# Register now to access 7 million high quality study materials (What's Course Hero?) Course Hero is the premier provider of high quality online educational resources. With millions of study documents, online tutors, digital flashcards and free courseware, Course Hero is helping students learn more efficiently and effectively. Whether you're interested in exploring new subjects or mastering key topics for your next exam, Course Hero has the tools you need to achieve your goals.

2 Pages

### multiple_ regression_overall

Course: PO 467, Fall 2009
School: SUNY Albany
Rating:

Word Count: 488

#### Document Preview

553 EPI/STA Principles of Statistical Inference II Fall 2006 Multiple regression -- the overall hypothesis October 17, 2006 Model The data consist of an independent random sample yi = o + 1 xi1 + L + k xik + i for k fixed values xi1 ,L, xik , i = 1,K, n , and deviation from mean i ~ N 0, 2 . So the mean of y is a linear function of the independent variables, i.e., yi | xi1 ,K, xik = o + 1 xi1 + L + k xik...

Register Now

#### Unformatted Document Excerpt

Coursehero >> New York >> SUNY Albany >> PO 467

Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.
553 EPI/STA Principles of Statistical Inference II Fall 2006 Multiple regression -- the overall hypothesis October 17, 2006 Model The data consist of an independent random sample yi = o + 1 xi1 + L + k xik + i for k fixed values xi1 ,L, xik , i = 1,K, n , and deviation from mean i ~ N 0, 2 . So the mean of y is a linear function of the independent variables, i.e., yi | xi1 ,K, xik = o + 1 xi1 + L + k xik . The k partial slope coefficients have the interpretation of mean changes in the dependent variable for a unit change in the respective independent variable, holding the others constant. Note that the simple linear regression model has k = 1 . ( ) Estimating parameters The regression parameters can be estimated according to the principle of least squares, i.e., minimizing ^ ^ ^ ^ ^ ^ the sum of square errors (residuals) SSE = ei2 = ( yi - yi )2 , where the yi = o + 1 xi1 + L + k xik are the ^ ^ ^ estimated mean values The least squares estimators , ,K, are weighted linear combinations of o, 1 k the yi , with weights determined by the independent values. Under the standard model assumptions, the 2 estimators are normally distributed, unbiased random variables, with sampling variances proportional ^ j to the error variance . They also have minimum variance over all unbiased estimators (BLUE). The actual formulas are best derived and written using techniques of linear (matrix) algebra. 2 Testing the overall hypothesis overall The hypothesis of interest is that there is a such a regression relation, i.e., that not all of the slope coefficients are zero. This can be tested using an analysis of variance, in which the total variation in y is decomposed into two parts, one due to error and the other due to the regression, namely ^ ^ ( yi - y ) = ( y i - yi ) + ( yi - y ) 2 2 2 , or SST = SSE + SSR . The ANOVA table has the following form: ANOVA for multiple regression Source df SS regression error total k n - k -1 n -1 SSR = SSE = ^ ( yi - yi ) 2 MS MSR = SSR k MSE = SSE (n - k - 1) F MSR SSR k = MSE SSE (n-k-1) ^ ( yi - yi ) 2 SST = ( yi - y ) 2 The hypothesis of no relation, and its alternative, can be written: H 0: 1 = 2 = K = k = 0 H1: at least one j 0 Test statistic and decision rule The test statistic F is the quotient of two mean squares. Under the null hypoth...

Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more. Course Hero has millions of course specific materials providing students with the best way to expand their education.

Below is a small sample set of documents:

