2

My statistics program plots vertical lines representing a t-ratio at 5% significance level along with its sorted parameter estimates when fitting a linear model. I don't understand how this line is calculated. Presumably if they could calculate it at 5% significance, I could calculate a t-ratio value at 10% significance. How would I do so?

t-ratio graphic

I know t Ratio = Estimate / Std Error, but I don't know how to relate either the estimate or standard error to a significance level. However, I think the standard error of the estimate is used as the standard deviation for the associated t-distribution that is then used to determine the p-value.

wdkrnls
  • 297
  • 1
  • 3
  • 11
  • 2
    A general answer to this question appears at http://stats.stackexchange.com/questions/34134/interpreting-t-value-as-a-measure-of-evidence/34135#34135. A specific answer is buried in the text at http://stats.stackexchange.com/questions/37768/t-test-and-p-value/37770#37770. A related discussion occurs at http://stats.stackexchange.com/questions/23800. – whuber Jul 26 '13 at 20:33
  • 1
    At heart you can get the value from t-tables, or from a program. What program are you using? – Glen_b May 12 '14 at 07:56

1 Answers1

1

You know that the p-value is $P(t<-|tRatio|)+P(t>|tRatio|)$ and its value is computed by the CDF of a $t$ distribution with the proper number of degrees of freedom, i.e. $n-p$ where $n$ is the number of observations and $p$ the number of parameters.

To get the absolute value of tRatio at a given significance you can use the quantile function. Let me use a toy example in R:

> x <- c(1, 2, 3)       # independent variable
> y <- c(2.9, 5.2, 6.9) # dependent variable, n = 3 observations
> linreg <- lm(y ~ x)
> summary(linreg)       # p = 2 parameters
[...]
Coefficients:
            Estimate Std. Error t value Pr(>|t|)  
(Intercept)   1.0000     0.3742   2.673    0.228  
x             2.0000     0.1732  11.547    0.055 .
[...]
> tRatio <- 11.547      # n - p = 1 degree of freedom
> p_value <- pt(-abs(tRatio), 1) +            # P(t < -|tRatio|)
+     pt(abs(tRatio), 1, lower.tail = FALSE)  # P(t > |tRatio|)
> p_value
[1] 0.0549957

You can compute the absolute value of tRatio at 5.49957% significance by the quantile function. Since it is a two sided test, you have to compute $F_t^{-1}(pvalue/2)$:

> abs(qt(p_value/2, 1))
[1] 11.547

You can compute the absolute value of tRatio at whatever significance by:

> abs(qt(0.05 / 2, 1)) # this is where those vertical lines come from
[1] 12.7062
> abs(qt(0.10 / 2, 1))
[1] 6.313752

So, if you are looking for 10% significance a tRatio > 6.314 or < -6.314 is enough.

Sergio
  • 5,628
  • 2
  • 11
  • 27