Question

Breakdowns of machines that produce steel cans are very costly. The more breakdowns, the fewer cans produced, and the smaller the company's profits. To help anticipate profit loss, the owners of a can company would like to find a model that will predict the number of breakdowns on the assembly line. The model is fit using the least squares procedure, the residuals are plotted against y-hat. There is no data set needed for this problem. A variance stabilizing transformation was performed and the following model was fit:

Where y is the number of breakdowns per 8-hour shift,

x1 = {1 if afternoon, 0 if not}

x2= {1 if midnight shift, 0 if not}

x3= temperature of the plant

x4= number of inexperienced personnel working on the assembly line.

y* = sqrt(y) =Beta0 + Beta1x1 + Beta2x2 + Beta3x3 + Beta4x4 + e

which produced the prediction equation:

y*-hat = 1.3 + .008x1 - .13x2 + .0025x3 + .26x4

a) Use the equation to predict the number of breakdowns (y not y*) during the midnight shift if the temperature of the plant at that time is 87 degrees and if there is only one inexperienced worker on the assembly line. Use 2 decimal places.

b) A 95% prediction interval for y* when x₁=0, x₂=0, x₃=90 and x₄=2 is (1.965, 2.125). For those same values of the independent variables, find a 95% prediction interval for y, the number of breakdowns per 8-hour shift.

c) A 95% confidence interval for y* when x₁=0, x₂=0, x₃=90 and x₄=2 is (1.965, 2.125). Using only the information in this problem, is it possible to find a 95% confidence interval for E(y)? If so, find it. If not, explain why it is not possible.

Answer 1