts2 >= FDUA$Sum) + 1) / (sims + 1)
pvalue2

[1] 3e-04

pvalue3 <- (sum(ts3 >= obs$obs_diff) + 1) / (sims + 1)
pvalue3

[1] 3e-04

There is enough evidence to suggest that the average delay times for United Airlines are statistically significant.

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.

# a.
FD %>%
  group_by(Carrier) %>%
  summarize(TrimMean = mean(Delay, trim = 0.25))

# A tibble: 2 x 2
  Carrier   TrimMean
   <fctr>      <dbl>
1      AA -2.5701513
2      UA -0.7957371

# b.
Trim <- FD %>%
  group_by(Carrier) %>%
  summarize(TrimMean = mean(Delay, trim = 0.25), n()) %>%
  summarize(obs_diff = diff(TrimMean))
sims <- 10^4 - 1
ts <- numeric(sims)
for(i in 1:sims) {
  index <- sample(4029, 1123, replace = FALSE)
  ts[i] <- mean(FD$Delay[index]) - mean(FD$Delay[-index])
}
hist(ts)

pvalue <- ((sum(ts >= Trim$obs_diff) + 1)/ (sims + 1)) * 2
pvalue

[1] 0.2312

\(H_0: TM_U - TM_A = 0\)

\(H_A: TM_U - TM_A\neq 0\)

There is no evidence to suggest that the difference in 25% trimmed means between United Airlines and American Airlines are different in some way.

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.

# a. 
Prop <- FD %>%
  group_by(Month) %>%
  summarize(Proportion = mean(Delay > 20), n()) %>%
  summarize(obs_diff = diff(Proportion))
Prop

# A tibble: 1 x 1
     obs_diff
        <dbl>
1 -0.02947582

sims <- 10^4 - 1
ts <- numeric(sims)
for(i in 1:sims){
  index <- sample(4029, 1999, replace = FALSE)
  ts[i] <- mean(FD$Delay[index] > 20) - mean(FD$Delay[-index] > 20)
}
hist(ts)

pvalue <- ((sum(ts <= Prop$obs_diff) + 1)/ (sims + 1)) * 2
pvalue

[1] 0.017

\(H_0: P_M - P_J = 0\)

\(H_A: P_M - P_J\neq 0\)

There is enough evidence to suggest that the proportions of times the flights in May and June were delayed more than 20 minutes are different in some way.

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.

# b. 
Var <- FD %>%
  group_by(Month) %>%
  summarize(Variance = var(Delay), n()) %>%
  summarize(obs_diff = diff(Variance))
Var

# A tibble: 1 x 1
   obs_diff
      <dbl>
1 -694.0982

sims <- 10^4 - 1
ts <- numeric(sims)
for(i in 1:sims) {
  index <- sample(4029, 1999, replace = FALSE)
  ts[i] <- var(FD$Delay[index]) - var(FD$Delay[-index])
}
hist(ts)

pvalue <- ((sum(ts <= Var$obs_diff) + 1)/ (sims + 1)) * 2
pvalue

[1] 0.0358

\(H_0: V_M - V_J = 0\)

\(H_A: V_M - V_J\neq 0\)

There is evidence to suggest that the variance of flight delay times in May and June are different in some way.

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.
2. Can the resluts be generalized to a population? Explain.

       LowFatDiet
Gender  Yes  No
  Women  35 146
  Men     8  97

\(H_0\): Gender and LowFatDiet are independent \(H_A\): Gender and LowFatDiet are dependent

DTT <- as.table(DT)
DTTDF <- as.data.frame(DTT)
DDF <- vcdExtra::expand.dft(DTTDF)
dim(DDF)

[1] 286   2

obs.stat <- chisq.test(xtabs(~Gender + LowFatDiet, data = DDF), correct = FALSE)$stat
obs.stat

X-squared 
 7.142724

sims <- 10^4 -1
ts <- numeric(sims)
for(i in 1:sims){
  ts[i] <- chisq.test(xtabs(~Gender + sample(LowFatDiet), data = DDF), 
                      correct = FALSE)$stat
}
hist(ts)

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

[1] 0.0081

There is evidence to suggest that gender and dieting are dependent variables that rely on each other for results. These results can be generalized to a population because the probability of women dieting is not the same as the probability of being a woman and dieting. This means that this sample is not independent which is exactly what we got in our chi-squared test.

