# Regression Help Needed!

1. Mar 4, 2013

1. The problem statement, all variables and given/known data
Okay, so the problem I have goes like this:

Shown below are the number of galleys for a manuscript (X) and the total dollar cost of correcting typographical errors (Y) in a random sample of recent orders handled by a rime specializing in technical manuscripts. Since Y involves variable costs only, an analyst wished to determine whether regression through the origin model is appropriate for studying the relation between the two variables.

Here is the data:

Y X
128.0 7.0
213.0 12.0
191.0 10.0
178.0 10.0
250.0 14.0
446.0 25.0
540.0 30.0
457.0 25.0
324.0 18.0
177.0 10.0
75.0 4.0
107.0 6.0

2. Relevant equations
The first part of the problem is to fit a regression model through the origin and state the estimated equation.
I used R and I got this as my equation:
a. Ŷ=18.02830X

The part that I am stuck on is part c and d of this problem which goes:

c. In estimating costs of handling prospective orders, management has used a standard of $17.50 per galley for the cost of correcting typographical errors. Test whether or not this standard should be revised; use alpha=.02. State the alternatives, decision rule, and conclusion. d. Obtain a prediction interval for the correction cost on a forthcoming job involving 10 galleys. Use a confidence coefficient of 98%. 3. The attempt at a solution I am really stuck on this and I don't know where to begin. Do I just plug in the$17.50 into my estimated equation and figure it out from there?
Neither the book or the notes that my professor has given are proving to be very useful in understanding what this problem is looking for and how to figure this out.
I really don't want this to be deleted because of my lack of attempt, but I am truly stuck as to what this problem is asking for and how to obtain a solution.

Thank you!

2. Mar 4, 2013

### Ray Vickson

Your estimated slope (= 18.02830) is different from the 'standard' slope (= 17.5). However, the difference could be largely due to random errors, and you want to assess whether or not that is the case. In other words, is the difference you are seeing (between 17.5 and 18.0) 'real', or is is it just due to chance?

Of course, you can never be 100% sure, but the question is asking whether you can be 98% sure. There are standard formulas for confidence intervals in regression that allow you to work this out. (However, most of them involve a model with an intercept as well as a slope, so you may need to look at slightly different formulas, or else follow the method in the book or notes and derive the appropriate results for yourself.)

3. Mar 4, 2013

So in other words, I am testing:
Ho: β1=17.50
Ha: β1≠17.50

?

4. Mar 4, 2013

Here are my numbers from my confidence interval:

From R: MSE=.9077103, s2{b1}= 3.104238e-05, s{b1}= 0.005571569
Confidence interval: b1±t(1-α/2,n-1) s{b1}
18.02830±t(1-.02/2,12-1)(0.005571569)
=18.02830±(2.718)(0.005571569)
(18.01316, 18.04344)

Could someone check to see if this is correct?

Thanks!!!