a study of bone density on 5 random women at ...

# Explanation: ## Step1: Calculate the means Let $x$ be the age and $y$ be the bone - density. $\bar{x}=\frac{45 + 49+53+57+65}{5}=\frac{269}{5}=53.8$ $\bar{y}=\frac{360 + 345+340+330+325}{5}=\frac{1700}{5}=340$ ## Step2: Calculate the sums $S_{xx}=\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}=(45 - 53.8)^{2}+(49 - 53.8)^{2}+(53 - 53.8)^{2}+(57 - 53.8)^{2}+(65 - 53.8)^{2}$ $=(-8.8)^{2}+(-4.8)^{2}+(-0.8)^{2}+(3.2)^{2}+(11.2)^{2}$ $=77.44 + 23.04+0.64 + 10.24+125.44=236.8$ $S_{yy}=\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}=(360 - 340)^{2}+(345 - 340)^{2}+(340 - 340)^{2}+(330 - 340)^{2}+(325 - 340)^{2}$ $=20^{2}+5^{2}+0^{2}+(-10)^{2}+(-15)^{2}$ $=400+25 + 0+100+225=750$ $S_{xy}=\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})=(45 - 53.8)(360 - 340)+(49 - 53.8)(345 - 340)+(53 - 53.8)(340 - 340)+(57 - 53.8)(330 - 340)+(65 - 53.8)(325 - 340)$ $=(-8.8)\times20+(-4.8)\times5+(-0.8)\times0+(3.2)\times(-10)+(11.2)\times(-15)$ $=-176-24 + 0-32-168=-400$ ## Step3: Calculate the correlation coefficient $r$ and $r^{2}$ The correlation coefficient $r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\frac{-400}{\sqrt{236.8\times750}}=\frac{-400}{\sqrt{177600}}\approx\frac{-400}{421.42615}\approx - 0.949$ The coefficient of determination $r^{2}=(-0.949)^{2}\approx0.901$ # Answer: $0.901$