Statistical significance of linear least squares

Question

Consider two data series, $X = (x_1, x_2, \dots, x_n)$ and $Y = (y_1, y_2, \dots, y_n),$ both with mean zero. We use linear regression (ordinary least squares) to regress $Y$ against $X$ (without fitting any intercept), as in $Y = aX + \epsilon$ where $\epsilon$ denotes a series of error terms.

It can be shown that $a=\frac{\rho_{XY} \sigma_Y}{\sigma_X}$.

Suppose that $\rho_{XY} = 0.01$. Is the resulting value of $a$ statistically significantly different from $0$ at the $95\%$ level if:
i. $n=10^2$
ii. $n = 10^3$
iii. $n = 10^4$

I know that I need to find a $p$-value for each of these, and I assume that it will be in turns of $\rho$ and $n$. However, everything I attempt leaves a lingering standard deviation somewhere from the definition of $a$. How do you test significance of a least squares in this case?

Answer 1

The following can be said about the statistical significance of $\rho_{XY}$:

$\rho_{XY}$ will be statistically significantly different from $0$ at the $95$ percent level if you perform a two-tailed $t$-test with $p<0.05$. The formula for $t$ in terms of $\rho_{XY}$ and $n$ is the following:

$$t = \rho_{XY}\sqrt{\frac{n-2}{1-\rho_{XY}^{2}}}$$

The numerator in the fraction is called the degree of freedom, sometimes denoted by $df$. Plugging $\rho_{XY}=0.01$ into this formula with each of the given $n$ values, we obtain the following:

a. $n=100$: $t=0.0989998995$

b. $n=1,000$: $t=0.315927177$

c. $n=10,000$: $t=0.999949994$

We can then look up the following critical $t$ values for a probability level of $0.05$ using a $t$-value calculator found here: https://www.danielsoper.com/statcalc/calculator.aspx?id=10

a. $n=100$, $df=n-2=98$. $t_{c}=+/- 1.98446745$

b. $n=1,000$, $df=n-2=998$, $t_{c}=+/- 1.98446745$

c. $n=10,000$, $df=n-2=9,998$, $t_{c}=+/- 1.96020126$

Now, comparing the three $t$ values we calculated with the $\rho_{XY}$ value of $0.01$, we see that in all three cases, $0, meaning that the $\rho_{XY}$ values is NOT statistically significantly different from $0$ in all three cases. However, if we move just one order of magnitude up, to $n=10^{5}=100,000$, we obtain the following: for $\rho_{XY}=0.01$, $t=3.16240416$ and $t_{c}=$ for $df=99,998$ $t_{c}=+/- 1.959988$ (another calculator is needed here: https://goodcalculators.com/student-t-value-calculator/). We then easily see that $t>|t_{c}|>0$, so we can conclude that the value of $\rho_{XY}$ IS statistically significantly different than $0$ using this $t$ test.

In regards to the regression coefficient, $a=\rho_{XY}\frac{\sigma_{Y}}{\sigma_{X}}$, I believe its statistical significance should correspond to that of $\rho_{XY}$, especially given that is very possible for $\sigma_{Y}=\sigma_{X}$, in which case we would actually have $a=\rho_{XY}$. While I don't have the details, I believe the reasoning for this should be from the fact that $\frac{\sigma_{Y}}{\sigma_{X}}$ is in some sense a dimensionless constant with regard to statistics and therefore the stat significance of $\rho_{XY}$ should be the same as any dimensionless multiple of it, e.g. $a=\rho_{XY}\frac{\sigma_{Y}}{\sigma_{X}}$.