Question

Question 3 (24 points) The article 'Drift in Posturography Systems Equipped with a Piezoelectric Force Platform: Analysis and Numerical Compensation' (L. Quagliarella, N. Sasanelli, and V. Monaco, IEEE Transactions on Instrumentation and Measurement, 2008:997-1004) reported the results of an experiment to determine the effect of load on the drift in signals derived from a piezoelectric force plate. The correlation coefficient, y, between output and time was computed for various loads, x, in kN. The data are shown in the following table. х у 0.600 -0.9603 0.368 -0.9468 0.608 -0.9703 0.550 -0.9451 0.586 -0.9612 0.377 -0.9645 0.488 -0.9516 0.619 -0.9482 0.460 -0.9582 0.625 -0.9406 a) (6 points) Create a scatterplot of y vs. x. Describe what you see. b) (3 points) Compute the least-squares line for predicting y from x. Report the fitted line. c) (9 points) Check that assumptions to use the t-distribution for statistical inference are reasonable. Be sure to include any plots used with this discussion. d) (6 points) Find a 98% confidence interval for the slope coefficient. Is there evidence of a linear relationship at the a = 0.02 level? Why?

I do not need help with a. Thank you for all of your help!

Answer 1

b)

x<-c(0.600,0.368,0.608,0.550,0.586,0.377,0.488,0.619,0.460,0.625)
y<-c(-0.9603,-0.9468,-0.9703,-0.9451,-0.9612,-0.9645,-0.9516,-0.9482,-0.9582,-0.9406)

scatter.smooth(y,x)
#install.packages("ggplot2")
library(ggplot2)

data <- data.frame("x" = x, "y" = y, "Name" = c("x","y"))

ggplot(data, aes(x=x, y=y)) +
geom_point()+
geom_smooth(method=lm, se=FALSE)

model<-lm(formula= y~x,data = data)
summary(model)
model

> summary (model) Cal1 1m (formula x, data = data) — у Residuals: Median 1Q -0.016050 0.006625 0.000071 Min 3Q Маx 0.008054 0.013559 Coefficients: Estimate Std. Error t value Pr(> t|) (Intercept) -0.957521 0.018420 -51.983 2.08e-11 *** 0.034343 0.005379 0.157 0.879 х 0.05 0.1''1 Signif. codes: 0 0.001 0.01 Residual standard error: 0.01017 on 8 degrees of freedom Multiple R-squared: 0.003057, Adjusted R-squared: -0.1216 F-statistic: 0.02453 on 1 and 8 DF, p-value: 0.8794 > model Call 1m (formula data) у ~ х, data Coefficients: (Intercept) -0.957521 х 0.005379 scatter.smoothCy,x)

-0.94 -0.95 -0.96 -0.97 0.55 0.60 0.40 0.45 0.50 X А

The regression equation =

$y= -0.957521 &plus; 0.005379 (x)$

c)

We want to see if the residuals follow a normal pattern for us to say that the data comes from a normal distribution and does not have heteroscedasticity.

100 0.0 0.5 1.0 -1.5 -1.0 -0.5 1.5 norm quantiles model$residuals -0.015 0.010 -0.005 0.000 0.005 0.010 о O

The residuals are normal and do not follow a pattern, so we can use t distribution for creating a confidence interval.

d)

Coef= 0.005379
SE= 0.034343
n=10
alpha=0.02
df= n-2

#Margin_of_error= critical value * standard error
critical_value= qt(1-alpha/2,df)
ME= critical_value*SE

#The lower and upper bounds are:

#Lower
Coef-ME

#Upper
Coef+ME

> Coef = 0.005 379 > SE= 0.034 343 > n=10 > alpha 0.02 > df= n-2 > #Margin_of_error= critical value > critical_value= qt (1-alpha/2,df) ME= critical_value*SE standard error > #The lower and upper bounds are: > #Lower > Coef-ME [1]0.09409411 > #Upper > Coef+ME [1] 0.10485 21