Calculating parameters

group_by(data, year) %>%
    summarise(mean=mean(fare), median=median(fare),
              q1=quantile(fare,.25), q3=quantile(fare,.75)) %>%
    knitr::kable(caption="Stat. values of fair for various years")

Flight’s distance for flight destination

data_table <- group_by(data, destin) %>%
    summarise(mean=mean(dist), sd=sd(dist))
knitr::kable(data_table, caption="mean and standard deviation for flight distances based on cities" )

Therefore, the average distance of flights from Detroit is:

result = grep("detroit", ignore.case = TRUE, data_table[["destin"]]) %>% 
    data_table[["mean"]][.]

570.5

Tabulation of filtered results

filter(data, year == "1997", fare < 55) %>% with(table(origin, destin)) %>%
    knitr::kable()

Correlation Table

filter(data, year == "2000") %>%
    select(dist, fare, passen) %>%
    cor() %>%
    knitr::kable()

The results match expectations. Firstly, there’s unit covariance between same elements. Then, dist has positive cov with fare as greater distance is more expensive. Also, dist has negative cor with passen since fewer passengers go on distant trips. Similarly for fare and pass.

Regressing the linear model

model <- lm( fare ~ dist + bmktshr + passen, data=data)
summary(model)

Call:
lm(formula = fare ~ dist + bmktshr + passen, data = data)

Residuals:
    Min      1Q  Median      3Q     Max 
-146.50  -43.81  -11.87   39.15  216.28 

Coefficients:
             Estimate Std. Error t value Pr(>|t|)    
(Intercept) 52.995247   4.488055  11.808  < 2e-16 ***
dist         0.087485   0.001645  53.177  < 2e-16 ***
bmktshr     69.430756   5.122536  13.554  < 2e-16 ***
passen      -0.004942   0.001057  -4.677 2.99e-06 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 57.13 on 4592 degrees of freedom
Multiple R-squared:  0.4183,    Adjusted R-squared:  0.4179 
F-statistic:  1101 on 3 and 4592 DF,  p-value: < 2.2e-16

To see significance for different values, we perform a t-test which also gives variances and answers other questions.

t.test(data$fare)

    One Sample t-test

data:  data$fare
t = 161.87, df = 4595, p-value < 2.2e-16
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
 176.6313 180.9622
sample estimates:
mean of x 
 178.7968 

fare

t.test(data$bmktshr)

    One Sample t-test

data:  data$bmktshr
t = 210.56, df = 4595, p-value < 2.2e-16
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
 0.6044344 0.6157955
sample estimates:
mean of x 
0.6101149 

bmktshr

t.test(data$passen)

    One Sample t-test

data:  data$passen
t = 53.168, df = 4595, p-value < 2.2e-16
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
 613.3426 660.3058
sample estimates:
mean of x 
 636.8242 

passen

t.test(data$dist)

    One Sample t-test

data:  data$dist
t = 109.67, df = 4595, p-value < 2.2e-16
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
  972.0519 1007.4381
sample estimates:
mean of x 
  989.745 

dist

For the case of testing dist for .01, one trick is to offset the null hypothesis by .01. This is done by setting the significance level to .99 as below:

t.test(data$dist, conf.level = 0.99)

    One Sample t-test

data:  data$dist
t = 109.67, df = 4595, p-value < 2.2e-16
alternative hypothesis: true mean is not equal to 0
99 percent confidence interval:
  966.4888 1013.0012
sample estimates:
mean of x 
  989.745 

dist with .99

These are the coefficients of the model:

summary(model)$coef

                Estimate  Std. Error   t value     Pr(>|t|)
(Intercept) 52.995247275 4.488054584 11.808067 1.015786e-31
dist         0.087485328 0.001645179 53.176794 0.000000e+00
bmktshr     69.430755685 5.122535725 13.553982 4.582865e-41
passen      -0.004942295 0.001056728 -4.676979 2.994895e-06

As we expected from the cor matrix, dist and passen have the respective correlation with fair. Same is for bmktshr.

To see how much of the variation is unxplained, we refer to \(1- R^2\) which is 0.5816954.

Table of values

This table was given in previous sections. For reference, it is again given here:

table1 <- summary(model)$coeff %>% as.data.frame %>%
    select("Estimate", "t value", "Pr(>|t|)")
table2 <- c("totalobserv"= summary(model)$res %>% length(), 
            "R-squared"= summary(model)$r.squared, 
            "F-statistic"= summary(model)$fstatistic[1])
knitr::kable(list(table1, table2))

estout package

There is an equivalent package for estout. The latex code however, does not display in html format.

library(estout)
eststo(model)
esttab(t.value=TRUE,round.dec=2,caption="Reg Table",label="Reg")

Using the outreg2

It seems that the r equivalent of esttab is actually an equivalent of outreg. What this function does is to wrap the data in a compherensive format that can be used for comparison between different models. This is in fact the very description given by the estout R package which is based on the stata estout.