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Question :

A die is marked on one side, and the number of times that the mark appears is recorded.

The mark appears once from 25 rolls, is the die biased?


working

$H_0 : $ The die is not biased

$H_1 : $ The die is biased

For this note that the die should follow $\sim Bin(n, p)$ as $\sim Bin(25, \frac{1}{6})$.

Testing at a $95\%$ significance level we need $P(x \leq 1)$, where $x = $ number of rolls with the mark atop. This is found from $P(0) + P(1)$ using:

\begin{aligned} P(0) = {25 \choose 0}\left( \frac{1}{6} \right)^{0}\left( 1 - \frac{1}{6} \right)^{25} \\ P(1) = {25 \choose 1}\left( \frac{1}{6} \right)^{1}\left( 1 - \frac{1}{6} \right)^{24} \\ \end{aligned}

Which gives

\begin{aligned} P(0) \approx 0.0104 \\ P(1) \approx 0.0524 \end{aligned}

Therefore the probability is $P(0) + P(1) = 0.0628 $, as we're testing at a $95\%$ level we have

$$ 0.0628 > 0.05 $$

Meaning that the probability of the observed outcome isn't less than $5\%$, and therefore we don't reject $H_0$.


edit

As stated the probability has been conducted for a one tailed test yet I've not been specific about whether I'm using two or one. So that I would change the hypothesis to read as:

$H_0$: The die is fair.

$H_1$: The die is biased such that there are less results of the marked side than there would be for a non biased die.

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  • 2
    $\begingroup$ Your hypothesis is two sided -- the die could be biased in either direction. Your reasoning is just fine for a one sided hypothesis. But we'd double the probability for a two sided hypothesis. (or, you could evaluate .0628 versus the .025 that's on the "low side" of the hypothesis). $\endgroup$ – zbicyclist Apr 29 '17 at 13:02
  • $\begingroup$ Note that you don't test "at the 95% significance level". See the definition of the significance level. You're testing at the 5% level. $\endgroup$ – Glen_b Apr 30 '17 at 16:14
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I'd grade you A, if you re-worded your hypo as "the die is biased against the marked side" because you're testing one sided hypo. The current wording is "the die is biased" allows for bias for the marked side, i.e. asks for a two sided hypo. For non symmetric distribution like binomial the two sided hypo is tricky.

The average is $np=25/6$, so how would you construct the tails here? I'd stick with one sided and submit it with an appropriate wording of the hypo

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  • $\begingroup$ Thanks - as said I didn't consider the two tails or word appropriately >.<. $np$ is the expected value yeah, so given that I'm not too sure how I would go about constructing the tails. I think that perhaps I'm able to something something assume symmetry and use a normal distribution? (There's certainly some blind hope there). I should consider this and perhaps make a separate post though. Cheers $\endgroup$ – baxx Apr 29 '17 at 14:13
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    $\begingroup$ @baxx But why would you assume symmetry when you can see it isn't so? How does making a plainly false assumption solve the problem resulting from the very asymmetry you're trying to assume away? $\endgroup$ – Glen_b Apr 30 '17 at 16:15
  • $\begingroup$ @Glen_b because I can't see i guess ;) Also - I thought that the binomial became symmetric under certain conditions, I wasn't sure if they were relevant here. $\endgroup$ – baxx Apr 30 '17 at 16:17
  • $\begingroup$ 1. Look up at the red parts in the diagram. See how it's not the same in each tail? That's the asymmetry in question. 2. If it were symmetric the problem Aksakal is discussing wouldn't be an issue in the first place, so there'd be nothing to solve by assuming symmetry. It's only a problem when it isn't close to symmetric. $\endgroup$ – Glen_b Apr 30 '17 at 16:20
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I think most of the reasoning is valid. Just in the end, we observe that the mark appears once, so the probability to observe it is $P(1) \approx 0.0524$. At that point, to compute $P(0)+P(1)$ is not meaningful to answer the question.

The question "the die is biased" isn't very clear, but one can assume that we want to construct a 95% confidence interval for the hypothesis, and that we take it two-sided. This means that the underlying random variable $X$ follows the distribution $Bin(25,1/6)$ under $H0$, and we want to find $x_0$ and $x_1$ such that: $P(X \leq x_0)=0.025$ and $P(X \geq x_1)=0.025$ (two-sided confidence interval), so that $P(x_0 < X < x_1)=0.95$.

Without expliciting the confidence interval, we can simply observe that $P(X=1)>0.05$, so it cannot be part of the 5% tail.

If we want to explicit the confidence interval, since the distribution $Bin(25,1/6)$ takes discrete values, I would choose $x_0 = 0$, $x_1=8$, and one can check that: $P(X \leq 0)=P(X=0) \approx 0.01$, $P(X \geq 8) \approx 0.04$.

The conclusion is the same: We cannot reject the hypothesis $H0$.

EDIT: I plot $P(X=x)$ here. With the two-sided interval hypothesis, I will not reject the hypothesis if $1 \leq x \leq 7$. enter image description here

[I also replaced 'symmetric' by 'two-sided' in my answer, because here, the binomial distribution is not symmetric.]

EDIT 2: I've just shown your edit. With this hypothesis: "The die is biased such that there are less results of the marked side than there would be for a non biased die.", then yes, your reasoning is correct!

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  • $\begingroup$ Q1 : ` to compute P(0) +P (1) is not meaningful to answer the question.` Why not? My thinking was that I have the cases of x <= 1 OR x > 1, and that I needed the probability of x <= 1 here. Are you saying that it's not meaningful logically or that it's not meaningful in the sense that the outcome of the calculation isn't affected? $\endgroup$ – baxx Apr 29 '17 at 14:33
  • $\begingroup$ Q2: we want to construct a 95% confidence interval for the hypothesis, and that we take it symmetric To be honest I didn't consider the tails for this test (as pointed out by others). I'm wondering what the justification for symmetry is here? Is it that we have $n = 25$? $\endgroup$ – baxx Apr 29 '17 at 14:33
  • $\begingroup$ Q3: I would choose x_1 = 8, I'm not sure what the reasoning is for this value? I mean, why $x_1$ was chosen such that $x_1 = 8$, is it just to demonstrate that we could have had the marked side appear 8 times and still not reject the hypothesis (therefore we're well within the required domain)? ... thanks $\endgroup$ – baxx Apr 29 '17 at 14:34
  • $\begingroup$ Q1: For example, if the mark appears 12 times, you would have $P(0)+...+P(12) \approx 0.9999464 > 0.05$, and you will conclude that you cannot reject $H0$. But in this case (the mark appears 12 times), the die is biased (you can try to prove it). $\endgroup$ – ahstat Apr 30 '17 at 5:37
  • $\begingroup$ Q2: In general, there are three choices to construct a confidence interval, depending of the assumptions on the problem. You can take: an unilateral test in one direction (you reject if $x>x_1$ where $P(X \geq x_1)=0.05$; An unilated test in the other direction (you reject if $x < x_0$ where $P(X \leq x_0)=0.05$; A two-sided test (what I wrote in my answer before). For the unilateral test in one direction, we only want to check if there is a bias making the mark appearing more times than usual. For the unilateral test in the other direction, check if the mark appearing less times than usual. $\endgroup$ – ahstat Apr 30 '17 at 5:45

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