F20 1201 HW8 Solutions

Homework 8 Solutions Fall 2020 Question 1 (a) Find the mean and standard deviations of both strengths. Find the corre- lation of the two. Cube strength: Mean = ¯ x = 10 X i =1 x i = 96 . 42 Std dev = s x = v u u t 1 9 10 X i =1 ( x i - ¯ x ) 2 = 8 . 26 Axial strength: Mean = ¯ y = 10 X i =1 y i = 63 . 36 Std dev = s y = v u u t 1 9 10 X i =1 ( y i - ¯ y ) 2 = 10 . 27 Correlation = 1 9 ∑ 10 i =1 ( x i - ¯ x )( y i - ¯ y ) s x · s y = 0 . 7938 (b) Build a linear model of using cube strength to explain axial strength. This means treat cube strength as x (explanatory variable) and axial strength as y (response). β 1 = r s y s x = ∑ 10 i =1 ( x i - ¯ x )( y i - ¯ y ) ∑ 10 i =1 ( x i - ¯ x ) 2 = 0 . 9870 β 0 = ¯ y - β 1 ¯ x = - 31 . 8037

So our linear model is: axial strength = 0 . 9870 * cube strength - 31 . 8037 Or, in point slope form: axial strength - 63 . 36 = 0 . 9870 * (cube strength - 96 . 42) Question 2 (a) Does a scatter plot of the data support the use of a simple linear regression? Yes, simple linear regression seems reasonable based on the scatter plot. (b) Find the mean and standard deviation of both your x and y. Find the correlation. Use those numbers to build a linear regression model. In this case, use rainfall for x and runoff for y. Rainfall: Mean = ¯ x = 15 X i =1 x i = 53 . 20 Std dev = s x = v u u t 1 14 15 X i =1 ( x i - ¯ x ) 2 = 38 . 35 Runoff:

Average residual = 0 Average(residual 2 ) = 26 . 61 (e) What is the total variance of the runoff volume? What proportion of the ob- served variation of runoff volume can be attributed to the linear relationship between rainfall and runoff? Total variance of runoff= 1 14 ∑ 15 i =1 ( y i - ¯ y ) 2 = 32 . 29 2 = 1048 . 84 The proportion of that variation explained by the linear relationship is given by: 1 - 1 14 ∑ 15 i =1 ( y i - ˆ y i ) 2 1 14 ∑ 15 i =1 ( y i - ¯ y ) 2 = 0 . 973 (fg) Plot the residuals vs. vs rainfall. Make a histogram of the residuals.

Question 3 For the provided Russia population data, transform population to log(population) and perform a linear regression. Interpret your results. Replacing the population values with log(population) and performing the same linear regression excercise as in Q1 and Q2, we get a model: If you took log to mean log base 10, you get: Log(population)=0.000965*year+6.3699 Or, in point slope form: Log(population)-8.2656=0.000965*(year-1964.1429) (fine if you rounded these numbers, e.g. rounding slope to 0.001) If you took log to mean natural log, you get: Log(population)=0.0022*year+14.6672 Or, in point slope form: Log(population)-19.0323=0.0022*(year-1964.1429) This model predicts that log(population) grows by 0.0022 for each 1 year increase (i.e. each year that passes). Equivalently, this means we predict that population increases by a multiplicative factor of e 0 . 0022 for each one year increase in time. If you took log to mean log base 10, your model predicts that log(population) grows by 0.000965 for each 1 year increase. Equivalently, that means you predict that population increases by a multiplicative factor of 10 0 . 000965 for each one year increase in time. In both cases, we see that the linear model is not a very good fit to the data. Why? The plot of log(population) vs. time shows a steady increase over time, until a sharp dropoff after 1991. This coincides with when the Soviet Union dissolved; we can guess that the population data for 1991 and earlier are for the Soviet Union, and that subsequent population numbers refer only to the Russian Federation.