Last updated: 2019-11-15
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Knit directory: STA_463_563_Fall2019/
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|---|---|---|---|---|
| Rmd | 1c951d9 | dleelab | 2019-11-15 | created |
data0120=read.table("http://www.stat.ufl.edu/~rrandles/sta4210/Rclassnotes/data/textdatasets/KutnerData/Chapter%20%201%20Data%20Sets/CH01PR20.txt")
colnames(data0120)=c("minutes","copiers_number")
fit=lm(minutes~copiers_number,data=data0120)
alphaF=1-0.9
n=nrow(data0120)
W=sqrt(2*qf((1-alphaF),2,(n-2)))#Or W=sqrt(2*qf((1-alphaF),length(fit$coefficients),fit$df.residual))
Yh_hat=predict(fit,newdata=data.frame(copiers_number=c(3,5,7)),data=data0120)
MSE=sum((fit$residuals)^2)/(n-2)
xbar=mean(data0120$copiers_number)
Sxx=sum((data0120$copiers_number-xbar)^2)
s_Yh_hat=sqrt(MSE*(1/n+(c(3,5,7)-xbar)^2/Sxx))
simulCI=cbind(Yh_hat,Yh_hat-W*s_Yh_hat,Yh_hat+W*s_Yh_hat)
colnames(simulCI)=c("fitted","lower","upper")
rownames(simulCI)=c("3","5","7")
simulCI fitted lower upper
3 44.52559 40.83265 48.21853
5 74.59608 71.66417 77.52800
7 104.66658 101.11278 108.22038The 90% simultaneous CI using W-H method is as above.
g=3
alpha=alphaF/g
predict(fit,new=data.frame(copiers_number=c(3,5,7)),interval = "conf",level=1-alpha,data=data0120) fit lwr upr
1 44.52559 40.84285 48.20832
2 74.59608 71.67227 77.51989
3 104.66658 101.12260 108.21056The 90% simultaneous CI using Bonferroni method is as above.
From result above, which simultaneous confidence interval calculation method do you prefer? Why? (Therefore, in practice, before compute the simultaneous CI or PI, you would compare the multiplier first, and use whichever method that yields a smaller multiplier)
Compare the results above, I would prefer the Bonferroni method. Because it provides a shorter confidence interval. Note: Therefore, before we compute the simultaneous CIs (PIs), we will compare the multipliers of different methods, then do the calculation. Here, in this example, multiplier for Bonferroni method is slightly smaller than the W-H method.
t=qt((1-alphaF/g/2),(n-2))
W[1] 2.204725t[1] 2.198632View the simple linear regression fit, and recall the discussion we had in class, adjust your model and fit a regression through the origin model (RTTO) instead. Obtain the estimated regression function.
summary(fit)
Call:
lm(formula = minutes ~ copiers_number, data = data0120)
Residuals:
Min 1Q Median 3Q Max
-22.7723 -3.7371 0.3334 6.3334 15.4039
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.5802 2.8039 -0.207 0.837
copiers_number 15.0352 0.4831 31.123 <2e-16 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 8.914 on 43 degrees of freedom
Multiple R-squared: 0.9575, Adjusted R-squared: 0.9565
F-statistic: 968.7 on 1 and 43 DF, p-value: < 2.2e-16fit2=lm(minutes~copiers_number-1,data=data0120)
summary(fit2)
Call:
lm(formula = minutes ~ copiers_number - 1, data = data0120)
Residuals:
Min 1Q Median 3Q Max
-22.4723 -3.6306 0.2111 6.3694 15.2639
Coefficients:
Estimate Std. Error t value Pr(>|t|)
copiers_number 14.9472 0.2264 66.01 <2e-16 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 8.816 on 44 degrees of freedom
Multiple R-squared: 0.99, Adjusted R-squared: 0.9898
F-statistic: 4358 on 1 and 44 DF, p-value: < 2.2e-16The estimated regression function is \(\hat{Y_i}=14.9472X_i\), (or estimated minutes=14.9472* copiers_number)
Check \(\sum_{i=1}^{n}e_i\neq0\) and \(\sum_{i=1}^{n}e_iX_i=0\).
res=fit2$residuals
sum(res)[1] -5.862797sum(data0120$copiers_number*res)[1] -1.461054e-13Based on the calculation above, \(\sum_{i=1}^{n}e_i=-5.86\) is not equal to 0 and \(\sum_{i=1}^{n}e_iX_i=-1.46*10^{-13}\), almost equal to 0.
Estimate \(\beta_1\) with a 95% confidence interval.
confint(fit2,,level=0.95) 2.5 % 97.5 %
copiers_number 14.4909 15.40356The 95% confidence interval for \(\beta_1\) is (14.49,15.40).
Predict the service time on a call in which six copiers are serviced. Use a 98% confidence level.
predict(fit2,new=data.frame(copiers_number=6),interval="prediction",level=0.98,data=data0120) fit lwr upr
1 89.68338 68.1491 111.2177The 98% prediction interval for the service time on a call in which 6 copiers are serviced is (68.15,111.22)
Estimate the number of minutes spent when there are 2,3,4,5 copiers to be serviced. Use a 95% family confidence coefficient and use Bonferroni method.
alphaF=1-0.95
g=4
alpha=alphaF/g
predict(fit2,new=data.frame(copiers_number=c(2,3,4,5)),interval = "pred",level=1-alpha,data=data0120) fit lwr upr
1 29.89446 6.90242 52.88650
2 44.84169 21.81186 67.87151
3 59.78892 36.70630 82.87154
4 74.73615 51.58583 97.88647The 95% simultaneous PI for the number of minutes spent when there’re 2,3,4,5 copiers to be serviced is as above.
sessionInfo()R version 3.6.1 (2019-07-05)
Platform: x86_64-apple-darwin15.6.0 (64-bit)
Running under: macOS Mojave 10.14.6
Matrix products: default
BLAS: /Library/Frameworks/R.framework/Versions/3.6/Resources/lib/libRblas.0.dylib
LAPACK: /Library/Frameworks/R.framework/Versions/3.6/Resources/lib/libRlapack.dylib
locale:
[1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8
attached base packages:
[1] stats graphics grDevices utils datasets methods base
loaded via a namespace (and not attached):
[1] workflowr_1.4.0 Rcpp_1.0.2 digest_0.6.20 rprojroot_1.3-2
[5] backports_1.1.4 git2r_0.26.1 magrittr_1.5 evaluate_0.14
[9] stringi_1.4.3 fs_1.3.1 whisker_0.3-2 rmarkdown_1.15
[13] tools_3.6.1 stringr_1.4.0 glue_1.3.1 xfun_0.9
[17] yaml_2.2.0 compiler_3.6.1 htmltools_0.3.6 knitr_1.24