2
$\begingroup$

The practical POV:

I sample 35 matches: 15 turn out as home win, 10 as tie and 10 as home lose. I want to conduct a test for the hypothesis "A home court advantage does exist." with a significance level of say 5%. What would be a reasonable test statistic / calculation of the p-value to accept or reject the hypothesis?

Clarification:

The matches I sample are independent of a specific team. I just take a number of matches and count for the home teams - wins, ties and defeats. Due to symmetry the guest teams will have deterministically corresponding results and hence can be ignored.


The generalized POV:

$A_1$ be the random variable for the number of matches the local team won in a total sample of $N$ matches. Naturally this is a bernoulli experiment (given we only assume the possibility of win and lose) and the p-value of the hypothesis "a home court advantage exists - i.e. the probability of a home win $p_1$ is larger than 0.5." given $a_1$ observed wins is:

$$\Pr[A_1 >= a_1 | p_1 = 0.5]=\sum_{k={a_1}}^N{N\choose{k}}(0.5)^k$$

So far so good (though please correct me if I am wrong or my wording is problematic) - now the aspect that puzzles me is how to factor in ties.

My idea would be to create a test statistic $\tilde{A}_1 := A_1 + \lfloor{A_0/2}\rfloor$ with $A_1$ and $A_0$ being the random variables for number of home wins and ties in $N$ matches.

$$\Pr[\tilde{A}_1 >= \tilde{a}_1 | \tilde{p}_1 = 0.5]=\sum_{k={a_1}}^N{N\choose{k}}(0.5)^k$$

Now in my mind this makes sense somehow because it reflects the indecisive and symmetric nature of a tie.

On the other hand I am not sure how to bend the bernoulli experiment to reflect the notion of $\tilde{p}_1$ being the probability of a home win in the second model (which means home win or "50% of a tie") while keeping up the pretence of formal correctness.

Does this make sense?

$\endgroup$
  • $\begingroup$ Some comments: 1) I find your use of $H_0$ and $H_1$ as denoting random variables to be very confusing 2) I'm curious why you have chosen this method to test for home court advantage rather than e.g. regression which could account for varying team strengths - this would seem critical to correctly quantify home advantage $\endgroup$ – M. Berk Nov 4 '13 at 9:53
  • $\begingroup$ 1) Good point - changed to A,a instead of H,h. 2) When significantly more matches are won by the respective home team - then this is already a valid result about whether or not this effect exists. Of course one can always choose a more complex way to factor in further aspects. Also the question addresses a theoretical problem not just the practical question. $\endgroup$ – Raffael Nov 4 '13 at 10:00
  • $\begingroup$ Are some teams appearing more than once in the data? $\endgroup$ – Michael M Dec 6 '13 at 17:24
1
$\begingroup$

First, the p-value will not tell you whether a home court advantage exists or not. It will only tell you the probability of the data given a hypothesis. Second, the data you have will not tell you anything about a home court advantage, because you have nothing to compare it to. What is the proportion of games won away or what is the proportion of games won overall? The only thing you try to find out is whether the proportion of games won is statistically significant different from .5.

Now if you have data on games played away, then you can set up your null hypothesis as:

$H_{0}: \; p_{home} = p_{away}$

where, $p_{home}=\frac{won_{home}}{total_{home}}$ and $p_{away}=\frac{won_{away}}{total_{away}}$ - if you want you can write a tie as a half win or exclude the ties all together - then what you have is categorical variable $home$ which can take a value of $1$ or $0$: $1$ for home and $0$ for away and two dependent proportions. For this problem you can then use the McNemar test or the paired t-test if your sample is large enough (see here: http://www.ats.ucla.edu/stat/stata/whatstat).

Hope this helps.

$\endgroup$
  • $\begingroup$ I did not write that the p-value tells me whether this effect exists. The hypothesis is "[effect] exists. $\endgroup$ – Raffael Nov 6 '13 at 13:06
  • $\begingroup$ The number of home wins is the number of outside defeats ... $\endgroup$ – Raffael Nov 6 '13 at 13:08
  • $\begingroup$ Actually you don't seem to grasp the question ... you cannot just say without justifying it - leave out ties or count them as .5 win - because the caculated p-value determining the significance changes with the depending calculation - and that is the core of the question. $\endgroup$ – Raffael Nov 6 '13 at 13:10
  • $\begingroup$ Raffael, it does not matter what you do with the ties. Either you count a win, loss, tie as 1,-1, and 0, or you exclude the ties, or you count the ties as .5 and win and loss as 1 or 0, the only thing that can change is the size of the effect and with that the power of the test. What you are interested in is whether the proportion of wins differs from the proportion of losses. If the proportion of wins at home is the proportion of losses away, then you can just test it against .5. It will however not tell you whether a home court advantage exists: that's not what you test. $\endgroup$ – Martin Nov 6 '13 at 13:25
  • 1
    $\begingroup$ Why does it not test the hypothesis "a HCA exists"? It tests whether there is an imbalance towards home wins which is significant - meaning assuming a random model with no imbalance will yield the observed result or extremer with probability lower than a chosen level - which leads to acceptance or rejection of the null-hypo ... this seems pretty straightforward to me. $\endgroup$ – Raffael Nov 6 '13 at 13:54

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.