Quiz 5
Kayla Bowen
- Q1 Import data
- Q2 Review data
- Q3 Visualize data
- Q4 Build a regression model to predict GDP per capita using life expectancy.
- Q5 Is the coefficient of life expectancy statistically significant at 5%?
- Q6 Interpret the coefficient of life expectancy.
- Q7 Your friend suggests that the more populous a country, the higher its standard living (GDP per capita) is. Create a new model below by adding an additional predictor to the regression model above to test this hypothesis. Is the new variable statistically significant? What would you say to your friend regarding his/her claim?
- Q8 Hide the messages, but display the code and its results on the webpage.
- Q9 Display the title and your name correctly at the top of the webpage.
- Q10 Use the correct slug.
Description of the data, data_quiz5
countrycontinentlifeExplife expectancy in yearpoptotal populationgdpPercapGDP per capita in U.S. dollar
Q1 Import data
Hint: The data is posted in Moodle. Look for data_quiz5.csv under the Data Files section.
myClusterData <- read.csv("data_quiz5 (1).csv")Q2 Review data
Hint: Use head() to display the first six rows.
head(myClusterData)
## 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 29796.048Q3 Visualize data
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)
data(myClusterData)
# simple scatterplot
ggplot(myClusterData,
aes(x = lifeExp,
y = gdpPercap)) +
geom_point()
Q4 Build a regression model to predict GDP per capita using life expectancy.
data(myClusterData)
gdpPercap_predict <- lm(gdpPercap ~ lifeExp,
data = myClusterData)
# View summary of model 1
summary(gdpPercap_predict)
##
## Call:
## lm(formula = gdpPercap ~ lifeExp, data = myClusterData)
##
## 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-16Q5 Is the coefficient of life expectancy statistically significant at 5%?
No because it is higher than 5%.
Q6 Interpret the coefficient of life expectancy.
Hint: Discuss both its sign and magnitude.
The coefficient of life expectancy is 3065.6.This means that an increase in 1 year of life expectancy would increase the gdpPercap by that amount.
Q7 Your friend suggests that the more populous a country, the higher its standard living (GDP per capita) is. Create a new model below by adding an additional predictor to the regression model above to test this hypothesis. Is the new variable statistically significant? What would you say to your friend regarding his/her claim?
Hint: Make your argument using the relevant test results, such as p-value.
data(myClusterData)
gdpPercap_predict <- lm(gdpPercap ~ lifeExp + pop,
data = myClusterData)
# View summary of model 1
summary(gdpPercap_predict)
##
## Call:
## lm(formula = gdpPercap ~ lifeExp + pop, data = myClusterData)
##
## Residuals:
## Min 1Q Median 3Q Max
## -17337 -4536 -82 3463 23993
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -2.151e+05 1.833e+04 -11.735 <2e-16 ***
## lifeExp 3.073e+03 2.408e+02 12.762 <2e-16 ***
## pop -6.064e-07 5.660e-06 -0.107 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: < 2.2e-16Yes, because the p value is less than 5%.