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Short-handed goals scored during the season. Power play goals against. Penalty killing percentage. Measures a team s ability to prevent goals while its opponent is on a power play. Opponent power play goals divided by opponent s opportunities. Short-handed goals against. Fit a multiple linear regression model relating wins to the other variables. Estimate 2 and find the standard errors of the regression coefficients.

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<% if request( posted )= true then Dim lineCount lineCount=1 Call printList ( mycar ,lineCount)

12-12.

where x1 g xi1 n and x2 g xi2 n. (a) Write out the least squares normal equations for this model. (b) Verify that the least squares estimate of the intercept in this model is g yi n y. 0 (c) Suppose that we use yi y as the response variable in the model above. What effect will this have on the least squares estimate of the intercept

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Call printList ( mytruck ,lineCount) Response.Write(lineCount & ) end if %> <script runat= server language= VBScript > sub printList(item,counter) Dim r r = Response r.Write( First list for & item & ) counter=counter + 1 r.Write( request(item) & ) counter=counter + 1 r.Write( Next list for & item & ) counter=counter + 1 Dim x for x=0 to request.QueryString.GetValues(item).Length-1 r.Write(request.QueryString.GetValues(item)(x) & ) counter=counter + 1 next end sub </script>

x1 2

x2 2

After making these changes and executing this page, another exception is thrown. It states Exception Details: System.NullReferenceException: Object reference not set to an instance of an object (see Figure 17.7).

12-2

In multiple linear regression problems, certain tests of hypotheses about the model parameters are useful in measuring model adequacy. In this section, we describe several important hypothesis-testing procedures. As in the simple linear regression case, hypothesis testing requires that the error terms i in the regression model are normally and independently distributed with mean zero and variance 2.

12-2.1

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The test for significance of regression is a test to determine whether a linear relationship exists between the response variable y and a subset of the regressor variables x1, x2, p , xk. The

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This exception is caused because the for-each loop returns an entry array under ASP, but now returns a null if there are no values. The revised code follows:

H0: H1:

0 (12-17)

p 0 implies that at least one of the regressor variables Rejection of H0: 1 2 k x1, x2, p , xk contributes significantly to the model. The test for significance of regression is a generalization of the procedure used in simple linear regression. The total sum of squares SST is partitioned into a sum of squares due to regression and a sum of squares due to error, say, SST SSR SSE

<% dim i if not Request.QueryString.GetValues( mycar ) is nothing then for each i in Request.QueryString.GetValues( mycar ) %> Car: <%=i%> <% Next end if %>

p 0 is true, SSR 2 is a chi-square random variable with k Now if H0: 1 2 k degrees of freedom. Note that the number of degrees of freedom for this chi-square random variable is equal to the number of regressor variables in the model. We can also show the SSE 2 is a chi-square random variable with n p degrees of freedom, and that SSE and SSR p 0 is are independent. The test statistic for H0: 1 2 k

(12-18)

This code checks to see if the GetValues( mycar ) returns anything prior to attempting to loop through the list. The code finally executes but does not quite operate the same as in the original ASP page.

We should reject H0 if the computed value of the test statistic in Equation 12-18, f0, is greater than f ,k,n p. The procedure is usually summarized in an analysis of variance table such as Table 12-9. We can find a computing formula for SSE as follows: SSE Substituting e y

(12-19)