Handout3Q5_solution

## # A tibble: 2 x 5 ## term estimate std.error statistic p.value ## <chr> <dbl> <dbl> <dbl> <dbl> ## 1 (Intercept) 25.9 1.09 23.8 3.94e-20 ## 2 Love 2.50 0.0176 142. 1.46e-41 ˆ MoneySpent = 25 . 95 + 2 . 5 × Love (d) Interpret the slope and intercept of the least-squares regression line in the context of this problem. Intercept: When Love= 0, MoneySpent is expected to equal 25.95. Slope: For each unit in Love, MoneySpent is expected to increase on average by 2.5. (e) Check all conditions and comment on your findings. model.aug <- augment (model) 1. Linearity: the relationship between the explanatory (Love) and response (MoneySpent) should be linear. ggplot (Love_and_Money, aes ( x = Love, y = MoneySpent)) + geom_point () + labs ( title = "MoneySpent versus Love" , x = "Love" , y = "MoneySpent" ) 3

0 100 200 300 100 200 Fitted Values Residuals Residuals vs. Fitted From the residual vs. fitted plot, it appears that residuals are not constant. However, the residuals have small magnitude compared to the magnitude of MoneySpent. Therefore, if you look at the scatterplot of Love vs. MoneySpent, ggplot (Love_and_Money, aes ( x = Love, y = MoneySpent)) + geom_point () + labs ( title = "MoneySpent versus Love" , x = "Love" , y = "MoneySpent" ) + geom_smooth ( method= "lm" , se= FALSE ) ## ‘geom_smooth()‘ using formula = ’y ~ x’ 6