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

\(H_0\): Age and Support are independent \(H_A\): Age and Support are dependent

chisq.test(T1, correct = FALSE)


    Pearson's Chi-squared test

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

obs <- chisq.test(T1, correct = FALSE)$stat
sims <- 10^4 - 1
ts <- numeric(sims)
for(i in 1:sims) {
  ts[i] <- chisq.test(xtabs(~Age + sample(Support), data = DF), correct = FALSE)$stat 
}
hist(ts)

pvalue <- (sum(ts >= obs) + 1)/ (sims + 1)
pvalue

[1] 0.0352

There is enough evidence to suggest that age and support of the use of marijuana are dependent variables and rely on each other.

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")
# a. 
T1 <- xtabs(~Age + Shelf, data = Cereals)
T1

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

# b.
chisq.test(T1, correct = FALSE)

Warning in chisq.test(T1, correct = FALSE): Chi-squared approximation may
be incorrect


    Pearson's Chi-squared test

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

# c.
chisq.test(T1, correct = FALSE)$exp

Warning in chisq.test(T1, correct = FALSE): Chi-squared approximation may
be incorrect

          Shelf
Age         bottom    middle      top
  adult    3.55814  7.511628 5.930233
  children 5.44186 11.488372 9.069767

Sends a warning because some observed and expected values are less than 5.

# d.
obs <- chisq.test(T1, correct = FALSE)$stat
obs

X-squared 
 28.62525

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

pvalue <- (sum(ts >= obs) + 1)/ (sims + 1)
pvalue

[1] 1e-04

There is enough evidence to suggest that Age and Shelf are dependent of each other.

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")
# a.
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

# b. 
chisq.test(T1, correct = FALSE)


    Pearson's Chi-squared test

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

# c. 
obs <- chisq.test(T1, correct = FALSE)$stat
sims <- 10^4 - 1
ts <- numeric(sims)
for(i in 1:sims) {
  ts[i] <- chisq.test(xtabs(~Gender + sample(Pres00), data = GSS2002), correct = FALSE)$stat 
}
hist(ts)

pvalue <- (sum(ts >= obs) + 1)/ (sims + 1)
pvalue

[1] 1e-04

The conclusion doesn’t change, we still have enough evidence to suggest that gender and the choice for president in 2000 election are dependent of each other.

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).

# a.
T1 <- xtabs(~Gender + Happy, data = GSS2002)
T1

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

# b.
obs <- chisq.test(T1, correct = FALSE)$stat
sims <- 10^4 - 1
ts <- numeric(sims)
for(i in 1:sims) {
  ts[i] <- chisq.test(xtabs(~Gender + sample(Happy), data = GSS2002), correct = FALSE)$stat 
}
hist(ts)

pvalue <- (sum(ts >= obs) + 1)/ (sims + 1)
pvalue

[1] 0.0041

There is enough evidence to suggest that gender and level of happiness are dependent of each other.

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).

# a.
T1 <- xtabs(~GunLaw + SpendMilitary, data = GSS2002)
T1

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

# b.
obs <- chisq.test(T1, correct = FALSE)$stat
sims <- 10^4 - 1
ts <- numeric(sims)
for(i in 1:sims) {
  ts[i] <- chisq.test(xtabs(~GunLaw + sample(SpendMilitary), data = GSS2002), correct = FALSE)$stat 
}
hist(ts)

pvalue <- (sum(ts >= obs) + 1)/ (sims + 1)
pvalue

[1] 0.2145

There isn’t sufficient evidence to suggest that support for gun laws and views on government spending on the military are dependent of each other.

Chapter Three Homework Solutions

Sara Hursa

Nov 02, 2017