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0̞X   X  RClick to edit Master text styles Second level Third level Fourth level Fifth level!     S  0xX  `` X  X*  0P zThe scatter diagram shows hourly earnings in 1994 plotted against highest grade completed for a sample of 570 respondents.&{ 2z { x D <A ??6qE H D 0޽h ? ̙33y___PPT10Y+D=' = @B +-  T L `8(  8R 8 s *|L 8 0:|  . reg EARNINGS S Source | SS df MS Number of obs = 570 ---------+------------------------------ F( 1, 568) = 65.64 Model | 3977.38016 1 3977.38016 Prob > F = 0.0000 Residual | 34419.6569 568 60.5979875 R-squared = 0.1036 ---------+------------------------------ Adj R-squared = 0.1020 Total | 38397.0371 569 67.4816117 Root MSE = 7.7845 ------------------------------------------------------------------------------ EARNINGS | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- S | 1.073055 .1324501 8.102 0.000 .8129028 1.333206 _cons | -1.391004 1.820305 -0.764 0.445 -4.966354 2.184347 ------------------------------------------------------------------------------ 0  '  %        w f 8 6J?  }'INTERPRETATION OF A REGRESSION EQUATION&((2' (   8 s *V|>P ZThis is the output from a regression of earnings on highest grade completed, using Stata. &[ 2Z [  8 0\Z `P 94 2 3f` 8 c $A ??0    H 8 0޽h ? ̙33y___PPT10Y+D=' = @B +    p  ( y0 R   s *X R  s *|L  0|  . reg EARNINGS S Source | SS df MS Number of obs = 570 ---------+------------------------------ F( 1, 568) = 65.64 Model | 3977.38016 1 3977.38016 Prob > F = 0.0000 Residual | 34419.6569 568 60.5979875 R-squared = 0.1036 ---------+------------------------------ Adj R-squared = 0.1020 Total | 38397.0371 569 67.4816117 Root MSE = 7.7845 ------------------------------------------------------------------------------ EARNINGS | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- S | 1.073055 .1324501 8.102 0.000 .8129028 1.333206 _cons | -1.391004 1.820305 -0.764 0.445 -4.966354 2.184347 ------------------------------------------------------------------------------ 0  '  %        w f  6l?  }'INTERPRETATION OF A REGRESSION EQUATION&((2' ( y  s *T|>P For the time being, we will be concerned only with the estimates of the parameters. The variables in the regression are listed in the first column and the second column gives the estimates of their coefficients.  2    0( `P 95 2 3f`   c $A ??0 `  H  0޽h ? ̙33y___PPT10Y+D=' = @B +      ( % R  s *X R  s *|L  0 |  . reg EARNINGS S Source | SS df MS Number of obs = 570 ---------+------------------------------ F( 1, 568) = 65.64 Model | 3977.38016 1 3977.38016 Prob > F = 0.0000 Residual | 34419.6569 568 60.5979875 R-squared = 0.1036 ---------+------------------------------ Adj R-squared = 0.1020 Total | 38397.0371 569 67.4816117 Root MSE = 7.7845 ------------------------------------------------------------------------------ EARNINGS | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- S | 1.073055 .1324501 8.102 0.000 .8129028 1.333206 _cons | -1.391004 1.820305 -0.764 0.445 -4.966354 2.184347 ------------------------------------------------------------------------------ 0  '  %        w f  6 ?  }'INTERPRETATION OF A REGRESSION EQUATION&((2' ( y  s *x|>P In this case there is only one variable, S, and its coefficient is 1.073. _cons, in Stata, refers to the constant. The estimate of the intercept is -1.391. P 2)t     0ԣ `P 96 2 3f`   c $A ?? P  H  0޽h ? ̙33y___PPT10Y+D=' = @B +  HW(  Hx H <A ??  [F  H 0$ `P 97 2 3f H s *d|>P BHere is the scatter diagram again, with the regression line shown.C 2C C  H 6c?  }'INTERPRETATION OF A REGRESSION EQUATION&((2' ( `  H c $A ??@   H H(p P 0 E^( 2H H 0޽h ? ̙33y___PPT10Y+D=' = @B +  <( .W- x   <A ??  [F   6@ ?  }'INTERPRETATION OF A REGRESSION EQUATION&((2' ( `  c $A w??@ w  H E p P 0 E^( 2  s *I |>P q'What do the coefficients actually mean?( 2( (    0 X  `P 98 2 3fH  0޽h ? ̙33y___PPT10Y+D=' = @B + 80 ( .W- x   <A ??  [F R   s *\   6$R ?  }'INTERPRETATION OF A REGRESSION EQUATION&((2' ( `  c $A y??@ y  HW p P 0 E^( 2  s *\\ |>P YTo answer this question, you must refer to the units in which the variables are measured.Z 2Z Z    0ha  `P 99 2 3fH  0޽h ? ̙33y___PPT10Y+D=' = @B +  o( .W- x   <A ??  [F R  s *\   6g ?  }'INTERPRETATION OF A REGRESSION EQUATION&((2' ( `  c $A {??@ {  Hl p P 0 E^( 2   s *q |>P IS is measured in years (strictly speaking, grades completed), EARNINGS in dollars per hour. So the slope coefficient implies that hourly earnings increase by $1.07 for each extra year of schooling. R 2=     0\y  `P :10 2 3fH  0޽h ? ̙33y___PPT10Y+D=' = @B + & ( T  R   s * . ` x  <A ??  [F   0  `P :11 2 3f0  s *胻 |>P We will look at a geometrical representation of this interpretation. To do this, we will enlarge the marked section of the scatter diagram. 2    6 ?  }'INTERPRETATION OF A REGRESSION EQUATION&((2' ( XB   0D  R   s *\ `   c $A ~??@ ~  H8 p P 0 E^( 2H  0޽h ? ̙33y___PPT10Y+D=' = @B +    | ( vw |x | <A )??Y  )F | s * |>P The regression line indicates that completing 12th grade instead of 11th grade would increase earnings by $1.073, from $10.413 to $11.486, as a general tendency.  2   | 6| ?  }'INTERPRETATION OF A REGRESSION EQUATION&((2' (  | 04  `P :12 2 3fdB  | <D8c$OL  | c $J j L | c $J `j L | c $J @j dB | <D? | Hh   >One year 2 dB | <D  | H 0 ;$1.07 2 | H R <$10.41 2^B | 6D  | H쯻 P <$11.49 2^B | 6DH | 0޽h ? ̙33y___PPT10Y+D=' = @B +g  `( H `x  ` <A ??  [F  ` 0  `P :13 2 3fR ` s *d |>P You should ask yourself whether this is a plausible figure. If it is implausible, this could be a sign that your model is misspecified in some way. 2&{     ` 6?  }'INTERPRETATION OF A REGRESSION EQUATION&((2' (   ` H p P 0 E^( 2R  ` s *\ `  ` c $A ??@ H ` 0޽h ? ̙33y___PPT10Y+D=' = @B +& ME x( pq xx  x <A ??  [F  x 0&  `P :14 2 3f x s *+ |>P mFor low levels of education it might be plausible. But for high levels it would seem to be an underestimate.n 2n n  x 6\/ ?  }'INTERPRETATION OF A REGRESSION EQUATION&((2' (   x H3 p P 0 E^( 2R  x s *\ `  x c $A ??@ H x 0޽h ? ̙33y___PPT10Y+D=' = @B +$ KC ( pq x   <A ??  [F R  s *3   0;  `P :15 2 3f  s *? |>P kWhat about the constant term? (Try to answer this question yourself before continuing with this sequence.)l 2l l   6C ?  }'INTERPRETATION OF A REGRESSION EQUATION&((2' (    HdH p P 0 E^( 2`   c $A ??@ H  0޽h ? ̙33y___PPT10Y+D=' = @B +@ g_  ( II@ x   <A ??  [F   0O  `P :16 2 3f+  s *S |>P Literally, the constant indicates that an individual with no years of education would have to pay $1.39 per hour to be allowed to work. 2    6hX ?  }'INTERPRETATION OF A REGRESSION EQUATION&((2' (    H\ p P 0 E^( 2R   s *3 `   c $A ??@ H  0޽h ? ̙33y___PPT10Y+D=' = @B + D<0 ( 0 x   <A ??  [F   0dd  `P :17 2 3f  s *h |>P dThis does not make any sense at all, an interpretation of negative payment is impossible to sustain.e 2e e   6l ?  }'INTERPRETATION OF A REGRESSION EQUATION&((2' (    H0q p P 0 E^( 2R   s *3 `   c $A ??@ H  0޽h ? ̙33y___PPT10Y+D=' = @B + @ i(  x  <A 1?? [E 1  0x  `P :18 2 3f_  s *| |>P A safe solution to the problem is to limit the interpretation to the range of the sample data, and to refuse to extrapolate on the ground that we have no evidence outside the data range.  