Find the value of the linear correlation coefficient r using the paired data below. x 57535961535660 y 156164163177159175151 0.214 -0.054 -0.078 0.109

We are given seven paired values. In this interpretation the data are - \(x\)-values (each two‐digit) \[ 57,\; 53,\; 59,\; 61,\; 53,\; 56,\; 60 \] - \(y\)-values (each three‐digit) \[ 156,\; 164,\; 163,\; 177,\; 159,\; 175,\; 151 \] We wish to compute the linear correlation coefficient using the formula \[ r=\frac{n\sum xy - (\sum x)(\sum y)}{\sqrt{\Bigl[n\sum x^2 - (\sum x)^2\Bigr]\Bigl[n\sum y^2 - (\sum y)^2\Bigr]}}. \] In our case, \(n=7\). ### Step 1. Compute the required sums 1. **Sum of the \(x\)–values** \[ \sum x = 57 + 53 + 59 + 61 + 53 + 56 + 60 = 399. \] 2. **Sum of the \(y\)–values** \[ \sum y = 156 + 164 + 163 + 177 + 159 + 175 + 151 = 1145. \] 3. **Sum of the squares of \(x\)** \[ \begin{array}{rcl} 57^2 &=& 3249,\\[1mm] 53^2 &=& 2809,\\[1mm] 59^2 &=& 3481,\\[1mm] 61^2 &=& 3721,\\[1mm] 53^2 &=& 2809,\\[1mm] 56^2 &=& 3136,\\[1mm] 60^2 &=& 3600. \end{array} \] Thus, \[ \sum x^2 = 3249+2809+3481+3721+2809+3136+3600=22805. \] 4. **Sum of the squares of \(y\)** \[ \begin{array}{rcl} 156^2 &=& 24336,\\[1mm] 164^2 &=& 26896,\\[1mm] 163^2 &=& 26569,\\[1mm] 177^2 &=& 31329,\\[1mm] 159^2 &=& 25281,\\[1mm] 175^2 &=& 30625,\\[1mm] 151^2 &=& 22801. \end{array} \] Thus, \[ \sum y^2 = 24336+26896+26569+31329+25281+30625+22801 = 187837. \] 5. **Sum of the products \(xy\)** \[ \begin{array}{rcl} 57\cdot156 &=& 8892,\\[1mm] 53\cdot164 &=& 8692,\\[1mm] 59\cdot163 &=& 9617,\\[1mm] 61\cdot177 &=& 10797