2.
保险公司统计的资料表明:居民住宅区到最近消防站的距离\(x(\)单位:千米\()\)和火灾所造成的损失数额\(y(\)单位:千元\()\)有如下的统计资料:如果统计资料表明\(y\)与\(x\)有线性相关关系,试求:
距消防站距离\(x(\)千米\()\) | \(1.8\) | \(2.6\) | \(3.1\) | \(4.3\) | \(5.5\) | \(6.1\) |
火灾损失费用\(y(\)千元\()\) | \(17.8\) | \(6\) | \(27.5\) | \(31.3\) | \(36.0\) | \(43.2\) |
\((\)Ⅰ\()\)求相关系数\(r(\)精确到\(0.01)\);
\((\)Ⅱ\()\)求线性回归方程\((\)精确到\(0.01)\);
\((III)\)若发生火灾的某居民区与最近的消防站相距\(10.0\)千米,评估一下火灾的损失\((\)精确到\(0.01)\).
参考数据:\(\sum\limits_{1}^{6}{{{y}_{i}}}=175.4\)
,\(\sum\limits_{1}^{6}{{{x}_{i}}{{y}_{i}}}=764.36\)
,\(\sum\limits_{i=1}^{6}{({{x}_{i}}-\bar{x}})({{y}_{i}}-\bar{y})=80.30\)
,\(\sum\limits_{i=1}^{n}{{{({{x}_{i}}-\bar{x})}^{2}}}=14.30\)
,\(\sum\limits_{i=1}^{n}{{{({{y}_{i}}-\bar{y})}^{2}}}\approx 471.65\)
,\(\sqrt{\sum\limits_{i=1}^{n}{{{({{x}_{i}}-\bar{x})}^{2}}}\sum\limits_{i=1}^{n}{{{({{y}_{i}}-\bar{y})}^{2}}}}\approx 82.13\)
参考公式:相关系数 \(r=\dfrac{\sum\limits_{i=1}^{n}{({{x}_{i}}-\bar{x})({{y}_{i}}-\bar{y})}}{\sqrt{\sum\limits_{i=1}^{n}{{{({{x}_{i}}-\bar{x})}^{2}}}\sum\limits_{i=1}^{n}{{{({{y}_{i}}-\bar{y})}^{2}}}}}\)
, 回归方程\(\overset{∧}{y}= \overset{∧}{a}+ \overset{∧}{b}t \) 中斜率和截距的最小二乘估计公式分别为:\(\hat{b}=\dfrac{\sum\limits_{i=1}^{n}{({{x}_{i}}-\bar{x})({{y}_{i}}-\bar{y})}}{\sum\limits_{i=1}^{n}{{{({{x}_{i}}-\bar{x})}^{2}}}}\),\(\hat{a}=\bar{y}-\hat{b}x\)