2    6 ?  }'INTERPRETATION OF A REGRESSION EQUATION&((2' ( dB   <D8c 55@ dB   <D8c PP@    HȆ p P 0 E^( 2R  s *3 `   c $A ??@ ^B  6DjJ p  H  0޽h ? ̙33y___PPT10Y+D=' = @B + P e( X\Wy0XW   0X  `P :19 2 3f[  s *X |>P With this explanation, the only function of the constant term is to enable you to draw the regression line at the correct height on the scatter diagram. It has no meaning of its own. 2    6X ?  }'INTERPRETATION OF A REGRESSION EQUATION&((2' ( dB   <D8c 55@ dB   <D8c PP@    HxX p P 0 E^( 2R  s *3 `   c $A ??@ x  <A ?? [E ^B  6DjJ p  H  0޽h ? ̙33y___PPT10Y+D=' = @B + ` "(    B| ?  ;MULTIPLE REGRESSION WITH TWO EXPLANATORY VARIABLES: EXAMPLE&<(2; < `F 6P p  P6 pZ  s *6P p  H X  - (EARNINGS = b1 + b2S + b3A+ u 2     0X  ^N 92 2 3f  s *֜| eSpecifically, we will look at an earnings function model where hourly earnings, EARNINGS, depend on years of schooling (highest grade completed), S, and a measure of cognitive ability, A.z 2P:&    s *u |@R qThe model has three dimensions, one each for EARNINGS, S, and A. The starting point for investigating the determination of EARNINGS is the intercept, b1.  2- =      s *|{ |2D  Literally the intercept gives EARNINGS for those respondents who have no schooling and who scored zero on the ability test. However, the ability score is scaled in such a way as to make it impossible to score zero. Hence a literal interpretation of b1 would be unwise.n 2      s * |   The next term on the right side of the equation gives the effect of variations in S. A one year increase in S causes EARNINGS to increase by b2 dollars, holding A constant. 2R       s *p |s   nSimilarly, the third term gives the effect of variations in A. A one point increase in A causes earnings to increase by b3 dollars, holding S constant. 2<     ]   s *L |9 K uThe final element of the model is the disturbance term, u. In this observation, u happens to have a positive value. ^v 28# v B  s *޽h ? ̙33v  &  p(  x  <A ?? x  <A ??DS   Bx ?  ;MULTIPLE REGRESSION WITH TWO EXPLANATORY VARIABLES: EXAMPLE&<(2; < /  s *@LJ |   The regression coefficients are derived using the same least squares principle used in simple regression analysis. The fitted value of Y in observation i depends on our choice of b1, b2, and b3. 2       0tև  ^N :11 2 3f  s *Dˇ |  7A sample consists of a number of observations generated in this way. Note that the interpretation of the model does not depend on whether S and A are correlated or not.^ 2    s * |   However we do assume that the effects of S and A on EARNINGS are additive. The impact of a difference in S on EARNINGS is not affected by the value of A, or vice versa. 2).!  H  0޽h ? ̙33 `X(  x  <A ?? x  <A ??DS x  <A ??Tc   B ?  ;MULTIPLE REGRESSION WITH TWO EXPLANATORY VARIABLES: EXAMPLE&<(2; <   s ** |>P 1]The residual ei in observation i is the difference between the actual and fitted values of Y.^ 2  ;&   O   03  ^N :12 2 3fH  0޽h ? ̙33{ +#(    0C | U "fZ  S A ? y    BN ?  ;MULTIPLE REGRESSION WITH TWO EXPLANATORY VARIABLES: EXAMPLE&<(2; <   s *Z |  ]gWe define RSS, the sum of the squares of the residuals, and choose b1, b2, and b3 so as to minimize it.h 2 6    h   0g  ^N :13 2 3fH  0޽h ? ̙33     (  R  s * p^  0v |  . reg EARNINGS S ASVABC Source | SS df MS Number of obs = 570 ---------+------------------------------ F( 2, 567) = 39.98 Model | 4745.74965 2 2372.87483 Prob > F = 0.0000 Residual | 33651.2874 567 59.3497133 R-squared = 0.1236 ---------+------------------------------ Adj R-squared = 0.1205 Total | 38397.0371 569 67.4816117 Root MSE = 7.7039 ------------------------------------------------------------------------------ EARNINGS | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- S | .7390366 .1606216 4.601 0.000 .4235506 1.054523 ASVABC | .1545341 .0429486 3.598 0.000 .0701764 .2388918 _cons | -4.624749 2.0132 -2.297 0.022 -8.578989 -.6705095 ------------------------------------------------------------------------------ 2t.%f  B@ ?  ;MULTIPLE REGRESSION WITH TWO EXPLANATORY VARIABLES: EXAMPLE&<(2; <   s *$ |>P JHere is the regression output for the earnings function using Data Set 21.&K 2J K   0  ^N :19 2 3fx  <A ??p 4MO  H  0޽h ? ̙33:     z (  R  s *$8 R  s *x J u R  s *x :u R  s * pY  0X |  . reg EARNINGS S A Source | SS df MS Number of obs = 570 ---------+------------------------------ F( 2, 567) = 39.98 Model | 4745.74965 2 2372.87483 Prob > F = 0.0000 Residual | 33651.2874 567 59.3497133 R-squared = 0.1236 ---------+------------------------------ Adj R-squared = 0.1205 Total | 38397.0371 569 67.4816117 Root MSE = 7.7039 ------------------------------------------------------------------------------ EARNINGS | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- S | .7390366 .1606216 4.601 0.000 .4235506 1.054523 A | .1545341 .0429486 3.598 0.000 .0701764 .2388918 _cons | -4.624749 2.0132 -2.297 0.022 -8.578989 -.6705095 ------------------------------------------------------------------------------ 2t)%f  B ?  ;MULTIPLE REGRESSION WITH TWO EXPLANATORY VARIABLES: EXAMPLE&<(2; <   0lІ  ^N :20 2 3fL   s *؆ |>P It indicates that earnings increase by $0.74 for every extra year of schooling and by $0.15 for every extra point increase in A.B 2~  x   <A ?? +  H  0޽h ? ̙33     ) (  R  s *x  u R  s *'  R  s * pY  0X |  . reg EARNINGS S A Source | SS df MS Number of obs = 570 ---------+------------------------------ F( 2, 567) = 39.98 Model | 4745.74965 2 2372.87483 Prob > F = 0.0000 Residual | 33651.2874 567 59.3497133 R-squared = 0.1236 ---------+------------------------------ Adj R-squared = 0.1205 Total | 38397.0371 569 67.4816117 Root MSE = 7.7039 ------------------------------------------------------------------------------ EARNINGS | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- S | .7390366 .1606216 4.601 0.000 .4235506 1.054523 A | .1545341 .0429486 3.598 0.000 .0701764 .2388918 _cons | -4.624749 2.0132 -2.297 0.022 -8.578989 -.6705095 ------------------------------------------------------------------------------ 2t)%f  Bb ?  ;MULTIPLE REGRESSION WITH TWO EXPLANATORY VARIABLES: EXAMPLE&<(2; <   0 b  ^N :21 2 3fU  s *`t |>P Literally, the intercept indicates that an individual who had no schooling and an A score of zero would have hourly earnings of -$4.62. B 2R6  x   <A ?? +  H  0޽h ? ̙33F      (  R  s *x  u R  s *'  R  s * pY  0pt |  . reg EARNINGS S A Source | SS df MS Number of obs = 570 ---------+------------------------------ F( 2, 567) = 39.98 Model | 4745.74965 2 2372.87483 Prob > F = 0.0000 Residual | 33651.2874 567 59.3497133 R-squared = 0.1236 ---------+------------------------------ Adj R-squared = 0.1205 Total | 38397.0371 569 67.4816117 Root MSE = 7.7039 ------------------------------------------------------------------------------ EARNINGS | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- S | .7390366 .1606216 4.601 0.000 .4235506 1.054523 A | .1545341 .0429486 3.598 0.000 .0701764 .2388918 _cons | -4.624749 2.0132 -2.297 0.022 -8.578989 -.6705095 ------------------------------------------------------------------------------ 2t)%f  Bt ?  ;MULTIPLE REGRESSION WITH TWO EXPLANATORY VARIABLES: EXAMPLE&<(2; <   0|t  ^N :22 2 3f  s *tt |>P XObviously, this is impossible. The lowest value of S in the sample was 6, and the lowest A score was 22. We have obtained a nonsense estimate because we have extrapolated too far from the data range.^ 24&n  x   <A ?? +  H  0޽h ? ̙33V      (  R  s *   0ڜ|  V & T  6 ?  ;t TEST OF A HYPOTHESIS RELATING TO A REGRESSION COEFFICIENT4<(2: < XB  0DP@ @   04   ~s.d. of b2 known> 2     H >  Mdiscrepancy between hypothetical value and sample estimate, in terms of s.d.:N 2N N `  c $A ??0p    H;     C5% significance test: reject H0: b2 = b2 if z > 1.96 or z < -1.96D    D     fH  ? Z  E0((2 ]   s *K |>P The diagram summarizes the procedure for performing a 5% significance test on the slope coefficient of a regression under the assumption that we know its standard deviation.& 2     0`  `P 91 2 3fB  s *޽h ? ̙33    V (  R  s *   0k |  V & T  6v ?  ;t TEST OF A HYPOTHESIS RELATING TO A REGRESSION COEFFICIENT4<(2: < XB  0DP@ @   0p   ~s.d. of b2 known> 2     H臞 >  Mdiscrepancy between hypothetical value and sample estimate, in terms of s.d.:N 2N N `  c $A ??0p    Hl     C5% significance test: reject H0: b2 = b2 if z > 1.96 or z < -1.96D    D R   s *    0   s.d. of b2 not known> 2       H0 >  Mdiscrepancy between hypothetical value and sample estimate, in terms of s.e.:N 2N N `   c $A ??0    fž  ? Z  E0((2 X  s *o |>P This is a very unrealistic assumption. We usually have to estimate it with the standard error, and we use this in the test statistic instead of the standard deviation.& 2    0  `P 92 2 3fB  s *޽h ? ̙33)    o (  R  s *   0蹍 |  V & T  6 ?  ;t TEST OF A HYPOTHESIS RELATING TO A REGRESSION COEFFICIENT4<(2: < XB  0DP@ @   0h   ~s.d. of b2 known> 2     H >  Mdiscrepancy between hypothetical value and sample estimate, in terms of s.d.:N 2N N `  c $A ??0p    H<     C5% significance test: reject H0: b2 = b2 if z > 1.96 or z < -1.96D    D R   s *    0x"   s.d. of b2 not known> 2       H41 >  Mdiscrepancy between hypothetical value and sample estimate, in terms of s.e.:N 2N N `   c $A ??0    f$  ? Z  E0((2 q  s *͍ |>P Because we have replaced the standard deviation in its denominator with the standard error, the test statistic has a t distribution instead of a normal distribution.B 2u/    0`s  `P 93 2 3fB  s *޽h ? ̙33 #(  R  s *   0P |  V & T  6 ?  ;t TEST OF A HYPOTHESIS RELATING TO A REGRESSION COEFFICIENT4<(2: < XB  0DP@ @   0   ~s.d. of b2 known> 2     Hle >  Mdiscrepancy between hypothetical value and sample estimate, in terms of s.d.:N 2N N `  c $A ??0p    H\'     C5% significance test: reject H0: b2 = b2 if z > 1.96 or z < -1.96D    D R   s *    0   s.d. of b2 not known> 2       HD/ >  Mdiscrepancy between hypothetical value and sample estimate, in terms of s.e.:N 2N N `   c $A ??0   Hb    E5% significance test: reject H0: b2 = b2 if t > tcrit or t < -tcritF @0         fb  ? Z  E0((2    fb  ? f  E0((2 A  s *=b |>P uAccordingly, we refer to the test statistic as a t statistic. In other respects the test procedure is much the same.Bv 21C v   0lCb  `P 94 2 3fB  s *޽h ? ̙33   f(  R  s *   0(Tt |  V & T  0t  `P 95 2 3f  6Zt ?  ;t TEST OF A HYPOTHESIS RELATING TO A REGRESSION COEFFICIENT4<(2: < XB  0DP@ @   s *t |>P *We look up the critical value of t and if the t statistic is greater than it, positive or negative, we reject the null hypothesis. If it is not, we do not.^ 2! m    0]t   ~s.d. of b2 known> 2      Ht >  Mdiscrepancy between hypothetical value and sample estimate, in terms of s.d.:N 2N N `   c $A ??0p    H@ף    C5% significance test: reject H0: b2 = b2 if z > 1.96 or z < -1.96D    D R   s *    0pۣ  s.d. of b2 not known> 2      H>  Mdiscrepancy between hypothetical value and sample estimate, in terms of s.e.:N 2N N `  c $A  ??0    HD   E5% significance test: reject H0: b2 = b2 if t > tcrit or t < -tcritF @0         f ? Z  E0((2    f ? f  E0((2 B  s *޽h ? ̙33]    0 (    0)|   t Distribution: Critical values of t Degrees of Two-tailed test 10% 5% 2% 1% 0.2% 0.1% freedom One-tailed test 5% 2.5% 1% 0.5% 0.1% 0.05% 1 6.314 12.706 31.821 63.657 318.31 636.62 2 2.920 4.303 6.965 9.925 22.327 31.598 3 2.353 3.182 4.541 5.841 10.214 12.924 4 2.132 2.776 3.747 4.604 7.173 8.610 5 2.015 2.571 3.365 4.032 5.893 6.869 & & & & & & & & & & & & & & 18 1.734 2.101 2.552 2.878 3.610 3.922 19 1.729 2.093 2.539 2.861 3.579 3.883 20 1.725 2.086 2.528 2.845 3.552 3.850 & & & & & & & & & & & & & & 120 1.658 1.980 2.358 2.617 3.160 3.373 1.645 1.960 2.326 2.576 3.090 3.291&" "(h  (h  09 `P :18 2 3f  6=?  ;t TEST OF A HYPOTHESIS RELATING TO A REGRESSION COEFFICIENT4<(2: < `  s *:|>P For this reason we need to refer to a table of critical values of t when performing significance tests on the coefficients of a regression equation.B 2BQ  `  c $A !?? 5  !B  s *޽h ? ̙33  a Y @ (  L  c $: `  0l_|   t Distribution: Critical values of t Degrees of Two-tailed test 10% 5% 2% 1% 0.2% 0.1% freedom One-tailed test 5% 2.5% 1% 0.5% 0.1% 0.05% 1 6.314 12.706 31.821 63.657 318.31 636.62 2 2.920 4.303 6.965 9.925 22.327 31.598 3 2.353 3.182 4.541 5.841 10.214 12.924 4 2.132 2.776 3.747 4.604 7.173 8.610 5 2.015 2.571 3.365 4.032 5.893 6.869 & & & & & & & & & & & & & & 18 1.734 2.101 2.552 2.878 3.610 3.922 19 1.729 2.093 2.539 2.861 3.579 3.883 20 1.725 2.086 2.528 2.845 3.552 3.850 & & & & & & & & & & & & & & 120 1.658 1.980 2.358 2.617 3.160 3.373 1.645 1.960 2.326 2.576 3.090 3.291&" "(h  (h  0p `P :19 2 3f  6?  ;t TEST OF A HYPOTHESIS RELATING TO A REGRESSION COEFFICIENT4<(2: < `  c $A "?? 5  "`  s *0|>P At the top of the table are listed possible significance levels for a test. For the time being we will be performing two-tailed tests, so ignore the line for one-tailed tests.& 2  B  s *޽h ? ̙33   x P (  L  c $ ~ L  c $@   0d|   t Distribution: Critical values of t Degrees of Two-tailed test 10% 5% 2% 1% 0.2% 0.1% freedom One-tailed test 5% 2.5% 1% 0.5% 0.1% 0.05% 1 6.314 12.706 31.821 63.657 318.31 636.62 2 2.920 4.303 6.965 9.925 22.327 31.598 3 2.353 3.182 4.541 5.841 10.214 12.924 4 2.132 2.776 3.747 4.604 7.173 8.610 5 2.015 2.571 3.365 4.032 5.893 6.869 & & & & & & & & & & & & & & 18 1.734 2.101 2.552 2.878 3.610 3.922 19 1.729 2.093 2.539 2.861 3.579 3.883 20 1.725 2.086 2.528 2.845 3.552 3.850 & & & & & & & & & & & & & & 120 1.658 1.980 2.358 2.617 3.160 3.373 1.645 1.960 2.326 2.576 3.090 3.291&" "(h  (h  0 `P :20 2 3f  6?  ;t TEST OF A HYPOTHESIS RELATING TO A REGRESSION COEFFICIENT4<(2: < `  c $A #?? 5  #%  s *|>P uHence if we are performing a (two-tailed) 5% significance test, we should use the column thus indicated in the table.