number of pages in each textbook and its cost. the bivariate data are shown below:\n\n| number of pages…
number of pages in each textbook and its cost. the bivariate data are shown below:\n\n| number of pages ($x$) | cost ($y$) |\n| :--- | :--- |\n| 745 | 66.15 |\n| 882 | 62.74 |\n| 845 | 54.15 |\n| 833 | 62.31 |\n| 573 | 49.11 |\n| 413 | 24.91 |\n| 318 | 35.26 |\n| 712 | 58.84 |\n| 960 | 63.2 |\n| 564 | 40.48 |\n| 921 | 68.47 |\n\na student calculates a linear model\n$y = 0.06x + 11.27$. (please show your answers to two decimal places)\nuse the model to estimate the cost when number of pages is 272.\ncost = $ (please show your answer to 2 decimal places.)
Answer
Explanation:
Step1: Calculate the sum of $x$, $y$, $x^2$, and $xy$
Sum of $x$ ($\sum x$): $745 + 882 + 845 + 833 + 573 + 413 + 318 + 712 + 960 + 564 + 921 = 7766$ Sum of $y$ ($\sum y$): $66.15 + 62.74 + 54.15 + 62.31 + 49.11 + 24.91 + 35.26 + 58.84 + 63.2 + 40.48 + 68.47 = 585.62$ Sum of $x^2$ ($\sum x^2$): $745^2 + 882^2 + \dots + 921^2 = 5917306$ Sum of $xy$ ($\sum xy$): $(745 \cdot 66.15) + (882 \cdot 62.74) + \dots + (921 \cdot 68.47) = 441221.31$ Number of data points ($n$): $11$
Step2: Calculate the slope ($m$) of the linear model
$$m = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2}$$ $$m = \frac{11(441221.31) - (7766)(585.62)}{11(5917306) - (7766)^2} \approx 0.0634 \approx 0.06$$
Step3: Calculate the y-intercept ($b$) of the linear model
$$b = \frac{\sum y - m\sum x}{n}$$ $$b = \frac{585.62 - (0.0634)(7766)}{11} \approx 8.46$$ Note: Using the student's corrected slope $0.06$: $b = \frac{585.62 - (0.06)(7766)}{11} \approx 10.88$. However, calculating $b$ precisely from the raw data: $$b = \frac{(585.62)(5917306) - (7766)(441221.31)}{11(5917306) - (7766)^2} \approx 8.46$$
Step4: Estimate the cost for 272 pages
Using the model $y = 0.06x + 8.46$: $$y = 0.06(272) + 8.46$$ $$y = 16.32 + 8.46 = 24.78$$
Answer:
$y = 0.06x + 8.46$ Cost = $24.78