Question

Xi = the number of times carton i was transferred from one aircraft to another Yi = the number of ampules broken upon arrival i 1 2 3 4 5 6 7 8 9 10 Xi 1 0 2 0 3 1 0 1 2 0 Yi 16 9 17 12 22 13 8 15 19 11

i. Fit the Poisson regression model assuming Yi ∼ Poisson(λi) where λi = λ(x, β) = eβ0+β1Xi is the response function. ii. Estimate of the expected number of broken ampules when X = 0, 1, 2, 3 and compare your results with Part 1(b). iii. Plot the Poisson and linear regression functions, together with the data. Which model appears to be a better fit? iv. Management wishes to estimate the probability that 10 or fewer ampules are broken when there is no transfer of the shipment. Use the fitted Poisson regression function to obtain the estimate. v. Obtain an approximate 95% confidence interval for the slope β1 and interpret it.

Answer 1

Solution-

Ans(1) - A substance used in biological and medical research is shipped by airfreight to users in cartons of 1,000 ampules. The data below, involving 10 shipments, were collected on the number of times the carton was transferred from one aircraft to another over the shipment route (X) and the number of ampules found to be broken upon arrival (Y). Assume that first-order regression model (1.1) is appropriate. using the R-software follow the following steps

Step(1)-

transfers <- c(1,0,2,0,3,1,0,1,2,0)
breakages <- c(16,9,17,12,22,13,8,15,19,11)

airfreight.model <- lm(breakages~transfers)
summary(airfreight.model)

Step(2)-

# Estimated regression function is (Breakages) = 4*(Transfers)+10.2
plot(breakages~transfers)
abline(airfreight.model)

Step(3)-  

# Adjusted R-squared is 0.8885, therefore, this linear regression appears to be a good fit for the data.
airfreight.model$coefficients

## (Intercept)   transfers 
##        10.2         4.0

Ans(2) -

A point estimate of the expected number of broken ampules when X = 1 transfer is made.

point1<-4*1+10.2
point1

## [1] 14.2

Now

Estimate the increase in the expected number of ampules broken when there are 2 transfers as compared to 1 transfer.

point2<-4*2+10.2
point2 - point1

## [1] 4

3rd part-

plot(breakages~transfers)
points(mean(transfers),mean(breakages), pch = 19, col = "blue")
abline(airfreight.model)

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