I have the following estimated model: $\hat{y} = 0.2857 + 0.8019x_1 - 0.0741x_2$ (the $t$-statistics are $1.8959$, $8.4198$, and $-3.7017$, respectively).

Furthermore, I know the sample size $N = 92$, the sum of squared residuals (SSR) [sum of squared errors (SSE)] $SSR = \sum_{i=1}^N[u_i^2]=39.3601$, and the (average sample) standard deviation of the dependent variable $\hat{\sigma}_y = 0.8861$.

Lastly, in an earlier question I have (correctly) calculated the standard error of the regression:

• Mean of Squared Residuals (MSR) [Mean Squared Error (MSE)]: $MSR = \frac{1}{89} \cdot 39.3601 \approx 0.4422$
• Standard Error of the Regression (Root MSR [Root MSE] / Residual Standard Error): $SE_R = \sqrt{0.4422} \approx 0.6650$

Central Question: how to calculate $R^2$? (The answer should be (approximately) $0.4491$)

This answer (https://math.stackexchange.com/questions/834681/when-residual-standard-error-is-equal-to-standard-deviation-of-dependent-variabl) suggests using the formula $SE_R=(1-R^2)\hat{\sigma}_y$, but using it yields $R^2 \approx 0.4995$.

Update 1:

Using https://people.duke.edu/~rnau/mathreg.htm, that is, the formulas $SE_R = \sqrt{1 - \bar{R}^2} \cdot \hat{\sigma}_y$ and $\bar{R}^2 = 1 - \frac{N-1}{N-k-1}(1 - R^2)$ I do arrive at $R^2 \approx 0.4491$, but it seems to me that this result can be reached more easily (that is, without computing $\bar{R}^2$), but how?

Update 2:

Yes, indeed, there is. We already know that $SSR = 39.3601$, so in order to compute $R^2$ using the simple formula $R^2 = 1 - \frac{SSR}{SST}$ we only have to determine $SST$.

We have that $\hat{\sigma}_y = 0.8861$ (average sample standard deviation (of the dependent variable)), so $MST = \hat{\sigma}_y^2 \approx 0.7852$ (average sample variance) and it then follows that $SST = 91 \cdot MST \approx 39.3601$ (total sample variance) and ultimately that $R^2 \approx 0.4491$ (which was the correct answer).

Remaining question:

Can the formula $SE_R=(1-R^2)\hat{\sigma}_y$ also be used to calculate $R^2$? How?