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Unformatted text preview: ily seen in scatterplots of the original data or by finding the regression model with and without the points.#######Œ##z#######R###Ú#######À## #### ###ŒÎv=######" ####C#a#l#i#b#r#i#########*#########ÿ####Œ##########z###,4# þ###Œ4#####$Ý4#ŒŒÎv= ##### ##########Œ############z###y## x## ###### ###$ŒÎv=ŒÎv=#########Œ##########z###,4# þ###Œ4#####$Ý4#ŒŒÎv= ##### ##########Œ### #########z###y## x## ###### ###$ŒÎv=ŒÎv=#########Œ######################z###,4# þ###Œ4#########$Ý4#ŒŒÎv= #b###If removing the point gives a very different model, we say that the point is an influential point. ###O###P###a############Œ############z###y## x## ###### ###$ŒÎv=ŒÎv=#########Œ##########z###,4# þ###Œ4#####$Ý4#ŒŒÎv= ##### ##########Œ### #########z###y## x## ###### ###$ŒÎv=ŒÎv=#########Œ##########z###,4# þ###Œ4#####$Ý4#ŒŒÎv= #Œ###E) Originally there were only four points here. Suppose instead that we had started with 50 points clustered in essentially the same region and displaying an association of roughly the same strength and direction. Would our fifth points sill be as influential? Where would you locate one additional point so influential that it changed the line as dramatically as (10, 0) did above?#########Œ######! #####z###y## x## ###### ###$ŒÎv=ŒÎv=#########Œ##########z###,4# þ###Œ4#####$Ý4#ŒŒÎv= ##### ##########Œ### #########z###y## x## ###### ###$ŒÎv=ŒÎv=#########Œ#### ####### #####z###,4# þ###Œ4#########$Ý4#ŒŒÎv= #ú###The fifty-first points would not be as influential, since the pattern is more established by 50 points than it was by 4. In order to be as influential as (10, 0) an additional point would have to be much further out along the x-axis, maybe (100, 0).####ã###ä############Œ######&#####z###y## x## ###### ###$ŒÎv=ŒÎv=#########Œ#### #####z###,4# þ###Œ4#####$Ý4#ŒŒÎv= ###### ##### ##########Œ######(#####z###y## x## ###$ŒÎv=ŒÎv=#########Œ##########z###,4# þ###Œ4#####$Ý4#ŒŒÎv= ##### ##########Œ### ###*#####z###y## x## ###### ###$ŒÎv=ŒÎv=#########Œ##########z###,4# þ###Œ4#####$Ý4#ŒŒÎv= ##### ##########Œ### ###,#####z###y## x## ###### ###$ŒÎv=ŒÎv=#########Œ##########z###,4# þ###Œ4#####$Ý4#ŒŒÎv= ####MORE EXAMPLES##############Œ######.#####z###y## x## ###### ###$ŒÎv=ŒÎv=#########Œ##############z###,4# þ###Œ4#########$Ý4#ŒŒÎv= #Œ###Example 3 The following is a scatterplot of the final exam score versus midterm score for 11 sections of an introductory statistics class:#### ############Œ######0#####z###y## x## ###### ###$ŒÎv=ŒÎv=#########Œ##########z###,4# þ###Œ4#####$Ý4#ŒŒÎv= #.###The correlation coefficient for these data is #############Œ######2#####z###y## x## ###### ###$ŒÎv=ŒÎv=#########Œ##########z###,4# þ###Œ4#####$Ý4#ŒŒÎv= #à###. If you had a scatterplot of the final exam score versus midterm score for all individual students in this introductory statistics course, would the correlation coefficient be weaker, stronger, or about the same? Explain.###########Œ### ###4#####z###y## x## ###### ###$ŒÎv=ŒÎv=#########Œ#### #####z###,4# þ###Œ4#####$Ý4#ŒŒÎv= #####Relationships based on averages have a higher correlation coefficients than relationships based on individual data. Therefore, a scatterplot of the final exam score versus midterm score for the individual students would show much more scatter and a weaker correlation coefficient.##########Œ######6#####z###y## x## ###### ###$ŒÎv=ŒÎv=#########Œ############û#####z###,4# þ###"##########$Ý4#ŒŒÎv= #0###E#x#a#m#p#l#e# #4# #A# #r#e#s#i#d#u#a#l#s# #p#l#o#t# #o#f# #t#h#e# #r#e#s#i#d#u#a#l#s# #v#e#r#s#u#s# #t#h#e# #f#i#t#t#e#d# #v#a#l#u#e#s# #f#o#r# #r#e#c#o#r#d#-#b#r#e#a#k#i#n#g# #t#i#m#e#s# #o#f# #f#e#m#a#l#e# #m#a#r#a#t#h#o#n# #r#u#n#n#e#r#s# #f#o#r# #t#h#e# #y#e#a#r#s# #1#9#9#8# ## #2#0#0#3# #i#s#:####### ###Œ##############Œ######8#####z###y## x## ###### ###$ŒÎv=ŒÎv=#########Œ##########z###,4# þ###Œ4#####$Ý4#ŒŒÎv= ##### ##########Œ### ###:#####z###y## x## ###### ###$ŒÎv=ŒÎv=#########Œ##########z###,4# þ###Œ4#####$Ý4#ŒŒÎv= ##### ##########Œ### ###<#####z###y## x## ###### ###$ŒÎv=ŒÎv=#########Œ##########z###,4# þ###Œ4#####$Ý4#ŒŒÎv= #h###Based on this residuals plot, does it seem reasonable to use linear regression for this model? Explain.###########Œ######>#####z###y## x## ###### ###$ŒÎv=ŒÎv=#########Œ##########z###,4# þ###Œ4#####$Ý4#ŒŒÎv= ##### ##########Œ### ###@#####z###y## x## ###### ###$ŒÎv=ŒÎv=#########Œ#### #####z###,4# þ###Œ4#####$Ý4#ŒŒÎv= #v###Since we see clear pattern in the residuals plot, it does not seem reasonable to use linear regression for this model.#############Œ######B#####z###y## x## ###### ###$ŒÎv=ŒÎv=#########Œ##############z###,4# þ###"##########$Ý4#ŒŒÎv= #æ###E#x#a#m#p#l#e# #5# #H#e#r#e# #i#s# #a# #s#c#a#t#t#e#r#p#l#o#t# #o#f# #w#e#i#g#h#t# #v#e#r#s#u#s# #h#e#i#g#h#t# #f#o#r# #s#t#u#d#e#n#t#s# #i#n# #a#n# #i#n#t#r#o#d#u#c#t#o#r#y# #s#t#a#t#i#s#t#i#c#s# #c#l#a#s#s#.# # #T#h#e# #m#e#n#...
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