. This problem will yield a standard formula given in elementary statistics for a least squares line, making use of the normal equations. (a) Given pairs of data points (x1, Y1), (x2, Y2), ..., (xn, Yn) consider approximating lines of the form y = mx+b. The error e; for the ith pair is the distance between y; and the height (y value) of the line at xi. This is ei = Yi — (mxi + b). If we consider the equations b + x;m = Yi for i n in the variables b and m we can = 1,2, = think of this as a system of equations Ax = 6 where A 1 x1 x2 = : [m] Хп Y1 Уп numbers. Here, note that the variables are m and b and the xi, Yi are given The least squares approximation for this system (which gives the intercept b and slope m of the best least squares line for the data) is the solution to the normal equations AT Ax = ATb. Determine ATA (a 2×2 matrix) and AT (a 2×1 matrix). The entries will be sums of terms involving the x; and y₁. Write these, first using Σ notation and then simplify the notation using…

= a) Recall that the formula for the projection p of vector & onto vector a is p = ±ªã. The plane through the origin in R³ given by ax+by+cz = 0 for real numbers a, b, c, is the set of all points (viewed as vectors) orthogonal to the normal n = (a, b, c) to the plane. To find the distance from a point w (xo, Yo, zo) to the plane ax+by+cz = 0 we can find the projection of w onto the normal ñ and then find the length of this projection. Do this to derive a generic formula for the distance in terms of xo, Yo, Zo, a, b, c. First find the length squared and then take a root. b) To find the distance from a point w (xo, Yo, zo) to a plane ax + by + cz = d, not necessarily not through the origin, one approach is to shift all z coordinates down by d. That is, replace (x, y, z) by (x, y, z — d). The relationship between the plane and w is unchanged so we can find the distance by finding the distance between (xo, Yo, zo - d) and the plane ax + by+cz = 0 through the origin. Do this to derive a…

Use Gram-Schmidt orthogonalization to find an orthonormal set of vectors with the same span as = 1 2 -0-08-0 a2 = 3 a3 = -2 Find also a QR factorization A = QR with the columns of Q the orthonormal set found above and R upper triangular and invertible and A with columns ā₁, ā2, ā3.

(a) Let Q be an orthonormal matrix. That is, the columns of Q have unit length and are pairwise orthogonal. Prove that the length of Qã is equal to the length of for any column vector x. Note that you can do this by showing that the lengths squared are equal. (b) Prove that if A and B are symmetric and πT Ax = x² Bã for all & then A = B. Make use of coordinate vectors ē;, which are 0 except for a 1 in position i. Explain why, for matrix A, entries can be found by arj = Aej. First show that diagonal = = entries aii bir. Then consider what (ē¿ + ē¿)ªA(ē¿ + ē¿) in terms of entries of A (and similarly for B) and make use of symmetry aij = aji and similarly for B. (c) Prove the converse: If for all x, ||Q|| ||||| then Q is orthonormal. One ap- proach is to use part (b) to show that QTQ = I. Recall that QTQ is symmetric (and trivially I is symmetric). Recall also that, written in matrix notation, vector length (squared) is given by ||7||2 = (this is the dot product written in matrix…

(a) What is wrong with the following statement and proof? A is invertible if and only if ATA is invertible. 'Proof': Since det(A) = det(AT), det(ATA) = det(A)². Either both are 0 (and both A and ATA are not invertible) or both are nonzero (and both A and ATA are invertible). (b) Prove that Null(A) C Null(ATA). That is, prove that if Ax = 0 then AT Ax = 0. (c) Prove that Null(ATA) C Null(A). That is, prove that if AT Añ = 0 then Aã = Ổ. Recall that the length (squared) of a vector v is V, which written in matrix multiplication notation is Tv. (d) Parts (b) and (c) imply that Null(A) = • Nul(ATA). Use this to prove that the columns of A are linearly independent if and only if ATA has an inverse.

Give a 99% confidence interval, for μ1 M2 given the following information. n₁ = 20, 1 = 2.31, s₁ = 0.47 n2 = 40, 2 = 2.21, 82 = 0.55 + Use Technology Rounded to 2 decimal places. Hint Question Help: Video Submit Question ch Η UP T SINK O DELL F2 F3 F4 F5 F6 F7 F8 144 ►11 4

Please could you provide solutions to the following questions. Thanks

Please could you provide solutions to the following questions. Thanks

or Beer and blood alcohol content: Many people believe that gender, weight, drinking habits, and many other factors are much more important in predicting blood alcohol content (BAC) than simply considering the number of drinks a person consumed. Here we examine data from sixteen student volunteers at Ohio State University who each drank a randomly assigned number of cans of beer. These students were evenly divided between men and women, and they differed in weight and drinking habits. Thirty minutes later, a police officer measured their blood alcohol content (BAC) in grams of alcohol per deciliter of blood (Malkevitc and Lesser, 2008). The scatterplot and regression table summarize the findings. to search BAC (grams per deciliter) 0.15- 0.10- 0.05- Part of Ti Calculator Output Ti Calculator y = ax+b Be and p = 0 t = 7.48 P 0.0000 df = 14 a = 0.0180 b = -0.0127 Part of Excel Output 2 4 0 0 6 Cans of beer (Intercept) beers Estimate -0.0127 0.0180 Std. Error t value 0.0126 P(>|t|) -1.00…