search for: fb3

...data.frame(height=ht2)) # OK plot(women, xlab = "Height (in)", ylab = "Weight (lb)") lines(ht2,ph2) 3) height <- women$height # 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 weight <- women$weight # 115 117 120 123 126 129 132 135 139 142 146 150 154 159 164 fb3 <- bs(height, df = 5) fm3 <- lm(weight ~ fb3) ht3 <- seq(57, 73, len = 200) ph3 <- predict(fm3, data.frame(height=ht3)) # Error message about newdata. Why ? plot(women, xlab = "Height (in)", ylab = "Weight (lb)") lines(ht3,ph3) # no line Thanks for the reason of th...

Dear All, I am analysing some data for a colleague (not my data, gotta be published so I cannot divulge). My response variable is the number of matings observed per day for some fruitlies. My factors are: Day: the observations were taken on 9 days Regime: 3 selection regimes Line: 3 replicates per selection regime. I have 81 observations in total The lines are coded A to I, so I do not need

...int2<-function(x2){ int_2<- (1/p+log(x2/lamda))*(-p/lamda) * j * choose(n,j) * (pweibull(x2,shape=p,scale=lamda))^(j-1) * (1 - pweibull(x2,shape=p,scale=lamda))^(n-j) * dweibull(x2,shape=p,scale=lamda) int_2 } v2[j]<-integrate(int2,lower=0,upper=t)$value } sum(v2) } #calculation for t3 fb3<-function(t,r){ v3<-numeric(0) for(j in 1:r){ int3<-function(x3){ int_3<- (p/lamda)^2 * j * choose(n,j) * (pweibull(x3,shape=p,scale=lamda))^(j-1) * (1 - pweibull(x3,shape=p,scale=lamda))^(n-j) * dweibull(x3,shape=p,scale=lamda) int_3 } v3[j]<-integrate(int3,lower=0,upper=t)$value }...

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