Pittsburgh - AEI - 1319
COMMISSION OF THE EUROPEAN COMMUNITIESSEC(94) 860finalBrussels, 15.06. 1994COMMUNICATION FROM THECOMM I 8\$1 ONFINANCING THE TRANS- EUROPEAN NETWORKS- 1-FINANCING THE TRANS- EUROPEAN NETWORKSINTRODUCTION1 The European Council meeting
SUNY Albany - PO - 467
EPI/STA 553 Principles of Statistical Inference II Fall 2006Multiple regression - testing simultaneous linear restrictionsOctober 17, 2006 Models In the multiple regression model yi = o + 1 xi1 + L + k xik + i , i = 1,K, n , one can test simult
Pittsburgh - AEI - 5622
Pittsburgh - AEI - 4187
COUNCIL OF THE EUROPEAN COMMUNITIES PRESS RELEASES PRESIDENCY: GERMANY JANUARY-JUNE 1983 Meetings and press releases June 1983Meeting number 848th 849th 850th 851st 852nd 853rd 854th 855th 856th 857th 858th 859th 860th 861st 862ndSubject Labour/S
Pittsburgh - AEI - 4176
SUNY Albany - PO - 467
Directions: choose values for and .= = 2.0 2.01.2 1.0e +xLogistic function1 + e +xlower upper units obs 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 210.01 -3.30 0.99 1.30 20 0.23 x -3.3 -3.1 -2.8 -2.6 -2.4 -2.1 -1.9 -1.7 -1.5 -
Pittsburgh - AEI - 960
14/78GREECEAND THEEUROPEAN COMMUNITYOn the occasion of the visit to Athens, on 28 and 29 September 1978, of Mr Roy JenkinsCommission of the European Communities, this document takes stock of the relations between Greece and the European Commu
Pittsburgh - AEI - 5075
SUNY Albany - PO - 467
EPI/STA 553 Principles of Statistical Inference II Fall 2006Review Problems: Populations, random variables, samplesSeptember 19, 2006 1. Table 2.12 (p. 38) gives frequencies for astigmatism. Assuming this represents the population of interest, com
Pittsburgh - AEI - 3496
SUNY Albany - PO - 467
Problem #1astigmatism A A lower 0.0 0.2 0.4 0.6 1.1 2.1 3.1 4.1 5.1 B B uppper 0.2 0.3 0.5 1.0 2.0 3.0 4.0 5.0 6.0 C (A+B)/2 midpoint 0.10 0.25 0.45 0.80 1.55 2.55 3.55 4.55 5.55 frequency D D 458 268 151 79 44 19 9 3 2 1,033 relative frequency E D/
Pittsburgh - AEI - 951
COMMISSION OF THEEUROPEANCOMMUNITIESBrussels, 4 November 1992SEC(92) 1986 finalINDUSTRIAL COMPETITIVENESS AND PROTECTION OF THE ENVIRONMENTCommunication of the Commissionto the Council and to the European ParliamentTABLE OF CONTENTS1.
SUNY Albany - PO - 467
Problem #1pop % sample expected if null is true test statistic OiEi (Oi - Ei)2/EiUS61.9% 86 92.85 0.51Canada8.8% 17 13.20 1.09England4.8% 15 7.20 8.45Ireland6.3% 10 9.45 0.03Germany13.6% 14 20.40 2.01Other4.6% 8 6.90 0.18Total
Pittsburgh - AEI - 3782
Pittsburgh - AEI - 7629
I~z LI.II:L9 LI.I&gt; LI.IQEUROPEAN COMMISSIONDE 90 . February , 1997Copyrighted photos have been removed._1&amp;tl~2~.The Cameroonian Economy Cameroon in FiguresII II IJ II.er.BIl1!lmr'qlt~_e.~t'~g~i'j&quot;~]fgmFrom the Treaty of Rome to
Pittsburgh - AEI - 7449
Pittsburgh - AEI - 3821
Pittsburgh - AEI - 3877
SUNY Albany - PO - 467
EPI/STA 553 Principles of Statistical Inference II Fall 2006Multiple regression - confidence and predictive intervalsOctober 18, 2006The problem In the simple regression model (one independent variable), one can derive a formula for the variance
SUNY Albany - PO - 467
