e*******a
发帖数: 514 | 1 以前无聊的时候琢磨过这个事情。
看了点数据后发现单场最高分总是出现职业高峰期，就是赛季平均分最高的那几年，并
且是那几个赛季平均分的2倍左右。
不知道有谁看过相关方面的统计研究没？ | p*********e
发帖数: 32207 | 2 贴点儿更详细的数据吧,这个研究挺有意思的,我这儿准备了一堆包子:D
【在 e*******a 的大作中提到】
: 以前无聊的时候琢磨过这个事情。
: 看了点数据后发现单场最高分总是出现职业高峰期，就是赛季平均分最高的那几年，并
: 且是那几个赛季平均分的2倍左右。
: 不知道有谁看过相关方面的统计研究没？
| e*******a
发帖数: 514 | 3 帮忙介绍个懂统计的朋友吧，我是受人歧视的文科生。。。
【在 p*********e 的大作中提到】
: 贴点儿更详细的数据吧,这个研究挺有意思的,我这儿准备了一堆包子:D
| x*d
发帖数: 1696 | 4 Without using raw data, it’s hard to directly support your hypothesis. But
an easier way is to build the statistical model assuming your hypothesis is
true, then test it against other data.
The simplest model: For single season, taking any player who played 80 games
in a season. The probability of the highest score is 1/80 (one sided). Set
mean Xb (the average), standard deviation Si, highest score Xh, then Prob((
Xh-Hb)/Si) ~ 1/80. Obtain Si = 0.446Xb, i.e. the deviation of a players
season score should be about half of the season average. So if your
hypothesis is true and if the model is correct, the following should also be
true:
(1) For any player, in one season, 80% (64 out of 80 games) scores should
be between 0.5 average and 1.5 of the average
(2) Can be extended to n seasons. For example, if a player played 10
years, 800 games, assuming the player variance is about the same, his 10-
year career average should be 82% of his best season score, from calculation
Prob((Xb – Xb_career)/squareroot(0.446Xb^2/10)) = 1/10.
(3) Can check how stable a player plays, say, Kobe in 2004 reached 81,
but the probability was 0.00020943. One thing to notice is that Kobe’s
score variance was higher in those years.
【在 e*******a 的大作中提到】
: 帮忙介绍个懂统计的朋友吧，我是受人歧视的文科生。。。
| c**********6
发帖数: 1173 | 5 kobe的81分怎么可能在04年呢
But
is
games
Set
((
be
【在 x*d 的大作中提到】
: Without using raw data, it’s hard to directly support your hypothesis. But
: an easier way is to build the statistical model assuming your hypothesis is
: true, then test it against other data.
: The simplest model: For single season, taking any player who played 80 games
: in a season. The probability of the highest score is 1/80 (one sided). Set
: mean Xb (the average), standard deviation Si, highest score Xh, then Prob((
: Xh-Hb)/Si) ~ 1/80. Obtain Si = 0.446Xb, i.e. the deviation of a players
: season score should be about half of the season average. So if your
: hypothesis is true and if the model is correct, the following should also be
: true:
|
|