&v 2u v H  0޽h ? ̙33    `A (  L  c $@   0D|   t Distribution: Critical values of t Degrees of Two-tailed test 10% 5% 2% 1% 0.2% 0.1% freedom One-tailed test 5% 2.5% 1% 0.5% 0.1% 0.05% 1 6.314 12.706 31.821 63.657 318.31 636.62 2 2.920 4.303 6.965 9.925 22.327 31.598 3 2.353 3.182 4.541 5.841 10.214 12.924 4 2.132 2.776 3.747 4.604 7.173 8.610 5 2.015 2.571 3.365 4.032 5.893 6.869 & & & & & & & & & & & & & & 18 1.734 2.101 2.552 2.878 3.610 3.922 19 1.729 2.093 2.539 2.861 3.579 3.883 20 1.725 2.086 2.528 2.845 3.552 3.850 & & & & & & & & & & & & & & 120 1.658 1.980 2.358 2.617 3.160 3.373 1.645 1.960 2.326 2.576 3.090 3.291&" "(h  (h  0  `P :21 2 3f  6?  ;t TEST OF A HYPOTHESIS RELATING TO A REGRESSION COEFFICIENT4<(2: < `  c $A $?? 5  $o  s *|>P The left hand vertical column lists degrees of freedom. The number of degrees of freedom in a regression is defined to be the number of observations minus the number of parameters estimated.& 2  3   f/|/|?P  gNumber of degrees of freedom in a regression = number of observations - number of parameters estimated.h 2hB  s *޽h ? ̙33    p) (  L  c $@   0p|   t Distribution: Critical values of t Degrees of Two-tailed test 10% 5% 2% 1% 0.2% 0.1% freedom One-tailed test 5% 2.5% 1% 0.5% 0.1% 0.05% 1 6.314 12.706 31.821 63.657 318.31 636.62 2 2.920 4.303 6.965 9.925 22.327 31.598 3 2.353 3.182 4.541 5.841 10.214 12.924 4 2.132 2.776 3.747 4.604 7.173 8.610 5 2.015 2.571 3.365 4.032 5.893 6.869 & & & & & & & & & & & & & & 18 1.734 2.101 2.552 2.878 3.610 3.922 19 1.729 2.093 2.539 2.861 3.579 3.883 20 1.725 2.086 2.528 2.845 3.552 3.850 & & & & & & & & & & & & & & 120 1.658 1.980 2.358 2.617 3.160 3.373 1.645 1.960 2.326 2.576 3.090 3.291&" "(h  (h  0& `P :22 2 3f  65?  ;t TEST OF A HYPOTHESIS RELATING TO A REGRESSION COEFFICIENT4<(2: < `  c $A %?? 5  %  s *A|>P 8In a simple regression, we estimate just two parameters, the constant and the slope coefficient, so the number of degrees of freedom is n - 2 if there are n observations.^ 2  B  s *޽h ? ̙33  F >   (  L  c $9> L  c $ > L  c $@   0tS|   t Distribution: Critical values of t Degrees of Two-tailed test 10% 5% 2% 1% 0.2% 0.1% freedom One-tailed test 5% 2.5% 1% 0.5% 0.1% 0.05% 1 6.314 12.706 31.821 63.657 318.31 636.62 2 2.920 4.303 6.965 9.925 22.327 31.598 3 2.353 3.182 4.541 5.841 10.214 12.924 4 2.132 2.776 3.747 4.604 7.173 8.610 5 2.015 2.571 3.365 4.032 5.893 6.869 & & & & & & & & & & & & & & 18 1.734 2.101 2.552 2.878 3.610 3.922 19 1.729 2.093 2.539 2.861 3.579 3.883 20 1.725 2.086 2.528 2.845 3.552 3.850 & & & & & & & & & & & & & & 120 1.658 1.980 2.358 2.617 3.160 3.373 1.645 1.960 2.326 2.576 3.090 3.291&" "(h  (h  0c `P :23 2 3f  6g?  ;t TEST OF A HYPOTHESIS RELATING TO A REGRESSION COEFFICIENT4<(2: < `  c $A &?? 5  &   s *}|>P CIf we were performing a regression with 20 observations, as in the price inflation/wage inflation example, the number of degrees of freedom would be 18 and the critical value of t for a 5% test would be 2.101.B 2  B  s *޽h ? ̙33  Q I   (  L  c $ 9J L  c $ P L  c $@   0Ȍ|   t Distribution: Critical values of t Degrees of Two-tailed test 10% 5% 2% 1% 0.2% 0.1% freedom One-tailed test 5% 2.5% 1% 0.5% 0.1% 0.05% 1 6.314 12.706 31.821 63.657 318.31 636.62 2 2.920 4.303 6.965 9.925 22.327 31.598 3 2.353 3.182 4.541 5.841 10.214 12.924 4 2.132 2.776 3.747 4.604 7.173 8.610 5 2.015 2.571 3.365 4.032 5.893 6.869 & & & & & & & & & & & & & & 18 1.734 2.101 2.552 2.878 3.610 3.922 19 1.729 2.093 2.539 2.861 3.579 3.883 20 1.725 2.086 2.528 2.845 3.552 3.850 & & & & & & & & & & & & & & 120 1.658 1.980 2.358 2.617 3.160 3.373 1.645 1.960 2.326 2.576 3.090 3.291&" "(h  (h  0t `P :24 2 3f  6?  ;t TEST OF A HYPOTHESIS RELATING TO A REGRESSION COEFFICIENT4<(2: < `  c $A '?? 5  '   s *Թ|>P NNote that as the number of degrees of freedom becomes large, the critical value converges on 1.96, the critical value for the normal distribution. This is because the t distribution converges on the normal distribution.B 23  B  s *޽h ? ̙33 ((  R  s *%  R  s *   0|  V & T  6?  ;t TEST OF A HYPOTHESIS RELATING TO A REGRESSION COEFFICIENT4<(2: < XB  0DP@ @   0h  ~s.d. of b2 known> 2     H>  Mdiscrepancy between hypothetical value and sample estimate, in terms of s.d.:N 2N N `   c $A (??0p (   H8    C5% significance test: reject H0: b2 = b2 if z > 1.96 or z < -1.96D    D R   s *    0L   s.d. of b2 not known> 2       H>  Mdiscrepancy between hypothetical value and sample estimate, in terms of s.e.:N 2N N `  c $A )??0 )  HT!   E5% significance test: reject H0: b2 = b2 if t > tcrit or t < -tcritF @0         f4 ? Z  E0((2    f? ? f  E0((2   s *T"|>P ;Hence, referring back to the summary of the test procedure,&< 2; <   00' `P :25 2 3fH  0޽h ? ̙33 e](  R  s *% p R  s *   0b|  V & T  6@n?  ;t TEST OF A HYPOTHESIS RELATING TO A REGRESSION COEFFICIENT4<(2: < XB  0DP@ @   0th  ~s.d. of b2 known> 2     H8>  Mdiscrepancy between hypothetical value and sample estimate, in terms of s.d.:N 2N N `   c $A *??0p *   H    C5% significance test: reject H0: b2 = b2 if z > 1.96 or z < -1.96D    D R   s *    0T  s.d. of b2 not known> 2       H>  Mdiscrepancy between hypothetical value and sample estimate, in terms of s.e.:N 2N N `  c $A +??0 +  Hй   E5% significance test: reject H0: b2 = b2 if t > 2.101 or t < -2.101F   F    f  ? Z  E0((2    f ? f  E0((2 "  s *|>P Vwe should reject the null hypothesis if the absolute value of t is greater than 2.101.BW 2> W   0T `P :26 2 3fH  0޽h ? ̙33Q      (   L   c $)~ L   c $@)   0,|   t Distribution: Critical values of t Degrees of Two-tailed test 10% 5% 2% 1% 0.2% 0 .1% freedom One-tailed test 5% 2.5% 1% 0.5% 0.1% 0.05% 1 6.314 12.706 31.821 63.657 318.31 636.62 2 2.920 4.303 6.965 9.925 22.327 31.598 3 2.353 3.182 4.541 5.841 10.214 12.924 4 2.132 2.776 3.747 4.604 7.173 8.610 5 2.015 2.571 3.365 4.032 5.893 6.869 & & & & & & & & & & & & & & 18 1.734 2.101 2.552 2.878 3.610 3.922 19 1.729 2.093 2.539 2.861 3.579 3.883 20 1.725 2.086 2.528 2.845 3.552 3.850 & & & & & & & & & & & & & & 120 1.658 1.980 2.358 2.617 3.160 3.373 1.645 1.960 2.326 2.576 3.090 3.291&" "(h  (h   0\ `P :27 2 3f   6L!?  ;t TEST OF A HYPOTHESIS RELATING TO A REGRESSION COEFFICIENT4<(2: <    s *|>P JIf instead we wished to perform a 1% significance test, we would use the column indicated above. Note that as the number of degrees of freedom becomes large, the critical value converges to 2.58, the critical value for the normal distribution.& 2  `   c $A ,?? 5  ,H   0޽h ? ̙33,     l (  L  c $9> L  c $)9 L  c $@)  04|   t Distribution: Critical values of t Degrees of Two-tailed test 10% 5% 2% 1% 0.2% 0 .1% freedom One-tailed test 5% 2.5% 1% 0.5% 0.1% 0.05% 1 6.314 12.706 31.821 63.657 318.31 636.62 2 2.920 4.303 6.965 9.925 22.327 31.598 3 2.353 3.182 4.541 5.841 10.214 12.924 4 2.132 2.776 3.747 4.604 7.173 8.610 5 2.015 2.571 3.365 4.032 5.893 6.869 & & & & & & & & & & & & & & 18 1.734 2.101 2.552 2.878 3.610 3.922 19 1.729 2.093 2.539 2.861 3.579 3.883 20 1.725 2.086 2.528 2.845 3.552 3.850 & & & & & & & & & & & & & & 120 1.658 1.980 2.358 2.617 3.160 3.373 1.645 1.960 2.326 2.576 3.090 3.291&" "(h  (h  0P> `P :28 2 3f  6E?  ;t TEST OF A HYPOTHESIS RELATING TO A REGRESSION COEFFICIENT4<(2: < +  s *D|>P _For a simple regression with 20 observations, the critical value of t at the 1% level is 2.878.B` 2D ` `   c $A -?? 