Pregunta
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Find the value of the linear correlation coefficient \( r \) using the paired data below. x 75359516356516 y 165146136177195157115 0.109 -0.078 -0.086 0.214

Ask by Patel Carrillo. in the United States
Mar 10,2025

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The linear correlation coefficient \( r \) is approximately 0.109.

Solución

We are given the following paired data (by splitting the digits into numbers): - \( x \): 75, 35, 95, 163, 56, 5, 16 - \( y \): 165, 146, 136, 177, 195, 157, 115 We wish to compute the linear correlation coefficient \[ r = \frac{n\sum xy - (\sum x)(\sum y)}{\sqrt{\Bigl[n\sum x^2 - (\sum x)^2\Bigr]\Bigl[n\sum y^2 - (\sum y)^2\Bigr]}}. \] Below are the step‐by‐step calculations. --- ### 1. Compute the sums There are \( n = 7 \) pairs. **Sum of \( x \):** \[ \sum x = 75 + 35 + 95 + 163 + 56 + 5 + 16. \] Step by step: \[ 75 + 35 = 110, \quad 110 + 95 = 205, \quad 205 + 163 = 368, \quad 368 + 56 = 424, \quad 424 + 5 = 429, \quad 429 + 16 = 445. \] Thus, \[ \sum x = 445. \] **Sum of \( y \):** \[ \sum y = 165 + 146 + 136 + 177 + 195 + 157 + 115. \] Step by step: \[ 165 + 146 = 311, \quad 311 + 136 = 447, \quad 447 + 177 = 624, \quad 624 + 195 = 819, \quad 819 + 157 = 976, \quad 976 + 115 = 1091. \] Thus, \[ \sum y = 1091. \] --- ### 2. Compute the sum of squares **Sum of squares for \( x \):** \[ \begin{array}{rcl} 75^2 &=& 5625,\\[1mm] 35^2 &=& 1225,\\[1mm] 95^2 &=& 9025,\\[1mm] 163^2 &=& 26569,\\[1mm] 56^2 &=& 3136,\\[1mm] 5^2 &=& 25,\\[1mm] 16^2 &=& 256. \end{array} \] Then, \[ \sum x^2 = 5625 + 1225

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Bonus Knowledge

To determine the linear correlation coefficient \( r \) from the given data, you would typically compute the means and standard deviations of both sets of values, and then apply the formula for \( r \). However, without performing the calculations, if you're looking at your provided choices, remember that \( r \) ranges from -1 to 1. A positive value indicates a positive correlation, while a negative value suggests a negative correlation. In your choices, the values are quite close to 0, suggesting weak correlations. If I had to pick one based on the process you would normally follow, I would encourage calculating it using a statistical program or calculator for accuracy! In practice, when you find \( r \), just keep in mind that the closer to ±1 it is, the stronger the correlation will be.

preguntas relacionadas

Question 12(Mulliple Choice Warth 5 points) \[ (04.06 \mathrm{HC}) \] A researcher wants to test the claim that the proportion of juniors who watch television regularly is greater than the proportion of seniors who watch television regularly She finds that 56 of 70 randomly selected juniors and 47 of 85 randomly selected seniors report watching television regularly. Construct \( 95 \% \) confidence intervals for each population proportion. Which of the statemente gives the correct outcome of the research or's tert of the dalim? The \( 95 \% \) confidence interval for juniors is (706, 894), and the \( 95 \% \) confidence interval for seniors is ( 447,659 ). Since the intervals overlap, there is not enough evidence to say the proportion of juniors who watch television regularly may be higher than that of seniors. The \( 95 \% \) confidence interval for juniors is (721, 879), and the \( 95 \% \) confidence interval for seniors is (464, 642). Since the interval for juniors is higher than the interval for seniors, there is evidence to say the proportion of juniors who watch television regularly may be higher than that of seniors. The \( 95 \% \) confidence interval for juniors is ( 706,894 ), and the \( 95 \% \) confidence interval for seniors is ( 447,659 ). Since the interval for juniors is higher than the interval for seniors, there is evidence to say the proportion of juniors who watch television regularly may be higher than that of seniors. The \( 95 \% \) confidence interval for juniors is ( \( 721, .879 \) ), and the \( 95 \% \) confidence interval for seniors is (464, 642). Since the intervals overlap, there is not enough evidence to say the proportion of juniors who watch television regularly may be higher than that of seniors.

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