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Refer to the REG procedure output:
An analyst has selected this model as a champion because it shows better model fit than a competing model with more predictors.
Which statistic justifies this rationale?
Suggested Answer: C
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26 comments · newest first
LE
Lenny
4 months ago
Agreed! It’s impressive when a simpler model outperforms a complex one.
VI
Virgina
4 months ago
I feel confident about that too. It shows efficiency in the model.
DE
Denna
4 months ago
Yes, exactly! A higher Adjusted R-squared means better fit with fewer predictors.
JA
Jaleesa
4 months ago
I think it's the Adjusted R-squared. It accounts for the number of predictors.
GE
Gerald
6 months ago
Don't forget about the BIC, it also supports the champion choice!
CH
Charolette
6 months ago
Surprised that fewer predictors can outperform a complex model!
IN
In
6 months ago
But wait, isn't it risky to choose a model with fewer predictors?
JE
Jeannine
6 months ago
Totally agree, simpler models can be better!
FR
Francis
6 months ago
The AIC value is lower for the champion model.
CE
Celia
6 months ago
The F-statistic, obviously. Can't beat that good old-fashioned significance test.
PE
Pearline
7 months ago
I'm just here for the jokes. This exam is making me sleepy already!
AN
Anisha
7 months ago
Nah, the Bayesian Information Criterion (BIC) is the way to go. Gotta love those information criteria!
HE
Herschel
7 months ago
Hmm, I'd go with the Akaike Information Criterion (AIC) on this one. Lower is better, right?
AL
Alease
7 months ago
The R-squared value looks pretty good, that's gotta be the one.
PA
Paris
7 months ago
I recall that the likelihood ratio test might be useful for comparing models, but I’m not confident if that’s the right approach for this specific question.
FR
Francesco
7 months ago
I practiced a question like this where we had to choose between models based on their fit statistics. I feel like the F-statistic could also play a role in justifying the champion model.
DA
Dallas
8 months ago
I'm not entirely sure, but I remember something about AIC or BIC being used to compare model fit. Could that be relevant here?
MA
Marguerita
8 months ago
I think the statistic we're looking for might be the adjusted R-squared, since it accounts for the number of predictors in the model.
SH
Sherita
9 months ago
The R-squared is high, but I'm not sure if that's the best way to compare these two models. I'll need to look at the other fit statistics, like the F-test and the significance of the individual predictors, to really justify why this model is the better choice.
KA
Kaitlyn
9 months ago
The R-squared is definitely the standout statistic here. If the competing model has more predictors but a lower R-squared, that's a pretty clear indication that this model is the better fit. I'll focus on explaining that in my answer.
HE
Hector
9 months ago
Okay, the R-squared is high, so that seems like a good sign. But I'm a little confused about how to interpret the other statistics in the output. I'll need to refresh my memory on which ones are most relevant for model selection.
JE
Jerry
9 months ago
The R-squared is definitely high, but I'm wondering if there are other important factors to consider, like the significance of the predictors or the overall model F-test. I'll need to dig into the details to really understand which statistic is the best justification.
HE
Heike
9 months ago
Hmm, the R-squared value looks promising, but I'm not sure if that's the best statistic to focus on here. I'll need to review the other model fit statistics to make a more informed decision.
CA
Caprice
3 months ago
Don't forget about AIC and BIC; they can help compare models too!
RO
Royce
3 months ago
True, adjusted R-squared accounts for the number of predictors.
BE
Ben
3 months ago
I agree, R-squared is important, but we should also check adjusted R-squared.