group 1 group 2 132 141 145 139 124 172 122 131 165 150 144 125 151 n=7 n=6 mean mean difference 140.43 143.00 2.57 std dev std dev pooled 15.44 16.60 15.98 t critical 0.44 2.20 do not reject
SUNY Albany - PO - 467
Chapter 5 Problem 7.3.20.0000 18.0000 16.0000 14.0000 12.0000 10.0000 8.0000 6.0000 4.0000 2.0000 0.0000 10.0 Column G Column H Column K Column JDIST (square root)20.030.040.0 MPH50.060.070.0For Chap 5 Prob 7.e., the formula iswhere
SUNY Albany - PO - 467
age X 6 7 8 9 10 11 12 13 14 15 16dry weight Y 0.029 0.052 0.079 0.125 0.181 0.261 0.425 0.738 1.130 1.882 2.812log(Y) Z -1.538 -1.284 -1.102 -0.903 -0.742 -0.583 -0.372 -0.132 0.053 0.275 0.449confidence interval for the correlation n 11 alpha
SUNY Albany - PO - 467
Y 47 38 47 39 44 64 58 49 55 52 49 47 40 42 63 40 59 56 76 67 57 57 42 54 60 33 55 36 36 42 41 42 39 27 31 39 56 40 58 43 40 46X1 287 236 255 135 121 171 260 237 261 397 295 261 258 280 339 161 324 171 265 280 248 192 349 263 223 316 288 256 318 27
SUNY Albany - PO - 467
Inference for the regression slope Obs. 1 2 3 sums means x 0 1 2 3 1 0.43 2.24 3.32 1.66 0.33 df two-sided p-value .05 critical points 1 0.80 -12.71 12.71 y 1 4 3 8.00 2.67x-x -1 0 1y-y -1.67 1.33 0.33( x - x )21 0 1 2( y - y)22.78 1.78 0.1
SUNY Albany - PO - 467
randomized complete block design treatments 1 2 3 5 3.33 2.78 total squares 9.00 4.00 0.00 source Treatments Blocks Interaction (Error) Total SS 16.67 28.67 2.67 48 3 2 3 4 8 5.00 0.00blocks 31 2 3 averages squares3 4 7 9 6.67 2.78averages 3.
CSU Stanislaus - HONS - 3050
Thursday, March 29, 2001Science - Taubes 288 (5470): 1319Page: 1THOMAS J CARTER | Change Password | Change User Info | CiteTrack Alerts | Subscription Help | Sign OutHYPERTENSION:A DASH of Data in the Salt DebateGary Taubes The controversy
UT Arlington - MATH - 3330
MATH 3330, Spring 20091MATRICES AND LINEAR ALGEBRAHomework 1.1: 3, 7, 10, 11, 16, 17, 32, 34 1.2: 3, 5, 8, 9, 11, 33, 34, 36 1.3: 1, 2, 3, 5, 9, 12, 14, 17, 18, 27, 28, 34, 36, 57 Chapter One Exercises (pp.3840): 1 10, odd numbers for the r
UT Arlington - MATH - 3330
Matrices and Linear AlgebraSolutions to Exam 2Problem 1. (25 pts) Find bases of the kernel and image of 1 2 2 -5 1 A = -1 -2 -1 4 8 5 -8 Solution: We start by computing RREF(A). 1 2 2 -5 - 1 2 2 -5 -2(II) 1 2 0 3 1 +(I) 0 0 1 -4 - 0 0
UT Arlington - MATH - 3330
Matrices and Linear AlgebraSolutions to Quiz 9Problem 1. (4 pts) Let 2 2 v1 = , 1 0 Compute the angle between v1 and v2 . Solution: Let the angle be . Then cos = So = , i.e., v1 and v2 are orthogonal. 2 Problem 2. (6 pts) Perform the Gram
UT Arlington - MATH - 3330
Matrices and Linear AlgebraSolutions to Quiz 10Problem 1. (5pts) Find out for which value(s) of the constant k the given matrix is invertible. 1 2 3 4 k 5 6 7 8 Solution: Use the fact that a square matrix is invertible if and only if its determ
UT Arlington - MATH - 3330
Matrices and Linear AlgebraSolutions to Quiz 11 Problem 1. (6pts) Find the derivative of the function f (t) = det Solution: Laplace expansion along 1 0 f (t) = t(-1)4+1 det 0 0 Therefore 1 9 9 t 7 1 0 0 1 0 2 2 0 2 0 3 3 3 9 0 4 4 4 1 4