Tuesday, July 26, 2011

Is Ryan Howard A Clutch Homerun Hitter?

This was inspired by a discussion at Baseball Think Factory about an article at Fangraphs called Ryan Howard and the RBI by Steve Slowinski (there is no shortage of great Polish baseball bloggers).

Some people say that Howard is a good "RBI guy" and that he is paid to drive in runs so maybe his overall stats and/or his OBP don't matter that much. But a good "RBI guy" would tend to hit better with runners on. In my opinion that means a higher AVG with runners in scoring position (RISP) than he normally gets and a higher SLG with runners on base (ROB) than normal

But this year Howard is only 47th in SLG (.463) with runners on base in all of baseball with guys with 150+ PA in that situation. Teixeira leads with .667.

Howard is 20th in AVG (.304) with runners in scoring position this year in all of baseball with guys with 100+ PA in that situation. Votto leads with .424.

So overall, nothing special on Howard's part.

But somehow Howard over his career has managed to hit better with ROB and RISP. His AVG/SLG with none on is .266/.528. With ROB it is .285/.593 and .281/.561 with RISP. The one difference that seems pretty big is the SLG with ROB.

From 1991-2000, ROB SLG was .422 in all of baseball and .411 with none on. Since 2006, it looks like SLG with ROB is about .009 higher than with none on. So Howard is way above that being .063 higher. Some of that might be because he is a lefty. But whether it is luck or some real clutch talent is hard to say.

So I thought I would try a statistical test on his HR% with none on (NONE) vs. his HR% with ROB. Here we calculate something called a "Z-score." To be significant at the 5% level (meaning there is less than a 5% chance of getting the difference between the two HR%s) the Z-score has to be at least 1.96 (plus or minus, since a guy could do worse with ROB, the clutch situation I am looking at). HR% will be highly correlated with SLG (also, it does not look like Howard hits very many more 2Bs or 3Bs with ROB than NONE). Some technical details on this are below.

Howard's HR% with NONE in his career is 6.857% while with ROB it is 8.221%. That may not seem like a big difference, but the ROB HR% is 19.9% higher (8.221/6.857 = 1.199). The question is whether or not it is statistically significant. This is where the Z-score comes in.

The Z-score takes into account the number of ABs and HR% in each situation (ROB, NONE). It also takes into account the normal major league difference (I used +.0009, the approximate difference from 2007-10). This is important because if the normal HR% was .01364, then Howard would not be clutch at all since this is the difference between his two percentages (.08221 - .06857). When I did this calculation for Howard, I got a Z-score of 1.655. That is significant at about the 10% level. I guess I would like to see it at the 5% level before considering that he is clutch in hitting HRs with ROB. But even if he had reached 1.96 or more in his Z-score, it is possible he got there by luck and not skill because 2.5% of the hitters will have a Z-score of +1.96 or more.

Z = (CLUTCH AVG – NONCLUTCH AVG + EXPECTED DIFFERENCE)/SD

NORMAL DIFFERENCE FOR CL = +.0009 (which is added in this case)

SD = STANDARD DEVIATION =

{[CLUTCH AVG*(1-CLUTCH AVG)]/CLUTCH AB + [NONCLUTCHAVG*(1-NONCLUTCH AVG)]/NONCLUTCH AB}

(FROM PETE PALMER, BY THE NUMBERS 3/90. He called it a “pooled” standard deviation)

In a normal distribution, 5% of the players would have a Z-score of at least +1.96 or less than –1.96.

2 comments:

JP said...

As I was reading this Howard hit a go ahead HR with a man on first. Ok...back to finding out if the z-score is statistically significant.

Cyril Morong said...

JP

Thanks for dropping by and commenting. As I read your comment I was listening to Jazz (91.7 FM, San Antonio).

So, I should keep updating my data everytime there is a new event? What do you think I am, some kind of Bayesian?

Cy