> 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
Question 1
1.1: This is a linear relationship model. X is the predictor and Y is the response.
The model shows how the predictor will affect the response in a straight line.
1.2: Given the regression formula, Y = a + bX +e
Looking at the model, the row labeled Intercept the Estimate is 19.206,
which means when x=0, y is about 19.206. (a)
In the row labeled X, the Estimate is 3.206, which is the slope, so for every 1
increase in x, y increases by 3.269. (b)
y = 19.206 + 3.269x
Question 2
> discharge <- c(3.600, 1.800, 3.333, 2.283, 4.533, 2.883) # Dataset for discharge
> waiting <- c(79, 54, 74, 62, 85, 55) # Dataset for waiting
> visit <- data.frame(discharge, waiting) # Create dataframe to group the two
> model <- lm(discharge ~ waiting, data = visit) # Create linear model
> summary(model) # Summary of the model
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
2.1: The formula for this relationship model is
discharge = a+b x waiting
This means that the discharge duration depends on the waiting time.
2.2: The estimated regression equation with the coefficient function is
discharge = -1.5332 + 0.0676 x waiting
2.3: Using the regression equation with the coefficient functions, plug in
x=80 because there is 80 minutes in wait time.
discharge = -1.5332 + 0.0676 x 80
discharge = -1.5332 + 5.408
discharge = 3.8748
When waiting 80 minutes, the predicted discharge time is around 3.87.
Question 3
> input <- mtcars[,c("mpg","disp","hp","wt")]
> lm(formula = mpg ~ disp + hp + wt, data = input)
Call:
lm(formula = mpg ~ disp + hp + wt, data = input)
Coefficients:
(Intercept) disp hp wt
37.105505 -0.000937 -0.031157 -3.800891
3.1: A Multi Regression Model is used in this example to see how mpg
(miles per gallon). The dependent variable. Is affected by disp, hp, and wt.
The coefficients are disp (displacement), hp (horsepower), wt (weight) these
coefficients tell how mpg is changed when the coefficients are changed. Also
known as independent variables or predictors.
Question 4
> library(ISwR)
> plot(metabolic.rate~body.weight,data=rmr)
> model <- lm(metabolic.rate ~ body.weight, data = rmr)
> summary(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
4.1: Intercept = 811.2267
body.weight = 7.0595
x = 70
Linear Regression Formula:
y = 8.11.2267 + 7.0595x
> coeff <- coefficients(model)
> predicted_rate <- coeff[1] + coeff[2] * 70
> predicted_rate
(Intercept)
1305.394
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