5  -H  0޽h ? ̙33 (  R  s * J R  s *% p R  s *   06|  V & T  6w?  ;t TEST OF A HYPOTHESIS RELATING TO A REGRESSION COEFFICIENT4<(2: < XB  0DP@ @   0q  ~s.d. of b2 known> 2      HP>  Mdiscrepancy between hypothetical value and sample estimate, in terms of s.d.:N 2N N `   c $A .??0p .   H    C5% significance test: reject H0: b2 = b2 if z > 1.96 or z < -1.96D    D R   s *    0Ԩ  s.d. of b2 not known> 2      H>  Mdiscrepancy between hypothetical value and sample estimate, in terms of s.e.:N 2N N `  c $A /??0 /  H   E1% significance test: reject H0: b2 = b2 if t > 2.878 or t < -2.878F   F    f` ? Z  E0((2    f ? f  E0((2   s *8|>P =So we should this figure in the test procedure for a 1% test.&> 2= >   0 `P :29 2 3fH  0޽h ? ̙338 ~(  q  0 <| A 3.10 A researcher with a sample of 50 individuals with similar education but differing amounts of training hypothesizes that hourly earnings, EARNINGS, may be related to hours of training, TRAINING, according to the relationship EARNINGS = b1 + b2 TRAINING + u He is prepared to test the null hypothesis H0: b2 = 0 against the alternative hypothesis H1: b2 0 at the 5 percent and 1 percent levels. What should he report 1. If b2 = 0.30, s.e.(b2) = 0.12? 2. If b2 = 0.55, s.e.(b2) = 0.12? 3. If b2 = 0.10, s.e.(b2) = 0.12? 4. If b2 = -0.27, s.e.(b2) = 0.12?B'    ,(E     6o?  ^EXERCISE& (2    0@p ^N 91 2 3f`  c $A 0??u  0B  s *޽h ? ̙33Q (  o  0x| q EARNINGS = b1 + b2 TRAINING + u H0: b2 = 0, H1: b2 0 n = 50, so 48 degrees of freedom br    *   6ب?  ^EXERCISE& (2  `  c $A 1??|P  1  s *|>P cThere are 50 observations and 2 parameters have been estimated, so there are 48 degrees of freedom.&d 2c d   0xy ^N 92 2 3fB  s *޽h ? ̙33 SK (      0D |  EARNINGS = b1 + b2 TRAINING + u H0: b2 = 0, H1: b2 0 n = 50, so 48 degrees of freedom tcrit, 5% = 2.01, tcrit, 1% = 2.68     %   ,m     6D ?  ^EXERCISE& (2  `   c $A 2??|P  2   s *|D |>P uThe table giving the critical values of t does not give the values for 48 degrees of freedom. We will use the values for 50 as a guide. For the 5% level the value is 2.01, and for the 1% level it is 2.68. The critical values for 48 will be slightly higher.B 2(     0:D  ^N 93 2 3fB   s *޽h ? ̙33 2* $(  $ $ 0? | \ EARNINGS = b1 + b2 TRAINING + u H0: b2 = 0, H1: b2 0 n = 50, so 48 degrees of freedom tcrit, 5% = 2.01, tcrit, 1% = 2.68 _______________________________________________ 1. If b2 = 0.30, s.e.(b2) = 0.12? t = 2.50.     %  @ ,m s  $ 6(D ?  ^EXERCISE& (2  ` $ c $A 3??|P  3 $ s *HD |>P +In the first case, the t statistic is 2.50.B, 2 ,  $ 0xD  ^N 94 2 3fB $ s *޽h ? ̙33+ 0(q(  (! ( 0L t | ; EARNINGS = b1 + b2 TRAINING + u H0: b2 = 0, H1: b2 0 n = 50, so 48 degrees of freedom tcrit, 5% = 2.01, tcrit, 1% = 2.68 _______________________________________________ 1. If b2 = 0.30, s.e.(b2) = 0.12? t = 2.50. Reject H0 at the 5% level but not at the 1% level.<    %  @=,m   ( 6b ?  ^EXERCISE& (2   ( 0,t  ^N 95 2 3f; ( s *t |>P oThis is greater than the critical value of t at the 5% level, but less than the critical value at the 1% level.Bp 2+C p ` ( c $A 4??|P  4B ( s *޽h ? ̙33> @,(  ,  , 0o| $ EARNINGS = b1 + b2 TRAINING + u H0: b2 = 0, H1: b2 0 n = 50, so 48 degrees of freedom tcrit, 5% = 2.01, tcrit, 1% = 2.68 _______________________________________________ 1. If b2 = 0.30, s.e.(b2) = 0.12? t = 2.50. Reject H0 at the 5%, but not at the 1%, level.%    %  @&,m   , 6H?  ^EXERCISE& (2  ` , c $A 5??|P  5e , s *l|>P In this case we should mention both tests. It is not enough to say "Reject at the 5% level", because it leaves open the possibility that we might be able to reject at the 1% level.& 2   , 0x ^N 96 2 3fB , s *޽h ? ̙33! P0g(  0  0 0D | $ EARNINGS = b1 + b2 TRAINING + u H0: b2 = 0, H1: b2 0 n = 50, so 48 degrees of freedom tcrit, 5% = 2.01, tcrit, 1% = 2.68 _______________________________________________ 1. If b2 = 0.30, s.e.(b2) = 0.12? t = 2.50. Reject H0 at the 5%, but not at the 1%, level.%    %  @&,m   0 6x?  ^EXERCISE& (2  ` 0 c $A 6??|P  6H 0 s *|>P Likewise it is not enough to say "Do not reject at the 1% level", because this does not reveal whether the result is significant at the 5% level or not.& 2   0 0 ^N 97 2 3fB 0 s *޽h ? ̙33| ,$`4(  4 4 0| Z EARNINGS = b1 + b2 TRAINING + u H0: b2 = 0, H1: b2 0 n = 50, so 48 degrees of freedom tcrit, 5% = 2.01, tcrit, 1% = 2.68 _______________________________________________ 2. If b2 = 0.55, s.e.(b2) = 0.12? t = 4.58.    %  @ ,m q  4 6.?  ^EXERCISE& (2  ` 4 c $A 7??|P  7 4 s *'|>P 'In the second case, t is equal to 4.58.B( 2 (  4 0?  ^N 98 2 3fB 4 s *޽h ? ̙33_ p8(  8 8 0|  EARNINGS = b1 + b2 TRAINING + u H0: b2 = 0, H1: b2 0 n = 50, so 48 degrees of freedom tcrit, 5% = 2.01, tcrit, 1% = 2.68 _______________________________________________ 2. If b2 = 0.55, s.e.(b2) = 0.12? t = 4.58. Reject H0 at the 1% level.    %  @,m   8 6g?  ^EXERCISE& (2  ` 8 c $A 8??|P  8 8 s *8s|>P @We report only the result of the 1% test. There is no need to mention the 5% test. If you do, you reveal that you do not understand that rejection at the 1% level automatically means rejection at the 5% level, and you look ignorant.& 2   8 0 ^N 99 2 3fB 8 s *޽h ? ̙33#  <i(  <? < 0G| ' EARNINGS = b1 + b2 TRAINING + u H0: b2 = 0, H1: b2 0 n = 50, so 48 degrees of freedom tcrit, 5% = 2.01, tcrit, 1% = 2.68 _______________________________________________ 2. If b2 = 0.55, s.e.(b2) = 0.12? t = 4.58. Reject H0 at the 0.1% level (tcrit, 0.1% = 3.50).6(    %  @  >m   < 6|?  ^EXERCISE& (2  ` < c $A 9??|P  9 < s *|>P Actually, given the large t statistic, it is a good idea to investigate whether we can reject H0 at the 0.1% level. It turns out that we can. The critical value for 50 degrees of freedom is 3.50. So we just report the outcome of this test. There is no need to mention the 1% test.n 2C    < 0| ^N :10 2 3fB < s *޽h ? ̙33j @(  @? @ 0| ' EARNINGS = b1 + b2 TRAINING + u H0: b2 = 0, H1: b2 0 n = 50, so 48 degrees of freedom tcrit, 5% = 2.01, tcrit, 1% = 2.68 _______________________________________________ 2. If b2 = 0.55, s.e.(b2) = 0.12? t = 4.58. Reject H0 at the 0.1% level (tcrit, 0.1% = 3.50).6(    %  @  >m   @ 65?  ^EXERCISE& (2  ` @ c $A :??|P  :[ @ s *<4|>P Why is it a good idea to press on to a 0.1% test, if the t statistic is large? Try to answer this question before looking at the next slide. B 29U   @ 0/ ^N :11 2 3fB @ s *޽h ? ̙33 xD(  D? D 0 I| ' EARNINGS = b1 + b2 TRAINING + u H0: b2 = 0, H1: b2 0 n = 50, so 48 degrees of freedom tcrit, 5% = 2.01, tcrit, 1% = 2.68 _______________________________________________ 2. If b2 = 0.55, s.e.(b2) = 0.12? t = 4.58. Reject H0 at the 0.1% level (tcrit, 0.1% = 3.50).6(    %  @  >m   D 6l?  c EXERCISE 3.10&(2   ` D c $A ;??|P  ; D s *tx|>P b The reason is that rejection at the 1% level still leaves open the possibility of a 1% risk of having made a Type I error (rejecting the null hypothesis when it is in fact true). So there is a 1% risk of the "significant" result having occurred as a matter of chance.