1.
下图是我国\(2008\)年至\(2014\)年生活垃圾无害化处理量\((\)单位:亿吨\()\)的折线图
\((\)Ⅰ\()\)由折线图看出,可用线性回归模型拟合\(y\)与\(t\)的关系,请用相关系数加以说明;
\((\)Ⅱ\()\)建立\(y\)关于\(t\)的回归方程\((\)系数精确到\(0.01)\),预测\(2016\)年我国生活垃圾无害化处理量.
附注:
参考数据:\(\sum\limits_{i=1}^{7}{{{y}_{i}}}=9.32\),\(\sum\limits_{i=1}^{7}{{{t}_{i}}{{y}_{i}}}=40.17\),\(\sqrt{\sum\limits_{i=1}^{7}{{{({{y}_{i}}-\bar{y})}^{2}}}}=0.55\),\(\sqrt{7}≈2.646\).
参考公式:相关系数\(r= \dfrac{ \sum\nolimits_{i=1}^{n}\left({t}_{i}- \overset{¯}{t}\right)\left({y}_{i}- \overset{¯}{y}\right)}{ \sqrt{ \sum\limits_{i=1}^{n}{\left({t}_{i}- \overset{¯}{t}\right)}^{2} \sum\limits_{i=1}^{n}{\left({y}_{i}- \overset{¯}{y}\right)}^{2}}}= \dfrac{ \sum\limits_{i=1}^{n}{t}_{i}{y}_{i}-n \overset{¯}{t} \overset{¯}{y}}{ \sqrt{ \sum\limits_{i=1}^{n}{\left({t}_{i}- \overset{¯}{t}\right)}^{2}} \sqrt{ \sum\limits_{i=1}^{n}{\left({y}_{i}- \overset{¯}{y}\right)}^{2}}} \)
回归方程\(\widehat{y}=\widehat{a}+\widehat{b}\,t\) 中斜率和截距的最小二乘估计公式分别为:\(\hat {b}= \dfrac{ \sum\nolimits_{i=1}^{n}\left({t}_{i}- \overset{¯}{t}\right)\left({y}_{i}- \overset{¯}{y}\right)}{ \sum\limits_{i=1}^{n}{\left(ti- \overset{¯}{t}\right)}^{2}} \widehat{a}{=}\bar{y}-\widehat{b}\,\bar{t}.\)