LIS 4273 Module #7 Assignment

#1.1 Define the relationship model between the predictor (x) and the response (Y) variable:

A simple linear regression can model the relationship between the predictor variable x and the response variable y.

#1.2 Calculate the coefficients?

> x <- c(16, 17, 13, 18, 12, 14, 19, 11, 11, 10)
> y <- c(63, 81, 56, 91, 47, 57, 76, 72, 62, 48)

> model <- lm(y ~ x)
> summary(model)

Call:
lm(formula = y ~ x)

Residuals:
    Min      1Q  Median      3Q     Max 
-11.435  -7.406  -4.608   6.681  16.834 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)  
(Intercept)   19.206     15.691   1.224   0.2558  
x              3.269      1.088   3.006   0.0169 *
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 10.48 on 8 degrees of freedom
Multiple R-squared:  0.5303,	Adjusted R-squared:  0.4716 
F-statistic: 9.033 on 1 and 8 DF,  p-value: 0.01693

#2.1 Define the relationship model between the predictor (x) and the response variable(y).

> visit <- data.frame(
+   discharge = c(3.600, 1.800, 3.333, 2.283, 4.533, 2.883),
+   waiting = c(79, 54, 74, 62, 85, 55)
+ )
> model_visit <- lm(discharge ~ waiting, data = visit)
> summary(model_visit)

Call:
lm(formula = discharge ~ waiting, data = visit)

Residuals:
      1       2       3       4       5       6 
-0.2039 -0.3149 -0.1331 -0.3724  0.3238  0.7005 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)  
(Intercept) -1.53317    1.12328  -1.365   0.2440  
waiting      0.06756    0.01623   4.162   0.0141 *
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.4724 on 4 degrees of freedom
Multiple R-squared:  0.8124,	Adjusted R-squared:  0.7655 
F-statistic: 17.32 on 1 and 4 DF,  p-value: 0.01413

#2.2 Extract the parameters of the estimated regression equation with the coefficients function.

> coefficients(model_visit)
(Intercept)     waiting 
-1.53317418  0.06755757 

#2.3 Determine the fit of the eruption duration using the estimated regression equation.

> predict(model_visit, newdata = data.frame(waiting = 80))
       1 
3.871431 

#3.1 Examine the relationship Multi Regression Model as stated above and its Coefficients using 4 different variables from mtcars (mpg, disp, hp and wt). Report on the result and explain what the multi-regression model and coefficients talk about in the data?

   
> input <- mtcars[, c("mpg", "disp", "hp", "wt")]
> multi_model <- lm(mpg ~ disp + hp + wt, data = input)
> summary(multi_model)

Call:
lm(formula = mpg ~ disp + hp + wt, data = input)

Residuals:
   Min     1Q Median     3Q    Max 
-3.891 -1.640 -0.172  1.061  5.861 

Coefficients:
             Estimate Std. Error t value Pr(>|t|)    
(Intercept) 37.105505   2.110815  17.579  < 2e-16 ***
disp        -0.000937   0.010350  -0.091  0.92851    
hp          -0.031157   0.011436  -2.724  0.01097 *  
wt          -3.800891   1.066191  -3.565  0.00133 ** 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 2.639 on 28 degrees of freedom
Multiple R-squared:  0.8268,	Adjusted R-squared:  0.8083 
F-statistic: 44.57 on 3 and 28 DF,  p-value: 8.65e-11

#4 According to the fitted model, what is the predicted metabolic rate for a body weight of 70 kg? 

> library(ISwR)
> data(rmr)
> plot(metabolic.rate ~ body.weight, data = rmr)
> rmr_model <- lm(metabolic.rate ~ body.weight, data = rmr)
> summary(rmr_model)

Call:
lm(formula = metabolic.rate ~ body.weight, data = rmr)

Residuals:
    Min      1Q  Median      3Q     Max 
-245.74 -113.99  -32.05  104.96  484.81 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 811.2267    76.9755  10.539 2.29e-13 ***
body.weight   7.0595     0.9776   7.221 7.03e-09 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 157.9 on 42 degrees of freedom
Multiple R-squared:  0.5539,	Adjusted R-squared:  0.5433 
F-statistic: 52.15 on 1 and 42 DF,  p-value: 7.025e-09

> predict(rmr_model, newdata = data.frame(body.weight = 70))
       1 
1305.394