In the following linear regression

$y_{t} = \beta_0 + \beta_1 x_{t} + \epsilon_t,$

I want to test whether $R^2 \geq 0.5$. In R, you can make use the CI.Rsq() function in the psychometric package. The formula it applies is:

$SE_{R^{2}} = \sqrt{\frac{4R^{2}(1-R^{2})^{2}(n-k-1)^{2}}{(n^2 - 1)(n+3)}}$

Then, the 95% CI is $R^{2} \pm 2 \cdot SE_{R^{2}}$.

Does this CI work for time series data?

  • 1
    $\begingroup$ You could look at the confidence intervals on the R2: stats.stackexchange.com/questions/175026/… $\endgroup$
    – mkt
    Aug 15 '17 at 14:05
  • $\begingroup$ As far as I understand, these confidence intervals are constructed under the null hypothesis that $c = 0$. In my case, under the null $c \neq 0$. $\endgroup$
    – user27808
    Aug 15 '17 at 16:48
  • $\begingroup$ There is nothing in the answers there that assumes c = 0, or that you can only use the resulting CIs to test some c other than 0. $\endgroup$ Aug 16 '17 at 16:13
  • 1
    $\begingroup$ I think this question is worth reopening, since it is now asking about whether a specific formula applies to time series data (an issue not covered in the other thread), and even for non-time-series data the formula was itself the subject of some discussion in the comments on the other thread about whether it really "worked". I'd far rather see this as a new question than as a discussion in the comments thread of an answer to another question. $\endgroup$
    – Silverfish
    Aug 17 '17 at 10:12
  • 3
    $\begingroup$ Simulations suggest the quoted formula is usually wrong even for independent data. For large $R^2$ it gives intervals with far too much coverage, even when $n$ is large. (I tested with bivariate normal data--for which this formula ought to work, if it works at all--up to $n=10000$ with $R^2$ up to $0.99$. E.g., with $R^2=0.8$ the interval is more than $40\%$ too wide.) $\endgroup$
    – whuber
    Aug 17 '17 at 13:31

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