Description of the data, data_quiz5
countrycontinentlifeExp life expectancy in yearpop total populationgdpPercap GDP per capita in U.S. dollaroptions(scipen=999)
Hint: The data is posted in Moodle. Look for data_quiz5.csv under the Data Files section.
myRegressionData <- read.csv("data_quiz5.csv")
Hint: Use head() to display the first six rows.
head(myRegressionData, 6)
## 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
Hint: Create a scatter plot to examine the relationship between GDP per capita (mapped to y-axis) and life expectancy (mapped to x-axis).
library(ggplot2)
library(tidyverse)
ggplot(myRegressionData,
aes(x = lifeExp,
y = gdpPercap)) +
geom_point() +
geom_smooth(method = "lm")
Regression_lm <- lm(gdpPercap ~ lifeExp,
data = myRegressionData)
summary(Regression_lm)
##
## Call:
## lm(formula = gdpPercap ~ lifeExp, data = myRegressionData)
##
## 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 <0.0000000000000002 ***
## lifeExp 3075.6 237.6 12.94 <0.0000000000000002 ***
## ---
## 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: < 0.00000000000000022
Yes the coefficient of life expectancy is statistically significant at 5% because the p value of the life expectancy is significantly less than 5.
Hint: Discuss both its sign and magnitude.
For an additional change in a unit of Life expectancy which is years the Gdp would increase by a total of 3076 dollars. This means that life expectancy has a positive effect on GDP.
Hint: Make your argument using the relevant test results, such as p-value.
Regression_lm <- lm(gdpPercap ~ lifeExp + pop,
data = myRegressionData)
summary(Regression_lm)
##
## Call:
## lm(formula = gdpPercap ~ lifeExp + pop, data = myRegressionData)
##
## Residuals:
## Min 1Q Median 3Q Max
## -17337 -4536 -82 3463 23993
##
## Coefficients:
## Estimate Std. Error t value
## (Intercept) -215081.8386328746 18328.1060846602 -11.735
## lifeExp 3072.6060571578 240.7532777810 12.762
## pop -0.0000006064 0.0000056600 -0.107
## Pr(>|t|)
## (Intercept) <0.0000000000000002 ***
## lifeExp <0.0000000000000002 ***
## pop 0.915
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 7625 on 80 degrees of freedom
## Multiple R-squared: 0.6741, Adjusted R-squared: 0.666
## F-statistic: 82.75 on 2 and 80 DF, p-value: < 0.00000000000000022
I would say to my friend that the how populous a country is not statistically significant at five percent because the p value of population is well about five percent or .05. I would tell my friend their prediction is wrong because of the p value of the pop variable is not statistically significant and has a very high P value.