1. Problem 4: Given \(\sum x = 90\), \(\sum x^2 = 308\), \(\sum y^2 = 27218\), \(\sum xy = 1617\), \(\sum y = 1010\), find the regression line of \(y\) on \(x\). Formula for regression line of \(y\) on \(x\): $$y = a + bx$$ where $$b = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2}$$ $$a = \frac{\sum y}{n} - b \frac{\sum x}{n}$$ Assuming \(n\) is the number of observations (not given explicitly), but from the context of the problem, let's assume \(n=10\) (since \(\sum y = 1010\) suggests 10 observations with average 101). Calculate slope \(b\): $$b = \frac{10 \times 1617 - 90 \times 1010}{10 \times 308 - 90^2} = \frac{16170 - 90900}{3080 - 8100} = \frac{-74730}{-5020} = 14.88$$ Calculate intercept \(a\): $$a = \frac{1010}{10} - 14.88 \times \frac{90}{10} = 101 - 14.88 \times 9 = 101 - 133.92 = -32.92$$ Regression line: $$y = -32.92 + 14.88x$$ Estimate ground moisture constant for an area equal to one chosen area (assuming \(x=1\)): $$y = -32.92 + 14.88 \times 1 = -18.04$$ 2. Problem 4 (continued): Calculate regression line of Final (F) on Initial (I) scores for 12 patients. Given data: \(n=12\), \(\sum I = 391\), \(\sum F = 268\), \(\sum I^2 = 14635\), \(\sum F^2 = 8346\), \(\sum IF = 9081\) (calculated from data). Calculate slope \(b\): $$b = \frac{n\sum IF - \sum I \sum F}{n\sum I^2 - (\sum I)^2} = \frac{12 \times 9081 - 391 \times 268}{12 \times 14635 - 391^2} = \frac{108972 - 104788}{175620 - 152881} = \frac{4184}{22739} \approx 0.184$$ Calculate intercept \(a\): $$a = \frac{\sum F}{n} - b \frac{\sum I}{n} = \frac{268}{12} - 0.184 \times \frac{391}{12} = 22.33 - 0.184 \times 32.58 = 22.33 - 6.0 = 16.33$$ Regression line: $$F = 16.33 + 0.184 I$$ Estimate improvement for initial score \(I=30\): $$F = 16.33 + 0.184 \times 30 = 16.33 + 5.52 = 21.85$$ Improvement = Final - Initial = \(21.85 - 30 = -8.15\) (negative means decrease). 3. Problem 5: Given \(\sum x=15\), \(\sum x^2=55\), \(\sum y=43\), \(\sum y^2=397\), \(\sum xy=145\), \(n=5\), find regression lines of \(y\) on \(x\) and \(x\) on \(y\). Calculate slope of \(y\) on \(x\): $$b_{yx} = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2} = \frac{5 \times 145 - 15 \times 43}{5 \times 55 - 15^2} = \frac{725 - 645}{275 - 225} = \frac{80}{50} = 1.6$$ Intercept \(a_{yx}\): $$a_{yx} = \frac{\sum y}{n} - b_{yx} \frac{\sum x}{n} = \frac{43}{5} - 1.6 \times \frac{15}{5} = 8.6 - 1.6 \times 3 = 8.6 - 4.8 = 3.8$$ Regression line \(y\) on \(x\): $$y = 3.8 + 1.6 x$$ Calculate slope of \(x\) on \(y\): $$b_{xy} = \frac{n\sum xy - \sum x \sum y}{n\sum y^2 - (\sum y)^2} = \frac{5 \times 145 - 15 \times 43}{5 \times 397 - 43^2} = \frac{725 - 645}{1985 - 1849} = \frac{80}{136} \approx 0.588$$ Intercept \(a_{xy}\): $$a_{xy} = \frac{\sum x}{n} - b_{xy} \frac{\sum y}{n} = 3 - 0.588 \times 8.6 = 3 - 5.06 = -2.06$$ Regression line \(x\) on \(y\): $$x = -2.06 + 0.588 y$$ 4. Problem 6: Given \(\sum x=567\), \(\sum y=552\), \(\sum x^2=36261\), \(\sum y^2=31779\), \(\sum xy=36112\), \(n=9\). (a) Find regression line of \(y\) on \(x\): Slope: $$b = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2} = \frac{9 \times 36112 - 567 \times 552}{9 \times 36261 - 567^2} = \frac{324, 1,008 - 312,984}{326,349 - 321,489} = \frac{11,024}{4,860} \approx 2.27$$ Intercept: $$a = \frac{\sum y}{n} - b \frac{\sum x}{n} = \frac{552}{9} - 2.27 \times \frac{567}{9} = 61.33 - 2.27 \times 63 = 61.33 - 143.01 = -81.68$$ Regression line: $$y = -81.68 + 2.27 x$$ (b) Estimate summer mark for Christmas mark \(x=70\): $$y = -81.68 + 2.27 \times 70 = -81.68 + 158.9 = 77.22$$ 5. Problem 7: Given data for units \(x\) and total cost \(y\), with \(\sum x^2 = 28740\), \(\sum xy = 38286\), \(n=12\). Calculate \(\sum x\) and \(\sum y\): $$\sum x = 14 + 29 + 55 + 74 + 11 + 23 + 47 + 69 + 18 + 36 + 61 + 79 = 516$$ $$\sum y = 35 + 50 + 73 + 93 + 31 + 42 + 65 + 80 + 38 + 54 + 81 + 96 = 738$$ Calculate slope \(b\): $$b = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2} = \frac{12 \times 38286 - 516 \times 738}{12 \times 28740 - 516^2} = \frac{459432 - 380808}{344880 - 266256} = \frac{78624}{78624} = 1$$ Calculate intercept \(a\): $$a = \frac{\sum y}{n} - b \frac{\sum x}{n} = \frac{738}{12} - 1 \times \frac{516}{12} = 61.5 - 43 = 18.5$$ Regression line: $$y = 18.5 + 1 x$$ (d) To find output where total income equals total cost: Total income = selling price \(\times\) output = \(1.6 x\) Set total income = total cost: $$1.6 x = 18.5 + x$$ $$1.6 x - x = 18.5$$ $$0.6 x = 18.5$$ $$x = \frac{18.5}{0.6} \approx 30.83$$ (e) Interpretation: At approximately 30.83 units of output, the company's total income equals total cost, meaning the break-even point where profit is zero.