data(women)
attach(women)
head(women)
## height weight
## 1 58 115
## 2 59 117
## 3 60 120
## 4 61 123
## 5 62 126
## 6 63 129
wmod<-lm(weight~height, data = women)
summary(wmod)
##
## Call:
## lm(formula = weight ~ height, data = women)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.7333 -1.1333 -0.3833 0.7417 3.1167
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -87.51667 5.93694 -14.74 1.71e-09 ***
## height 3.45000 0.09114 37.85 1.09e-14 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.525 on 13 degrees of freedom
## Multiple R-squared: 0.991, Adjusted R-squared: 0.9903
## F-statistic: 1433 on 1 and 13 DF, p-value: 1.091e-14
confint(wmod,level = .9)
## 5 % 95 %
## (Intercept) -98.030599 -77.002734
## height 3.288603 3.611397
newdata <- data.frame(height = 48)
weight.wmod <- lm(weight ~ height)
(predy <- predict(weight.wmod, newdata, interval="predict") )
## fit lwr upr
## 1 78.08333 73.3104 82.85627
#predict the weight given that the height is 48in.
(confy <- predict(weight.wmod, newdata, interval="confidence") )
## fit lwr upr
## 1 78.08333 74.62983 81.53684
confy %*% c(0, -1, 1)
## [,1]
## 1 6.907015
predy %*% c(0, -1, 1)
## [,1]
## 1 9.545874
confy[1] == predy[1]
## [1] TRUE
#Both intervals are centered at the same place but the predict interval is wider than the confidence interval
weight <- rnorm(136.7, sd=15.49869)
height <- rnorm(65, sd=4.472136)
var.test(weight, height)
##
## F test to compare two variances
##
## data: weight and height
## F = 15.647, num df = 135, denom df = 64, p-value < 2.2e-16
## alternative hypothesis: true ratio of variances is not equal to 1
## 95 percent confidence interval:
## 10.07854 23.51869
## sample estimates:
## ratio of variances
## 15.64706
#This is the F test to compare two vairances. There is a strong relationship between the weight and height