1. a)
data<-read.table("hospital.txt", sep="\t", dec=".", header=TRUE)
head(data)
## ID Stay Age InfctRsk Culture Xray Beds MedSchool Region Census Nurses
## 1 5 11.20 56.5 5.7 34.5 88.9 180 2 1 134 151
## 2 10 8.84 56.3 6.3 29.6 82.6 85 2 1 59 66
## 3 11 11.07 53.2 4.9 28.5 122.0 768 1 1 591 656
## 4 13 12.78 56.8 7.7 46.0 116.9 322 1 1 252 349
## 5 18 11.62 53.9 6.4 25.5 99.2 133 2 1 113 101
## 6 23 9.78 52.3 5.0 17.6 95.9 270 1 1 240 198
## Facilities
## 1 40.0
## 2 40.0
## 3 80.0
## 4 57.1
## 5 37.1
## 6 57.1
I<-data$InfctRsk
S<-data$Stay
a<-lm(I ~ S)
plot(S, I)
abline(a)
1. b)
summary(a)
##
## Call:
## lm(formula = I ~ S)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.6145 -0.4660 0.1388 0.4970 2.4310
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1.15982 0.95580 -1.213 0.23
## S 0.56887 0.09416 6.041 1.3e-07 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.024 on 56 degrees of freedom
## Multiple R-squared: 0.3946, Adjusted R-squared: 0.3838
## F-statistic: 36.5 on 1 and 56 DF, p-value: 1.302e-07
Intercept estimate=-1.15982, std.error= 0.95580
If you would be 0 days in hospital infection risk would be -1.16%
but 0 days in hospital is not realistic. Every day that you are in
hospital risk increases by 0.57%.
1. c)
confint(a)
## 2.5 % 97.5 %
## (Intercept) -3.0745094 0.7548714
## S 0.3802334 0.7574995
{0.3802334, 0.7574995}
We can tell there is significant statistical linear connection
between days in hospital and infection risk.
1. d)
summary(a)$sigma^2
## [1] 1.049577
=1.049577
2. a)
Data<-read.table("muscle.txt", sep="\t", dec=".", header=FALSE)
DATA <- read.table(text = Data$V1, header = FALSE)
head(DATA)
## V1 V2
## 1 106 43
## 2 106 41
## 3 97 47
## 4 113 46
## 5 96 45
## 6 119 41
hist(DATA$V2)
hist(DATA$V1)
boxplot(DATA$V2)
boxplot(DATA$V1)
b<-lm(V1 ~ V2, data=DATA)
plot(DATA$V2, DATA$V1)
abline(b)
2. b)
coef(b)
## (Intercept) V2
## 156.346564 -1.189996
´=-1.189996 One year decreases muscle mass 1.19 units.
156.346564-1.189996*60
## [1] 84.9468
Estimate for muscle mass for women at 60 is 84.95
residuals(b)[8]
## 8
## 4.443252
=4.443252
2.c)
predict(b,
newdata=data.frame(V2=60),
interval="confidence",
level=0.95)
## fit lwr upr
## 1 84.94683 82.83471 87.05895
=84.95 82.83 87.06
predict(b,
newdata=data.frame(V2=60),
interval="prediction",
level=0.95)
## fit lwr upr
## 1 84.94683 68.45067 101.443
=84.95 68.45 101.44
2. d)
summary(b)
##
## Call:
## lm(formula = V1 ~ V2, data = DATA)
##
## Residuals:
## Min 1Q Median 3Q Max
## -16.1368 -6.1968 -0.5969 6.7607 23.4731
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 156.3466 5.5123 28.36 <2e-16 ***
## V2 -1.1900 0.0902 -13.19 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 8.173 on 58 degrees of freedom
## Multiple R-squared: 0.7501, Adjusted R-squared: 0.7458
## F-statistic: 174.1 on 1 and 58 DF, p-value: < 2.2e-16
There is statistically significant evidence of a negative linear
association between age and muscle mass at the 5% significance
level.
2. e)
par(mfrow = c(2,2))
plot(b)
3. a)
data3<-read.table("Stopdist.txt", sep="\t", dec=".", header=TRUE)
head(data3)
## StopDist Speed
## 1 4 4
## 2 2 5
## 3 8 5
## 4 8 5
## 5 4 5
## 6 6 7
c <- lm(StopDist ~ Speed, data = data3)
summary(c)
##
## Call:
## lm(formula = StopDist ~ Speed, data = data3)
##
## Residuals:
## Min 1Q Median 3Q Max
## -25.141 -7.300 -2.141 6.044 35.946
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -20.2734 3.2384 -6.26 4.25e-08 ***
## Speed 3.1366 0.1517 20.68 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 11.8 on 61 degrees of freedom
## Multiple R-squared: 0.8752, Adjusted R-squared: 0.8731
## F-statistic: 427.7 on 1 and 61 DF, p-value: < 2.2e-16
plot(data3$Speed, data3$StopDist)
abline(c)
3. b)
par(mfrow = c(2,2))
plot(c)
3. c)
data3$StopDist <- sqrt(data3$StopDist)
d<-lm(StopDist~Speed, data=data3)
summary(d)
##
## Call:
## lm(formula = StopDist ~ Speed, data = data3)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.4879 -0.5487 0.0098 0.5291 1.5545
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 0.918283 0.197406 4.652 1.82e-05 ***
## Speed 0.252568 0.009246 27.317 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.7193 on 61 degrees of freedom
## Multiple R-squared: 0.9244, Adjusted R-squared: 0.9232
## F-statistic: 746.2 on 1 and 61 DF, p-value: < 2.2e-16
par(mfrow = c(2,2))
plot(d)
3. d)
newdata <- data.frame(Speed = seq(10, 50, by = 10))
ennustus <- predict(d,
newdata,
interval = "prediction",
level = 0.95)
ennustus
## fit lwr upr
## 1 3.443966 1.984875 4.903057
## 2 5.969649 4.519884 7.419414
## 3 8.495332 7.031415 9.959249
## 4 11.021015 9.520132 12.521898
## 5 13.546698 11.987657 15.105739
3. e)
aenn<- ennustus^2
aenn
## fit lwr upr
## 1 11.86090 3.93973 24.03997
## 2 35.63671 20.42935 55.04771
## 3 72.17067 49.44080 99.18664
## 4 121.46277 90.63292 156.79793
## 5 183.51303 143.70392 228.18335