&  2    D 0 ^N :12 2 3fB D s *޽h ? ̙33E H(  H? H 08| ' EARNINGS = b1 + b2 TRAINING + u H0: b2 = 0, H1: b2 0 n = 50, so 48 degrees of freedom tcrit, 5% = 2.01, tcrit, 1% = 2.68 _______________________________________________ 2. If b2 = 0.55, s.e.(b2) = 0.12? t = 4.58. Reject H0 at the 0.1% level (tcrit, 0.1% = 3.50).6(    %  @  >m   H 6?  ^EXERCISE& (2  ` H c $A <??|P  < H 0 ^N :13 2 3f6 H s * |>P If you can reject at the 0.1% level, you reduce that risk to one tenth of 1%. This means that the result is almost certainly genuine.& 2  B H s *޽h ? ̙33} -%L(  L L 0z| Z EARNINGS = b1 + b2 TRAINING + u H0: b2 = 0, H1: b2 0 n = 50, so 48 degrees of freedom tcrit, 5% = 2.01, tcrit, 1% = 2.68 _______________________________________________ 3. If b2 = 0.10, s.e.(b2) = 0.12? t = 0.83.    %  @ ,m q  L 6T+?  ^EXERCISE& (2  ` L c $A =??|P  = L 0 ^N :14 2 3f L s *(|>P 'In the third case, t is equal to 0.83. B( 2 ( B L s *޽h ? ̙33p  P(  P P 0L|  EARNINGS = b1 + b2 TRAINING + u H0: b2 = 0, H1: b2 0 n = 50, so 48 degrees of freedom tcrit, 5% = 2.01, tcrit, 1% = 2.68 _______________________________________________ 3. If b2 = 0.10, s.e.(b2) = 0.12? t = 0.83. Do not reject H0 at the 5% level.    %  @,m   P 6h?  ^EXERCISE& (2  ` P c $A >??|P  > P 0,t ^N :15 2 3f P s *~|>P IWe report only the result of the 5% test. There is no need to mention the 1% test. If you do, you reveal that you do not understand that not rejecting at the 5% level automatically means not rejecting at the 1% level, and you look ignorant. & 2  B P s *޽h ? ̙33 0(T(  T T 0В| \ EARNINGS = b1 + b2 TRAINING + u H0: b2 = 0, H1: b2 0 n = 50, so 48 degrees of freedom tcrit, 5% = 2.01, tcrit, 1% = 2.68 _______________________________________________ 4. If b2 = -0.27, s.e.(b2) = 0.12? t = -2.25.    %  @ ,m s  T 6?  ^EXERCISE& (2  ` T c $A ???|P  ? T s *|>P (In the fourth case, t is equal to -2.25.B) 2 )  T 0 ^N :16 2 3fB T s *޽h ? ̙33\  X(  X$ X 0| > EARNINGS = b1 + b2 TRAINING + u H0: b2 = 0, H1: b2 0 n = 50, so 48 degrees of freedom tcrit, 5% = 2.01, tcrit, 1% = 2.68 _______________________________________________ 4. If b2 = -0.27, s.e.(b2) = 0.12? t = -2.25. Reject H0 at the 5% level but not at the 1% level.?    %  @>,m   X 6t!?  lEXERCISE& (2  ` X c $A @??|P  @Z X s *|>P The absolute value of the t statistic is between the critical values for the 5% and 1% tests. So we mention both tests, as in the first case.B 2s   X 0 ^N :17 2 3fB X s *޽h ? ̙33t $\(  \ \ 04| U "f \ 63?  ~F TESTS OF GOODNESS OF FIT4(2   \ 0C ^N 91 2 3fh \ s *h|>P This sequence describes two F tests of goodness of fit in a multiple regression model. The first relates to the goodness of fit of the equation as a whole.B 2  ` \ c $A A??@0 A` \ c $A B??P` BB \ s *޽h ? ̙33 {`(  ` ` 0| U "f ` 6 ?  ~F TESTS OF GOODNESS OF FIT4(2   ` 0 |>P mWe will consider the general case where there are k - 1 explanatory variables. For the F test of goodness of fit of the equation as a whole, the null hypothesis, in words, is that the model has no explanatory power at all.^ 22$  ` ` c $A C??@0 C` ` c $A D??P` DB ` s *޽h ? ̙33  d^(  d d 0 | U "f d 6R ?  ~F TESTS OF GOODNESS OF FIT4(2   d 08E  ^N 93 2 3f  d s *l |>P \Of course we hope to reject it and conclude that the model does have some explanatory power.&] 2\ ] ` d c $A E??@0 E` d c $A F??P` FB d s *޽h ? ̙33 0hX(  h h 0P The model will have no explanatory power if it turns out that Y is unrelated to any of the explanatory variables. Mathematically, therefore, the null hypothesis is that all the coefficients b2, ..., bk are zero. 2>   &    ` h c $A G??@0 G` h c $A H??P` HB h s *޽h ? ̙337 @l}(  l l 0 3 | U "f l 6 ?  ~F TESTS OF GOODNESS OF FIT4(2  ` l c $A I??@0 I` l c $A J??P` J l 0  ^N 95 2 3f+ l s *ˬ |%7 _The alternative hypothesis is that at least one of these b coefficients is different from zero.B` 29% ` B l s *޽h ? ̙33 PpM(  p p 0 | U "f p 6 ?  ~F TESTS OF GOODNESS OF FIT4(2  ` p c $A K??@0 K` p c $A L??P` L p 0%  ^N 96 2 3f p s * |<N In the multiple regression model there is a difference between the roles of the F and t tests. The F test tests the joint explanatory power of the variables, while the t tests test their explanatory power individually. 2P D1  B p s *޽h ? ̙33 D<`t(  t t 0 | U "f t 64 ?  ~F TESTS OF GOODNESS OF FIT4(2  ` t c $A M??@0 M` t c $A N??P` N t 0,  ^N 97 2 3f t s *b |<N .In the simple regression model the F test was equivalent to the (two-tailed) t test on the slope coefficient because the "group" consisted of just one variable.^ 2#)R  B t s *޽h ? ̙33| ,$p x(  x x 0\ | U "f x 6 ?  ~F TESTS OF GOODNESS OF FIT4(2  x x <A O??H  O` x c $A P??@0 P` x c $A Q??P` QL x c $   x 0 b  ^N 98 2 3f  x s *b |>P BThe F statistic for the test was defined in the last sequence in Chapter 3. ESS is the explained sum of squares and RSS is the residual sum of squares.z 2H%   B x s *޽h ? ̙33F  |(  | | 0b | U "f | 6|b ?  ~F TESTS OF GOODNESS OF FIT4(2  x | <A R??H  R` | c $A S??@0 S` | c $A T??P` T | 08b  ^N 99 2 3ff | s *b |>P nIt can be expressed in terms of R2 by dividing the numerator and denominator by TSS, the total sum of squares.no 2  . o L  | c $  B | s *޽h ? ̙33f (    0$b | U "f  6b ?  ~F TESTS OF GOODNESS OF FIT4(2    0Mb  ^N :10 2 3fx  <A U??H  U`  c $A V??@0 V`  c $A W??P` W  s *LDt |>P oESS / TSS is equal to R2 and RSS / TSS is equal to (1 - R2). (For proofs, see the last sequence in Chapter 3.)p 2   5 p B  s *޽h ? ̙33H (    6hjb ?  ~F TESTS OF GOODNESS OF FIT4(2    0t  ^N :11 2 3f`  c $A X??P@ X  s *H |>P The educational attainment model will be used as an example. We will suppose that S depends on ASVABC, the ability score, and SM, and SF, the highest grade completed by the mother and father of the respondent, respectively. 2S W  R  s * ` H  0޽h ? ̙33/ o(    6 ?  ~F TESTS OF GOODNESS OF FIT4(2    0  ^N :12 2 3f`  c $A Y??px` Y  s *8 |>P %The null hypothesis for the F test of goodness of fit is that all three slope coefficients are equal to zero. The alternative hypothesis is that at least one of them is non-zero.B 2  R  s * ` `  c $A Z??P@ ZH  0޽h ? ̙33     ! (  R  s *v]  0pV| m . reg S ASVABC SM SF Source | SS df MS Number of obs = 570 ---------+------------------------------ F( 3, 566) = 110.83 Model | 1278.24153 3 426.080508 Prob > F = 0.0000 Residual | 2176.00584 566 3.84453329 R-squared = 0.3700 ---------+------------------------------ Adj R-squared = 0.3667 Total | 3454.24737 569 6.07073351 Root MSE = 1.9607 ------------------------------------------------------------------------------ S | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- ASVABC | .1295006 .0099544 13.009 0.000 .1099486 .1490527 SM | .069403 .0422974 1.641 0.101 -.013676 .152482 SF | .1102684 .0311948 3.535 0.000 .0489967 .1715401 _cons | 4.914654 .5063527 9.706 0.000 3.920094 5.909214 ------------------------------------------------------------------------------ 0nl  +  %         f  6n?  ~F TESTS OF GOODNESS OF FIT4(2    0g ^N :13 2 3f  s *|>P 0Here is the regression output using Data Set 21.