b. There appears to be a positive relationship between X and Y; that is, as X increases, so does Y.
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1 .46. a. The frst varable (X) s the frst umber the par ad s plotted o the horzotal axs, whle the secod varable (Y) s the secod umber the par ad s plotted o the vertcal axs. The scatterplot s show the fgure below: b. There appears to be a postve relatoshp betwee X ad Y; that s, as X creases, so does Y..48. a-b. The scatterplot s show the fgure below. Notce that there s a egatve relatoshp betwee X ad Y.
2 .3. Two pots are eeded to graph a straght le. Whe x, y. Whe x, y/. The graph s show below... a. From the MINITAB prtout, ˆβ.4736 ad ˆβ 63.86, so that the least squares le s y ˆ β + ˆ β x ˆ x Note that the lecture otes, we use the otato α for the tercept ad β for the slope coeffcet. The graph of the least-squares le ad the te data pots s show below. I order to test for lear relatoshp betwee X ad Y, we test the hypothess H : ˆ β H a : ˆ β ad the test statstc s ˆ β β t s.75
3 The p-value s P (t >.75).475 whch we ca see the MINITAB output. ce the observed level of sgfcace s so large, H s ot rejected. We caot coclude that X ad Y are learly related (sce β mght be zero). b. From the MINITAB prtout, ˆβ.3773 ad ˆβ 6.5, so that the least squares le s y ˆ β + ˆ β x ˆ x The graph of the least-squares le ad the te data pots s show below. I order to test for lear relatoshp betwee X ad Y, the hypothess to be tested s H : ˆ β H a : ˆ β ad the test statstc s t ˆ β β s 3.5 The p-value s P (t > 3.5).6, take from the prtout. ce the observed level of sgfcace (.e., p-value) s so smaller tha.5, H s rejected. We ca coclude that there s evdece that X ad Y are learly related. c. From the result of part a ad b, X meda home prce appears to be the better predctor of the housg dex..9. a. The mea (average) crease Y for a oe-ut chage X s gve by β ( β, the otato of Prof. Wag). The 9% cofdece terval s ˆ β ± t. 5 (std. error of ˆβ ) 7.46 ±.8(.88) 7.46 ± 3.3, takg the stadard error value from the prtout.
4 or 4.94 < β < Itervals costructed ths maer wll eclose β 9% of the tme repeated samplg. Hece, we are farly cofdet that ths partcular terval ecloses β. b. From the last le of the prtout, the 95% cofdece terval for E(y) whe x s < E(y) < 9736 c. For each house the sample, the prce per square foot s calculated as Z Y / X, ad the results are show below The average cost per square foot s z z 5.4 Ths s ot the same as ˆβ 4.46 ad should ot be, sce they are calculated totally dfferet ways. d. From the last le of the prtout, the 95% predcto terval for y whe x s 9,83 < x <,33.3. a. From the MINITAB prtout, ˆβ 49.5 ad ˆβ 648. The least squares le s the yˆ x b. The hypothess of terest s H : β H a : β ad the test statstc s t ˆ β β s 8.9 The p-value s P (t > 8.9) <.5 <. (the sgfcace level). Therefore, we ca reject H. There s suffcet evdece to dcate that the explaatory varable X does help predctg values of the depedet varable Y. c. The data pots are show the fgure below. Note the strog lear relatoshp.
5 d. Whe x 8, the estmate of E(y), s y ˆ (8) 37,539 ad the 95% cofdece terval s gve the last le of prtout to be 35,65 < E(y) < 39,48 e. Whe x 8, the estmate of y s ŷ 37,539 ad the 95% predcto terval s 34,63 < y < 4, We wll perform the calculatos by had. a. Prelmary calculato: x y 5 x 7 The, xy yy y x y 5 ( x )( y ) x y 7 7 ( ) x x ( ) y y 5 6 ˆ xy 7 β.7 ad ˆ α y ˆ β x.7(). ad the least squares le s Y ˆ ˆ α + ˆ β X +. 7X
6 b. The least-squares le s graphed as follows: * whe x, yˆ +.7(). * whe x, yˆ +.7().7. The graph s show above. c. Calculate E T R ( XY ) 49 YY 6 XX. ad. s E.3667 ( ) 3 To test H : β, H a : β, the value of the test statstc s ˆ β β.7 t s.3667 d. The 9% cofdece terval for β s ˆ s.3667 β ± t.5 >.7 ± ±. 45 or.49 < β <. 5 e. Whe x, the estmate of E(y) s yˆ +.7(). 7 ad the 9% cofdece terval s ( x x) ( ) yˆ ± s [ + ] >.7 ± [ + ] 5.7 ±. 78 or.9 < E(y) <.48
7 f. Calculate xy 7 r. 94 (6) yy Because r (.94).87, we kow that 8.7% of the total varato Y ca be explaed by the varable X. There s a hgh postve correlato betwee X ad Y..54 a. The hypothess of terest s H β H β : : ad the test statstc s ˆ β β 5.8 t s The p-value s P (t > 5.48) < (.5).5, whch s smaller tha the sgfcace level ad we ca reject H. There s evdece to dcate that the explaatory varable X does help predctg values of the depedet varable Y. b. Whe x, the estmate of E(y) s yˆ () 38. ad the 9% cofdece terval s ( x ) p x (.5) yˆ ± s [ + ] > 38. ± [ + ] ± 4.63 or < E(y) < Whe x, yˆ () 54. ad the 9% predcto terval s ( x ) p x (.5) yˆ ± s [ + ] > 54. ± [ + + ] ±.86 or 4.94 < y <
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