Description of the data, data_quiz5
country
continent
lifeExp
life expectancy in yearpop
total populationgdpPercap
GDP per capita in U.S. dollarlibrary(tidyverse)
library(scales)
data_quiz5 <- read.csv("data_quiz5.csv")
head(data_quiz5)
## country continent lifeExp pop gdpPercap
## 1 Albania Europe 76.423 3600523 5937.030
## 2 Algeria Africa 72.301 33333216 6223.367
## 3 Argentina Americas 75.320 40301927 12779.380
## 4 Australia Oceania 81.235 20434176 34435.367
## 5 Austria Europe 79.829 8199783 36126.493
## 6 Bahrain Asia 75.635 708573 29796.048
library(ggplot2)
data(data_quiz5)
ggplot(data_quiz5,
aes(x = gdpPercap,
y = lifeExp)) +
geom_point()
houses_lm <- lm(gdpPercap ~ lifeExp,
data = data_quiz5)
summary(houses_lm)
##
## Call:
## lm(formula = gdpPercap ~ lifeExp, data = data_quiz5)
##
## Residuals:
## Min 1Q Median 3Q Max
## -17319.8 -4512.4 -63.2 3443.1 24014.4
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -215340.5 18057.2 -11.93 <2e-16 ***
## lifeExp 3075.6 237.6 12.94 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 7578 on 81 degrees of freedom
## Multiple R-squared: 0.6741, Adjusted R-squared: 0.6701
## F-statistic: 167.5 on 1 and 81 DF, p-value: < 2.2e-16
No, it is not significant at 5%.
Hint: Discuss both its sign and magnitude.
Because the co efficient of life expectancy has a p value of -2e16. this means that it doesn not have any stasticial significance at 5 percent.
Hint: Make your argument using the relevant test results, such as p-value.
library(ggplot2)
data(data_quiz5)
ggplot(data_quiz5,
aes(x = pop,
y = gdpPercap)) +
geom_point()
I would tell him it is not statistically significant.