0
$\begingroup$

The question asks that when the case is $X_1 = 1$ (when I am an asian instead of other ethnicity, a dummy variable), then what is the value of $Y$? As the $b_1$ has a P-value much larger than $0.05$, which indicates an insignificant result, I assume the result of $Y$ be $500$ only (Intercept: $b_0 = 500$) as $b$ should be no different from zero. However, the answer is $500 + 100(1) = 600$. May I know why my interpretation is incorrect? Thank you very much for your time.

$\endgroup$
0
$\begingroup$
 I'm sorry, your question is not entirely clear. 

I infer that the model is:

$Y = b_0 + b_1X + \epsilon$

and that the model with estimated parameters is:

$\hat Y = 500 + 100X$

The question is, if the estimate for $b_1$ is "not significant," then should we not use $0$ as the estimated value for $b_1$? The American Statistical Association says that the concept of "statistical significance" is utterly meaningless:

https://www.tandfonline.com/doi/full/10.1080/00031305.2019.1583913

This statement only amplifies a 2016 statement:

http://valjhun.fmf.uni-lj.si/~mihael/ul/ps/pdfzanimivosti/asa_pvalues.pdf

The ASA view, then, is that the state of an estimate being "statistically significant" should have no bearing on your interpretation of results. Given that supporting assumptions hold, the parameter estimate (for $b_1$, apparently, the value of $100$) is the best guess for the unknown and unknowable value of the parameter. Yes, your significance test says that the estimate is "not significantly different from $0$." But the estimate is also not significantly different from a range of other values, which are all just as plausible as $0$, against that criterion. The only value that has anything special to recommend it is the parameter estimate, again given that supporting assumptions hold.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.