Question

In R please

Problem 6.4 Water in a cylindrical object will be ejected out the side of the water in the cylinder is above the hole. We expect that the distance the water shoots out the side will increase as the depth of the water above the hole increases. In this experiment, the cylinder is a 2 liter plastic soda bottle with a hole in the side. We fill the bottle (with the hole covered) 30 times and measure the dis- tance the water is ejected after the hole is uncovered. The runs are randomly assigned to 10 depths (6, 7, 8, ... 15 cm). Data are shown in the following table (from X. Meng, pers. comm., data set cfcdae: :WaterEjection). 15 14 13 12 11 10 9 8 7 6 16.4 14.4 13.0 12.7 12.0 10.8 10.8 11.2 6.7 3.4 16.4 15.3 14.0 14.0 11.8 12.0 10.5 10.0 4.2 5.0 16.2 15.0 14.3 13.5 12.2 12.0 11.0 10.0 6.5 5.5 (a) Find a good polynomial fit for these data. Are there outliers or other problems? Does your model make physical sense (for example, what does it predict for the distance at a depth of 1 cm)? (b) My (perhaps naive) intuition suggests that distance should be directly pro- portional to ejection velocity, and ejection velocity should be directly propor- tional to depth. Thus my intuition suggests that a model with a linear term and no intercept should fit the data. How well does this model work? Which model do you prefer?

Answer 1

R-commands and outputs:

x=15:6
x
length(x)
xx=rep(x,each=3)
xx
[1] 15 15 15 14 14 14 13 13 13 12 12 12 11 11 11 10 10 10 9 9 9 8 8 8 7 7 7 6 6 6
y=c(16.4,16.4,16.2,14.4,15.3,15,13,14,14.3,12.7,14,13.5,12,11.8,12.2,10.8,12,12,10.8,10.5,11,11.2,10,10,6.7,4.2,6.5,3.4,5,5.5)

# 'lm()' command is used to fit linear model in R.

fit=lm(y~xx)
fit
Call:
lm(formula = y ~ xx)
Coefficients:
(Intercept) xx
-0.9939 1.1766

summary(fit)
Call:
lm(formula = y ~ xx)
Residuals:
Min 1Q Median 3Q Max
-3.0420 -0.5260 -0.1631 0.8974 2.7814
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.99394 0.86123 -1.154 0.258
xx 1.17657 0.07911 14.872 8.09e-15 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 1.245 on 28 degrees of freedom
Multiple R-squared: 0.8876, Adjusted R-squared: 0.8836
F-statistic: 221.2 on 1 and 28 DF, p-value: 8.09e-15

#From the output of 'lm()' command, we get, intercept= -0.9939 and slope=1.1766
#Regression equation is: yhat= -0.9939 + 1.1766*(xx)

par(mfrow=c(2,2))
plot(fit)

Normal Q-Q Residuals vs Fitted 220 3 2 1 1 Residuals Standardized residuals 0 1 -2 -1 3 тттттт 6 8 10 12 14 16 -2 -1 0 1 2 Fitted values Theoretical Quantiles Residuals vs Leverage Scale-Location 028°220 1.5 1.0 VIStandardized residuals Standardized residuals -1 0 1 2 O doo 80000 0.5 --- Cook's distance 280.5 0.0 -3 TTTT 8 10 12 14 6 16 0.00 0.04 0.08 Fitted values Leverage

#From Residuals VS Leverage graph, observation number '22,26,28' appears to be outliers.

#Model value when xx=depth=1 cm
ypred=-0.9939+1.1766*(1)
ypred
#[1] 0.1827
#Yes this model makes physical sense.

####### MODEL WITHOUT INTERCEPT
fit1=lm(y~xx-1)
fit1
Call:
lm(formula = y ~ xx - 1)
Coefficients:
xx
1.088

summary(fit1)
Call:
lm(formula = y ~ xx - 1)

Residuals:
Min 1Q Median 3Q Max
-3.4195 -0.7197 0.0438 0.8794 2.4920
Coefficients:
Estimate Std. Error t value Pr(>|t|)
xx 1.08850 0.02099 51.85 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 1.252 on 29 degrees of freedom
Multiple R-squared: 0.9893, Adjusted R-squared: 0.989
F-statistic: 2688 on 1 and 29 DF, p-value: < 2.2e-16

#From the output of 'lm()' command, we get,slope=1.088
#Regression equation is: yhat= 1.088*(xx)
y1pred=1.088*(1)
y1pred
#[1] 1.088par(mfrow=c(2,2))
plot(fit1)

Normal Q-Q Residuals vs Fitted 220 2 - 2 29 207 1 - 0 0 - Residuals Standardized residuals - -2 -2 -1 ...o Presentacdoo 7°28026 - -4 8 16 -2 -1 0 1 2 10 12 14 Fitted values Theoretical Quantiles Scale-Location Residuals vs Leverage 02828 1.5 1 220 220 OSOO 1.0 to 8 8 V Standardized residuals 0 -1 0 1 2 0 Standardized residuals 0 0.5 0.0 -3 -- Gook's distance ----0.5 TTTTTTT 0.00 0.02 0.04 0.06 8 10 12 14 16 Fitted values Leverage

#From Residuals VS Leverage graph, observation number '22,26,28' appears to be outliers.

### Comparing Models
# We get values of R-square and Residual standard error from output of 'summary()' command.
# Residual error is almost similar. [Residual error for Model with Intercept=1.245 and that for model without intercept= 1.252]

# R-squared is the proportion of variance explained by the regressor(s) in the model.
# R-square (Model With Intercept)=0.8876
# R-square (Model WithOUT Intercept)= 0.9893 [98.93% i.e. about 99% of variation is explained by the regressor (xx) in the model.]
# The model with Higher value of R-square is preferred. [Higher the value of R-square, BETTER is the model].
# Thus, model without intercept is better.