&1 20 1 R  s * ` `  c $A [??P@ [`   c $A \??px` \H  0޽h ? ̙33  a Y   (  R  s *v]  0| m . reg S ASVABC SM SF Source | SS df MS Number of obs = 570 ---------+------------------------------ F( 3, 566) = 110.83 Model | 1278.24153 3 426.080508 Prob > F = 0.0000 Residual | 2176.00584 566 3.84453329 R-squared = 0.3700 ---------+------------------------------ Adj R-squared = 0.3667 Total | 3454.24737 569 6.07073351 Root MSE = 1.9607 ------------------------------------------------------------------------------ S | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- ASVABC | .1295006 .0099544 13.009 0.000 .1099486 .1490527 SM | .069403 .0422974 1.641 0.101 -.013676 .152482 SF | .1102684 .0311948 3.535 0.000 .0489967 .1715401 _cons | 4.914654 .5063527 9.706 0.000 3.920094 5.909214 ------------------------------------------------------------------------------ 0nl  +  %         f  6?  ~F TESTS OF GOODNESS OF FIT4(2  R  s * ` `  c $A ]??P@ ]`  c $A ^??px` ^  0Գ ^N :14 2 3f`   c $A _??L  < _   s *|>P 2In this example, k - 1, the number of explanatory variables, is equal to 3 and n - k, the number of degrees of freedom, is equal to 566.z 2=4  `   c $A `??M 0 `L   c $` `@H  0޽h ? ̙33l     (  R  s * R  s *E ` R  s *v]  0$Զ| m . reg S ASVABC SM SF Source | SS df MS Number of obs = 570 ---------+------------------------------ F( 3, 566) = 110.83 Model | 1278.24153 3 426.080508 Prob > F = 0.0000 Residual | 2176.00584 566 3.84453329 R-squared = 0.3700 ---------+------------------------------ Adj R-squared = 0.3667 Total | 3454.24737 569 6.07073351 Root MSE = 1.9607 ------------------------------------------------------------------------------ S | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- ASVABC | .1295006 .0099544 13.009 0.000 .1099486 .1490527 SM | .069403 .0422974 1.641 0.101 -.013676 .152482 SF | .1102684 .0311948 3.535 0.000 .0489967 .1715401 _cons | 4.914654 .5063527 9.706 0.000 3.920094 5.909214 ------------------------------------------------------------------------------ 0nl  +  %         f  6`Ͷ?  ~F TESTS OF GOODNESS OF FIT4(2  R  s * ` `  c $A a??P@ a`   c $A b??px` b`   c $A c??L  < c`   c $A d??M 0 d   0 ^N :15 2 3f   s *|>P 9The numerator of the F statistic is the explained sum of squares divided by k - 1. In the Stata output these numbers are given in the Model row.^ 26D&[  2 L  c $` `@H  0޽h ? ̙33    . (  R  s *  R  s *P +R  s *v]  0| m . reg S ASVABC SM SF Source | SS df MS Number of obs = 570 ---------+------------------------------ F( 3, 566) = 110.83 Model | 1278.24153 3 426.080508 Prob > F = 0.0000 Residual | 2176.00584 566 3.84453329 R-squared = 0.3700 ---------+------------------------------ Adj R-squared = 0.3667 Total | 3454.24737 569 6.07073351 Root MSE = 1.9607 ------------------------------------------------------------------------------ S | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- ASVABC | .1295006 .0099544 13.009 0.000 .1099486 .1490527 SM | .069403 .0422974 1.641 0.101 -.013676 .152482 SF | .1102684 .0311948 3.535 0.000 .0489967 .1715401 _cons | 4.914654 .5063527 9.706 0.000 3.920094 5.909214 ------------------------------------------------------------------------------ 0nl  +  %         f  6?  ~F TESTS OF GOODNESS OF FIT4(2  R  s * ` `  c $A e??P@ e`   c $A f??px` f`   c $A g??L  < g`   c $A h??M 0 h   0  ^N :16 2 3f   s *> |>P eThe denominator is the residual sum of squares divided by the number of degrees of freedom remaining.&f 2e f L  c $` @H  0޽h ? ̙33  / '   (  R  s * ]F R  s *0R  s *v]  0|> | m . reg S ASVABC SM SF Source | SS df MS Number of obs = 570 ---------+------------------------------ F( 3, 566) = 110.83 Model | 1278.24153 3 426.080508 Prob > F = 0.0000 Residual | 2176.00584 566 3.84453329 R-squared = 0.3700 ---------+------------------------------ Adj R-squared = 0.3667 Total | 3454.24737 569 6.07073351 Root MSE = 1.9607 ------------------------------------------------------------------------------ S | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- ASVABC | .1295006 .0099544 13.009 0.000 .1099486 .1490527 SM | .069403 .0422974 1.641 0.101 -.013676 .152482 SF | .1102684 .0311948 3.535 0.000 .0489967 .1715401 _cons | 4.914654 .5063527 9.706 0.000 3.920094 5.909214 ------------------------------------------------------------------------------ 0nl  +  %         f  6D> ?  ~F TESTS OF GOODNESS OF FIT4(2  `  c $A i??P@ i`  c $A j??px` j`   c $A k??L  < k`   c $A l??M 0 l   0 ^N :17 2 3fT   s *|>P Hence the F statistic is 110.8. All serious regression packages compute it for you as part of the diagnostics in the regression output.B 2 }  H  0޽h ? ̙33$     d (  R  s * ~ R  s * ]F R  s *v]  0"| m . reg S ASVABC SM SF Source | SS df MS Number of obs = 570 ---------+------------------------------ F( 3, 566) = 110.83 Model | 1278.24153 3 426.080508 Prob > F = 0.0000 Residual | 2176.00584 566 3.84453329 R-squared = 0.3700 ---------+------------------------------ Adj R-squared = 0.3667 Total | 3454.24737 569 6.07073351 Root MSE = 1.9607 ------------------------------------------------------------------------------ S | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- ASVABC | .1295006 .0099544 13.009 0.000 .1099486 .1490527 SM | .069403 .0422974 1.641 0.101 -.013676 .152482 SF | .1102684 .0311948 3.535 0.000 .0489967 .1715401 _cons | 4.914654 .5063527 9.706 0.000 3.920094 5.909214 ------------------------------------------------------------------------------ 0nl  +  %         f  62?  ~F TESTS OF GOODNESS OF FIT4(2  `  c $A m??P@ m`  c $A n??px` n`   c $A o??M 0 o   0F ^N :18 2 3f   s *LQ|>P The critical value for F(3,566) is not given in the F tables, but we know it must be lower than F(3,120), which is given. At the 0.1% level, this is 5.78. Hence we easily reject H0 at the 0.1% level. 2+S   `   c $A p?? p  pH  0޽h ? ̙33)     i (  R  s * ~ R  s * ]F R  s *v]  0G| m . reg S ASVABC SM SF Source | SS df MS Number of obs = 570 ---------+------------------------------ F( 3, 566) = 110.83 Model | 1278.24153 3 426.080508 Prob > F = 0.0000 Residual | 2176.00584 566 3.84453329 R-squared = 0.3700 ---------+------------------------------ Adj R-squared = 0.3667 Total | 3454.24737 569 6.07073351 Root MSE = 1.9607 ------------------------------------------------------------------------------ S | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- ASVABC | .1295006 .0099544 13.009 0.000 .1099486 .1490527 SM | .069403 .0422974 1.641 0.101 -.013676 .152482 SF | .1102684 .0311948 3.535 0.000 .0489967 .1715401 _cons | 4.914654 .5063527 9.706 0.000 3.920094 5.909214 ------------------------------------------------------------------------------ 0nl  +  %         f  64?  ~F TESTS OF GOODNESS OF FIT4(2  `  c $A q??P@ q`  c $A r??px` r`   c $A s??M 0 s`   c $A t?? p  t   0p ^N :19 2 3f   s *t|>P This result could have been anticipated because both ASVABC and SF have highly significant t statistics. So we knew in advance that both b2 and b4 were non-zero. 25.    