基于t_value <- summary(model)$coefficients["Fe", "t value"]的答案不正确。该t_value将用于H0: Fe = 0的测试，而不是用于H0: Hypothesis: (Intercept) + 1.5 Fe >= 95的测试。
这里的困难在于linearHypothesis()函数不做单侧测试。
执行单侧检验的一种方法是使用单侧检验和双侧检验的p值之间的关系，当参数估计值具有正态分布误差时，该关系适用于线性模型。
如果H0: param = x与H1: param != x的检验的p值为p，则H0: param >= x与H1: param < x的检验的p值将为p/2或1-p/2。当参数估计值小于x时，p值将为p/2;当参数估计值大于x时，p值将为1-p/2。
在您的示例中，参数为(intercept) + 1.5*Fe，其估计值可以从summary(model)中找到，为129.787 - 1.5*24.02 = 93.757，小于95，因此您应该使用p/2 = 0.3097/2 = 0.15485。
编辑以添加：上面的答案实际上是对统计问题的回答，而不是对编程问题的回答。下面是编程的方法：

library(faraway)
library(car)
#> Loading required package: carData
#> 
#> Attaching package: 'car'
#> The following objects are masked from 'package:faraway':
#> 
#>     logit, vif

# Load the data
data(corrosion)

# Simple linear regression
model <- lm(loss ~ Fe, data = corrosion)

# Two-sided test of H0: (intercept) + 1.5 Fe = 95

test <- linearHypothesis(model, c(1, 1.5), rhs = 95)
test
#> Linear hypothesis test
#> 
#> Hypothesis:
#> (Intercept)  + 1.5 Fe = 95
#> 
#> Model 1: restricted model
#> Model 2: loss ~ Fe
#> 
#>   Res.Df    RSS Df Sum of Sq      F Pr(>F)
#> 1     12 113.45                           
#> 2     11 102.85  1    10.603 1.1341 0.3097

p <- test$"Pr(>F)"[2]
p
#> [1] 0.3097301

estimate <- coef(model) %*% c(1, 1.5)
estimate
#>          [,1]
#> [1,] 93.75676

onesided <- if (estimate < 95) p/2 else 1 - p/2
onesided
#> [1] 0.1548651

创建于2023年3月5日，使用reprex v2.0.2

阿德里安·里奥，一个出路可能是...
首先，从拟合模型中提取t-value：

# Simple linear regression
model <- lm(loss ~ Fe, data = corrosion)

检验零假设H0：（截距）+1.5 Fe〉= 95，我们需要使用估计的系数及其标准误差计算t值：

# first, extract the estimated coefficients and their standard errors
b <- coef(model)
SE_of_b <- sqrt(diag(vcov(model)))

# the, calculate the t-value for `(intercept) + 1.5 Fe = 95`
t_value <- (b[1] + 1.5 * b[2] - 95) / sqrt(SE_of_b[1]^2 + (1.5 * SE_of_b[2])^2)

然后计算p值：

# finally, calculate the 'p-values' (for the one-sided test where the 
# null hypothesis is '(intercept) + 1.5 Fe >= 95')

p_value <- pt(t_value, df = nrow(corrosion) - 2, lower.tail = TRUE)

最后，

# if the p-value is less than 0.5 (significance level) and t_value is less than 0, 
# reject Null Hypothesis (H0), otherwise, we fail to reject H0

ifelse((p_value < 0.5 & t_value < 0), "Reject H0: loss is less than 95mg/dm2/day", "Failed to reject the H0: weight loss is at least 95mg/dm2/day")

这里，我假设零假设定义为：

Null hypothesis (H0): (intercept) + 1.5*Fe >= 95

and,

Alternative hypothesis: (intercept) + 1.5*Fe < 95

如果你有什么想法就告诉我...