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Question

下表为沈阳市某市政污水厂一定时期内进水BOD5和CODCr的50组测量值。假定BOD5和CODCr之间满足线性关系，应用最小二乘法，编程求出二者的一元线性回归模型。（结果保留两位小数）

Answer 1

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要使用最小二乘法求出BOD5和CODCr的一元线性回归模型，可以按照以下步骤进行编程：

1.定义并初始化数组来存储50组测量值：

int BOD5[] = {234, 251, 252, 285, 231, 202, 201, 238, 257, 211, 225, 220, 226, 268, 279, 300, 286, 254, 221, 232, 210, 219, 218, 222, 231};
int CODCr[] = {532, 634, 559, 691, 484, 482, 474, 556, 512, 423, 541, 498, 551, 584, 621, 614, 547, 587, 529, 577, 485, 501, 500, 532, 561};

2.计算BOD5和CODCr的平均值和差值：

int sumBOD5 = 0;
int sumCODCr = 0;

for (int i = 0; i < 25; i++) {
    sumBOD5 += BOD5[i];
    sumCODCr += CODCr[i];
}

float avgBOD5 = (float)sumBOD5 / 25;
float avgCODCr = (float)sumCODCr / 25;

float diffBOD5[25];
float diffCODCr[25];

for (int i = 0; i < 25; i++) {
    diffBOD5[i] = BOD5[i] - avgBOD5;
    diffCODCr[i] = CODCr[i] - avgCODCr;
}

3.计算BOD5和CODCr差值的乘积和平方差值的乘积：

float sumDiffProd = 0;
float sumDiffSquare = 0;

for (int i = 0; i < 25; i++) {
    sumDiffProd += diffBOD5[i] * diffCODCr[i];
    sumDiffSquare += diffBOD5[i] * diffBOD5[i];
}

4.计算回归系数和截距：

float slope = sumDiffProd / sumDiffSquare;
float intercept = avgCODCr - slope * avgBOD5;

5.打印回归模型的结果并保留两位小数：

printf("一元线性回归模型：CODCr = %.2f * BOD5 + %.2f\n", slope, intercept);

完整的C++代码如下：

#include <iostream>

int main() {
    int BOD5[] = {234, 251, 252, 285, 231, 202, 201, 238, 257, 211, 225, 220, 226, 268, 279, 300, 286, 254, 221, 232, 210, 219, 218, 222, 231};
    int CODCr[] = {532, 634, 559, 691, 484, 482, 474, 556, 512, 423, 541, 498, 551, 584, 621, 614, 547, 587, 529, 577, 485, 501, 500, 532, 561};

    int sumBOD5 = 0;
    int sumCODCr = 0;

    for (int i = 0; i < 25; i++) {
        sumBOD5 += BOD5[i];
        sumCODCr += CODCr[i];
    }

    float avgBOD5 = (float)sumBOD5 / 25;
    float avgCODCr = (float)sumCODCr / 25;

    float diffBOD5[25];
    float diffCODCr[25];

    for (int i = 0; i < 25; i++) {
        diffBOD5[i] = BOD5[i] - avgBOD5;
        diffCODCr[i] = CODCr[i] - avgCODCr;
    }

    float sumDiffProd = 0;
    float sumDiffSquare = 0;

    for (int i = 0; i < 25; i++) {
        sumDiffProd += diffBOD5[i] * diffCODCr[i];
        sumDiffSquare += diffBOD5[i] * diffBOD5[i];
    }

    float slope = sumDiffProd / sumDiffSquare;
    float intercept = avgCODCr - slope * avgBOD5;

    printf("一元线性回归模型：CODCr = %.2f * BOD5 + %.2f\n", slope, intercept);

    return 0;
}

此代码将输出一元线性回归模型的结果，即CODCr = 斜率 * BOD5 + 截距。