Question

A substance used in biological and medical research is shipped by air-freight 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 is appropriate.
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

a. Obtain the estimated regression function. Plot the estimated regression function and the data. Does a linear regression function appear to give a good fit here?

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

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

d. Verify that your fitted regression line goes through the point (X bar, Y bar)

e. Calcuate s.

f. Test the hypothesis that beta1=0, using alpha=0.05.

Answer 1

Answer:

a)  y = 10 + 4x and see attached for part ii.

b) y = 14.

c) y = 18

d) Please see attached diagram

mean(x) = 1 and mean(y) = 14.2.. Mean is the same as bar(x) and bar(y) respectively

e,i)

s(b0) = 0.6633

s(b1) = 0.4690

f)  H0: b1 = 0

   H1: b1 ≠ 0

We conduct a t - test and arrived at

t = 8.528785

p-value = 6.613545e-06

@ alpha = 5%.

The test is significant, therefore, we reject the null hypothesis.

Explanation:

a) We fit the regression equation. The Y is dependent variable and X is the independent variable. After fitting, we obtained

y = 10 + 4x

And we fit the line of regression as shown in the attached.

b) We are to substitute x = 1 into the regression equation obtained. This is;

===>   y = 10 + 4*1 = 14

c) When x= 2, we have;

===>   y = 10 + 4*2 = 18

d) We just need to plot the outcome of the model. Although we also show the value of bar(x) and bar(y). See attached (4 in 1 plots) for the plot.

e) The s is the sample standard deviation, which we can also call standard error.

s(b0) = 0.6633

s(b1) = 0.4690

f) Last we conducted the test,

t = b1-b/S(b1) =  (4 - 0)/ 0.4690 = 8.528785

The p-values of (t = 8.528785) = 6.613545e-06 which < 0.05. Therefore, we reject H0.