如何计算 R 中线性回归模型中斜率的 95% 置信区间

问题描述

这是一个来自 R 的介绍性统计练习:

Here is an exercise from Introductory Statistics with R:

使用 rmr 数据集，绘制代谢率与体重的关系图.将线性回归模型拟合到该关系.根据拟合模型，70 公斤体重的预测代谢率是多少?给出直线斜率的 95% 置信区间.

With the rmr data set, plot metabolic rate versus body weight. Fit a linear regression model to the relation. According to the fitted model, what is the predicted metabolic rate for a body weight of 70 kg? Give a 95% confidence interval for the slope of the line.

rmr 数据集位于ISwR"包中.它看起来像这样:

rmr data set is in the 'ISwR' package. It looks like this:

> rmr
   body.weight metabolic.rate
1         49.9           1079
2         50.8           1146
3         51.8           1115
4         52.6           1161
5         57.6           1325
6         61.4           1351
7         62.3           1402
8         64.9           1365
9         43.1            870
10        48.1           1372
11        52.2           1132
12        53.5           1172
13        55.0           1034
14        55.0           1155
15        56.0           1392
16        57.8           1090
17        59.0            982
18        59.0           1178
19        59.2           1342
20        59.5           1027
21        60.0           1316
22        62.1           1574
23        64.9           1526
24        66.0           1268
25        66.4           1205
26        72.8           1382
27        74.8           1273
28        77.1           1439
29        82.0           1536
30        82.0           1151
31        83.4           1248
32        86.2           1466
33        88.6           1323
34        89.3           1300
35        91.6           1519
36        99.8           1639
37       103.0           1382
38       104.5           1414
39       107.7           1473
40       110.2           2074
41       122.0           1777
42       123.1           1640
43       125.2           1630
44       143.3           1708

我知道如何计算给定 x 处的预测 y，但如何计算斜率的置信区间?

I know how to calculate the predicted y at a given x but how can I calculate the confidence interval for the slope?

推荐答案

让我们拟合模型:

> library(ISwR)
> fit <- lm(metabolic.rate ~ body.weight, rmr)
> summary(fit)

Call:
lm(formula = metabolic.rate ~ body.weight, data = rmr)

Residuals:
    Min      1Q  Median      3Q     Max
-245.74 -113.99  -32.05  104.96  484.81

Coefficients:
            Estimate Std. Error t value Pr(>|t|)
(Intercept) 811.2267    76.9755  10.539 2.29e-13 ***
body.weight   7.0595     0.9776   7.221 7.03e-09 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 157.9 on 42 degrees of freedom
Multiple R-squared: 0.5539, Adjusted R-squared: 0.5433
F-statistic: 52.15 on 1 and 42 DF,  p-value: 7.025e-09

斜率的 95% 置信区间是估计系数 (7.0595) ± 两个标准误差 (0.9776).

The 95% confidence interval for the slope is the estimated coefficient (7.0595) ± two standard errors (0.9776).

这可以使用 confint 计算:

This can be computed using confint:

> confint(fit, 'body.weight', level=0.95)
               2.5 % 97.5 %
body.weight 5.086656 9.0324

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