Description of the data, data_quiz5
country
continent
lifeExp
life expectancy in yearpop
total populationgdpPercap
GDP per capita in U.S. dollarHint: The data is posted in Moodle. Look for data_quiz5.csv under the Data Files section.
Hint: Use head()
to display the first six rows.
## 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).
##
## Call:
## lm(formula = lifeExp ~ gdpPercap, data = MyClusterData)
##
## Residuals:
## Min 1Q Median 3Q Max
## -4.718 -1.451 -0.026 1.547 4.731
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 7.194e+01 3.790e-01 189.79 <2e-16 ***
## gdpPercap 2.192e-04 1.693e-05 12.94 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.023 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
Yes it is significant because the life expectancy is less than 5%
Hint: Discuss both its sign and magnitude. Its sign is positive and its magnitude of which is the pr value are both significant.
Hint: Make your argument using the relevant test results, such as p-value.