[英]How to calculate the 95% confidence interval for the slope in a linear regression model in R
Here is an exercise from Introductory Statistics with R: 以下是R的入门统计练习:
With the rmr data set, plot metabolic rate versus body weight. 使用rmr数据集,绘制代谢率与体重的关系。 Fit a linear regression model to the relation.
将线性回归模型拟合到关系中。 According to the fitted model, what is the predicted metabolic rate for a body weight of 70 kg?
根据拟合模型,体重70公斤的预测代谢率是多少? Give a 95% confidence interval for the slope of the line.
给出该线斜率的95%置信区间。
rmr data set is in the 'ISwR' package. rmr数据集位于“ISwR”包中。 It looks like this:
它看起来像这样:
> 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
I know how to calculate the predicted y at a given x but how can I calculate the confidence interval for the slope? 我知道如何计算给定x的预测y,但我如何计算斜率的置信区间?
Let's fit the model: 让我们适合这个模型:
> 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
The 95% confidence interval for the slope is the estimated coefficient (7.0595) ± two standard errors (0.9776). 斜率的95%置信区间是估计系数(7.0595)±两个标准误差(0.9776)。
This can be computed using confint
: 这可以使用
confint
计算:
> confint(fit, 'body.weight', level=0.95)
2.5 % 97.5 %
body.weight 5.086656 9.0324
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.