H  0޽h ? ̙33    0 X (  R  s * ~ R  s * ]F R  s *v]  0\| m . reg S ASVABC SM SF Source | SS df MS Number of obs = 570 ---------+------------------------------ F( 3, 566) = 110.83 Model | 1278.24153 3 426.080508 Prob > F = 0.0000 Residual | 2176.00584 566 3.84453329 R-squared = 0.3700 ---------+------------------------------ Adj R-squared = 0.3667 Total | 3454.24737 569 6.07073351 Root MSE = 1.9607 ------------------------------------------------------------------------------ S | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- ASVABC | .1295006 .0099544 13.009 0.000 .1099486 .1490527 SM | .069403 .0422974 1.641 0.101 -.013676 .152482 SF | .1102684 .0311948 3.535 0.000 .0489967 .1715401 _cons | 4.914654 .5063527 9.706 0.000 3.920094 5.909214 ------------------------------------------------------------------------------ 0nl  +  %         f  6?  ~F TESTS OF GOODNESS OF FIT4(2  `  c $A u??P@ u`  c $A v??px` v`   c $A w??M 0 w`   c $A x?? p  x   0 ^N :20 2 3f   s *|>P It is unusual for the F statistic not to be significant if some of the t statistics are significant. In principle it could happen though. Suppose that you ran a regression with 40 explanatory variables, none being a true determinant of the dependent variable.^ 20  H  0޽h ? ̙33Q    @  (  R  s * ~ R  s * ]F R  s *v]  0| m . reg S ASVABC SM SF Source | SS df MS Number of obs = 570 ---------+------------------------------ F( 3, 566) = 110.83 Model | 1278.24153 3 426.080508 Prob > F = 0.0000 Residual | 2176.00584 566 3.84453329 R-squared = 0.3700 ---------+------------------------------ Adj R-squared = 0.3667 Total | 3454.24737 569 6.07073351 Root MSE = 1.9607 ------------------------------------------------------------------------------ S | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- ASVABC | .1295006 .0099544 13.009 0.000 .1099486 .1490527 SM | .069403 .0422974 1.641 0.101 -.013676 .152482 SF | .1102684 .0311948 3.535 0.000 .0489967 .1715401 _cons | 4.914654 .5063527 9.706 0.000 3.920094 5.909214 ------------------------------------------------------------------------------ 0nl  +  %         f  6?  ~F TESTS OF GOODNESS OF FIT4(2  `  c $A y??P@ y`  c $A z??px` z`   c $A {??M 0 {`   c $A |?? p  |   0 ^N :21 2 3f&   s *"|>P Then the F statistic should be low enough for H0 not to be rejected. However, if you are performing t tests on the slope coefficients at the 5% level, with a 5% chance of a Type I error, on average 2 of the 40 variables could be expected to have "significant" coefficients. 2 $ 5  H  0޽h ? ̙33    P % (  R  s * ~ R  s * ]F R  s *v]  02| m . reg S ASVABC SM SF Source | SS df MS Number of obs = 570 ---------+------------------------------ F( 3, 566) = 110.83 Model | 1278.24153 3 426.080508 Prob > F = 0.0000 Residual | 2176.00584 566 3.84453329 R-squared = 0.3700 ---------+------------------------------ Adj R-squared = 0.3667 Total | 3454.24737 569 6.07073351 Root MSE = 1.9607 ------------------------------------------------------------------------------ S | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- ASVABC | .1295006 .0099544 13.009 0.000 .1099486 .1490527 SM | .069403 .0422974 1.641 0.101 -.013676 .152482 SF | .1102684 .0311948 3.535 0.000 .0489967 .1715401 _cons | 4.914654 .5063527 9.706 0.000 3.920094 5.909214 ------------------------------------------------------------------------------ 0nl  +  %         f  6'?  ~F TESTS OF GOODNESS OF FIT4(2  `  c $A }??P@ }`  c $A ~??px` ~`   c $A ??M 0 `   c $A ?? p     0\ ^N :22 2 3f   s *g|>P `The opposite can easily happen, though. Suppose you have a multiple regression model which is correctly specified and the R2 is high. You would expect to have a highly significant F statistic.n 2{ 9   H  0޽h ? ̙33    ` 4 (  R  s * ~ R  s * ]F R  s *v]  0z| m . reg S ASVABC SM SF Source | SS df MS Number of obs = 570 ---------+------------------------------ F( 3, 566) = 110.83 Model | 1278.24153 3 426.080508 Prob > F = 0.0000 Residual | 2176.00584 566 3.84453329 R-squared = 0.3700 ---------+------------------------------ Adj R-squared = 0.3667 Total | 3454.24737 569 6.07073351 Root MSE = 1.9607 ------------------------------------------------------------------------------ S | Coef. Std. Err. t P>|t| [95% Conf. Interval] ---------+-------------------------------------------------------------------- ASVABC | .1295006 .0099544 13.009 0.000 .1099486 .1490527 SM | .069403 .0422974 1.641 0.101 -.013676 .152482 SF | .1102684 .0311948 3.535 0.000 .0489967 .1715401 _cons | 4.914654 .5063527 9.706 0.000 3.920094 5.909214 ------------------------------------------------------------------------------ 0nl  +  %         f  6\^?  ~F TESTS OF GOODNESS OF FIT4(2  `  c $A ??P@ `  c $A ??px` `   c $A ??M 0 `   c $A ?? p     0 ^N :23 2 3f   s *|>P oHowever, if the explanatory variables are highly correlated and the model is subject to severe multicollinearity, the standard errors of the slope coefficients could all be so large that none of the t statistics is significant.B 2&_  t H  0޽h ? ̙33أ,xtͦ"B4)I IhAC HAĊذwP`GEgvfEy{s;'=lXkG|ftYah+"؁eyeB T?PNm z_6R%VS%ZJmU)J}Pi4VR&JTIS)JsRiVQ(mvJ{%CtT:)JgrUtWJOg߶^JV(JWW(AJ2X U)Õe2R9AVNTNRNV(cSS+$HLQ4e2CRJT)Sʕʩi< L,leRᴽDʵ-4rMˈhK|q}ky9 afx[&g8KϺcχ뒬hgj]v_[7@s;'Ku'8a?]~sN]nf>_v;^qgkOqcߜN,-)+\wĢJJ;{Y禎'wu'gh3]WkWGjp1u?R(o+ٖ{ܚ\yK,wro,wem0sQ\2s|Yjn)E<Mᴡ"EVu{kotMųeÖ5]yR[ڹyJP9_Y\,V.T.R.V.Q.U.˕++kkfV6eYdN.nWO_y@yPYԐOhh.?ړw7c>p*t6%uD;Ź[s|h2F ;`3G0tNUW8CЋth7Ywݑ֬MfYcxkMvIcGH~2Rʹ;#d*h;BSm ۺӪMkdMCl U0 t֜2/R=^'A3  HW`=K]*kL@ϘV)ަ:71]O5FJqHffQ.icL?걺~Z, fMBv}hehBD/lt93m54-hZ&E*f)xDf]jUwSMw^7úmZ՝[׶uZOI}ٟԣ{G?+CuouJUUZg Y"ϤqV.53ּ DX}Ǘw01PЊE-WnΈ.|mݖ^s9m{/YS8hJiQYԒYf1jbF 0%Y-y ]7Kg)/gnẙק[ }G<((Ex+Gr] mMNyY^X>Fd-oyc?VprG<Ӎ?cvmkfi!v8=YrOAFQyWt+kMl53ԳU_zgjӊ2a=}PzNg']zˇ~)20u+sig3|Zf%4Čcq-_쫳oGM5qx?ʝ98*t^==әy[Z| msa֏Yof]~cݦev } 4E*46}ߧUa&w3398GB4s.|5T'A:#9׼FZq~.k{7[V+œPxZǔ#z/wXy:h|YQ.R YC&9kpl}zO~kmurbu53:A}@If"Py?>I;Ԣ6Oi:ϼXҥLw|QQQ{iU'˼x|H$#<[nVFcGévLLɔݍtlziawfN#cY/'Qڝb4*hfƒUu3F/ԩ@ϼ̷Z@ϡ3˳r-`ВKa/: ?[0-yih򵐃!VC`i^D3#ۆ~s}^@JKUfķUU|m_5gYW\33iGb}eq6}cījNͼ4otJƎhk}zef};{4+~&S|)|qkkZg5 Y-`=5ۓoOzv6Mu՚BuoB]ҵx7 3o؛_c̯uY.b%d-o ͲX-|[ж_7o/TxL[Növ'm1븎m5Op5ӾVyOմgnœqjnr|:3$g4_̡!ΙtWU}gXw; ks׫;b Ը v+6* =b ?P+6*J8T8T8TTTT$T$T$TTTT4T4T4T T T T,T,T,TTTT|JWlTT>T>`P@ 5j(PaPàA 5jP#FB:FA5j4hN:$N:d@5j,XSN:j8qPC5jDP&AMUUU5j2d)PS@CCCM 5j4iPӡCM5jLP3fA͂UUU5j6l9Ps@BBBAAACCCͅ 5TSN: 4ӠN:tyPAuPgB u&YPgAu6PgCu9PC͇ZjԹPB uyPAUBUBUB-Z|·ZjP@]j1b .".b.K.2ˠ.Zj PC]uPW@] u%ԕPWA]uPWC] u 5P@] u-ԵPA]uPC]u P7@u#ԍP7AuP7C u -P@ u+ԭPAuRPKnv;N; .n{j2ePB u/}PAu?PC=PB= rPˡz!zGzQGz 1ǡz 'zI'z )zgzYgz9P+V@Z y硞Z jjP@Z P/@"ԋPkBz %^zeW^zUW^z 5נ^zuuPAZj Po@FP6Am&ԛPoB[PoC 6fP@m;P@ .ԻPAPCP@}!ԇPA}PC} 'P@} )ԧP[Bm 3Ϡ>smP۠Am /K/ +o[o;P;v@}=P?@ԏP?Bj'N~ g~_vAj7n=P{@ j>}PCW_~ 7ߡ~PB P%ToPņl(By¡¡¡""""""bbbbbb|P>(T-ZPjCՆm^E !NB $!B{oA,(QĎX@,błsZYgf|wʺ;{{f3P Tm* :@P@E T]Ub*Xb*8⁊*JPT"P@%U@*IP 9H$ T2P)@T*P@TPi@T:P@eTL2* ,jTC T6P@5Pj TcTP9@5 PM*\*|)PMj TP@ f@59ṔjT Z@%PjT+ *@5PmjT@PjT{UTPE@uPTG: N@u3P T@u+P݀T7@P=@'PT/z@P} T_OP8H?P5@ jP`5!@ jPC P5 a@ j8PÁF5@jPhF51@j Pc X5q@jPZ@-jPZb%@-j PKZ R e@-j9PˁZ V@jPZjV5@j PkZ Z*u@jPu1P@..R.2.2.r.jP@]@]) "WuPW@mjjjPf6kk뀺끺끺 u Pu Pu+PuPuP[ VnvnjPۀ@@mj;Pہvur9H]@@ @=@Խ@ Խ@}@Nvzzzzz]@jP najP{#@=#@= ԣ@= c@=c@=@=@=@= ԓ@= S@=S@j/P{zz}@jP P,P,PPs> K++k[;;u3@}@}@@ԏ@O@O@,Pg___߀ ߀߁??? 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