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If anyone is interested, this question is prompted by the recent NBA Lottery results. In the past 20 years, the team with the worst record in the NBA has won the lottery exactly 3 times. The team with the worst record every year has a 25% chance of winning the number one overall pick.

Of course, many people will talk about how the lottery is rigged and cite this statistic as an example of why. However, it doesn't seem like this is a statistically significant deviation from the expected value of 5 wins.

Does anyone know which test I should use to find out whether the worst record team winning the overall pick only 3 or fewer times in the past 20 years is statistically significantly unlikely?

Mostly interested in the process here, since I should be familiar with this. My first instinct tells me to use the binomial probability function and thus, $\sum_{x=0}^3 {n \choose x} (.25)^3 (.75)^{(20-x)}$, which yields around 22.5%, which is greater than an $\alpha$ of say, 0.05, so this isn't statistically significant.

However, I feel like there is a better way to do this involving the expected value of 5. Does anyone have a clue on what I'm looking for here?

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Your intuition is spot on. What you calculated with $$ \sum_{x = 0}^3 { 20 \choose x } (0.25)^x (1-0.25)^{20-x} = 22.5 \% $$ is just the probability of this strange event or more extreme occurring given the null hypothesis that the probability of getting the first overall pick is equal to $0.25$ for the worst team. This is by definition the p-value for this likelihood (Binomial), hypothesis (probability being $0.25$), and data ($3$ in $20$) for the one-sided extreme or half the p-value for the two-sided extreme (i.e., a p-value of $45 \%$!)

This is the exact p-value, and not the approximation that the Normal approximation to the Binomial would give you (and the approximation would be poor with a low sample size and $5$ so close to $0$). Although that approach might give you nice interpretability about how far $3$ is from the expected value of $5$ in a statistical sense.

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