注：年份代码1～7分别对应年份2008～2014.
(1)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； (2)建立y关于t的回归方程(系数精确到0.01)，预测2019年我国生活垃圾无害化处理量． 附注：
7
7
7
参考数据：?yi＝9.32，?tiyi＝40.17，i＝1
i＝1
? ?yi－yi＝1
?＝0.55，7≈2.646.
2
n? ?ti－t??yi－yi＝1
?
，
参考公式：相关系数r＝
n2
n? ?ti－t?? ?yi－yi＝1
i＝1
^
^
^
?
2
回归方程y ＝a ＋b t中斜率和截距的最小二乘估计公式分别为：
n? ?ti－t??yi－y^
?
^
^
i＝1
b ＝
n，a ＝y－b t.
? ?ti－ti＝1
?
2
解 (1)由折线图中数据和附注中参考数据得
7
7
2
t＝4，? (ti－t)＝28,
i＝1
? ?yi－yi＝1
7
?＝0.55.
2
77
? (ti－t)(yi－y)＝?tiyi－t?yi＝40.17－4×9.32＝2.89，
i＝1
i＝1
i＝1
2.89所以r≈≈0.99.
0.55×2×2.646
因为y与t的相关系数近似为0.99，说明y与t的线性相关程度相当高，从而可以用线性回归模型拟合y与t的关系． 9.32
(2)由y＝≈1.331及(1)得
7
1
7
? ?ti－t??yi－y?
^
i＝1
b ＝
＝2.89≈0.10， 7
28
? ?ti－t?
2
i＝1
^
^
a ＝y－b t≈1.331－0.103×4≈0.92.
^
所以y关于t的线性回归方程为y ＝0.10t＋0.92. 将2019年对应的t＝12代入线性回归方程，得
^
y ＝0.92＋0.10×12＝2.12.
所以预测2019年我国生活垃圾无害化处理量将约为2.12亿吨．
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1
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