Problem 19.6 Easy Free 1 variation

Error Rates of a Binomial Test

A coin is tossed $36$ times, and $X$ denotes the number of heads observed. To test the null hypothesis that the coin is fair ( $p=1/2$ ) against the alternative that it is not, you adopt the rejection region $|X-18|\geq 4$ . Find the significance level $\alpha$ of this test (the probability of rejecting $H_0$ when the coin is in fact fair), to the nearest thousandth.

Work it out, then compare with the answer.

SolutionShow the worked solution

Under $H_0$ we have $X\sim\mathrm{Bin}(36,\tfrac12)$ . The test rejects when $X\le 14$ or $X\ge 22$ , so it is simplest to compute the complement:

\alpha=\mathbb{P}(|X-18|\ge 4)=1-\mathbb{P}(15\le X\le 21)=1-2^{-36}\sum_{k=15}^{21}\binom{36}{k}.

The distribution is symmetric about $18$ . The coefficients follow from $\binom{36}{18}=9075135300$ by the ratio $\binom{36}{k-1}=\binom{36}{k}\frac{k}{37-k}$ :

\binom{36}{17}=8597496600,\qquad \binom{36}{16}=7307872110,\qquad \binom{36}{15}=5567902560 .

Hence

\sum_{k=15}^{21}\binom{36}{k}=\binom{36}{18}+2\Bigl[\binom{36}{17}+\binom{36}{16}+\binom{36}{15}\Bigr]=52021677840 .

Since $2^{36}=68719476736$ ,

\alpha=\frac{68719476736-52021677840}{2^{36}}=\frac{16697798896}{2^{36}}=\frac{1043612431}{2^{32}}\approx 0.24299 .

\boxed{\alpha=\frac{1043612431}{2^{32}}\approx 0.243}

Key idea. The significance level is the $H_0$ -probability of the rejection region. The normal approximation with continuity correction gives $2\,\mathbb{P}(Z\ge 3.5/3)\approx 0.243$ and confirms the result.

Variations

The same idea with a twist. Each solution gives only the changed step.

Variation 19.6a

A coin is tossed $36$ times, and $X$ denotes the number of heads observed. To test the null hypothesis that the coin is fair against the alternative that it is biased, you use the rejection region $|X-18|\geq 4$ . Suppose the true heads probability is actually $p=0.7$ . Find the probability $\beta$ of failing to reject $H_0$ (i.e. $|X-18|<4$ ) in this case, to the nearest ten-thousandth.

Work it out, then compare with the answer.

Solution

Now $X\sim\mathrm{Bin}(36,0.7)$ . We fail to reject exactly when $15\le X\le 21$ , so

\beta=\sum_{k=15}^{21}\binom{36}{k}(0.7)^k(0.3)^{36-k}.

The largest term is $p_{21}=\binom{36}{15}(0.7)^{21}(0.3)^{15}\approx 0.0446241$ .

Each smaller term follows from the ratio $p_{k-1}/p_k=\frac{k}{37-k}\cdot\frac37$ . Writing every term as a multiple of $p_{21}$ , the multipliers for $k=21,20,\dots,15$ are

1,\ 0.5625,\ 0.28361,\ 0.12830,\ 0.05209,\ 0.01898,\ 0.00620 .

They sum to $2.05168$ , so

\beta\approx 0.0446241\times 2.05168\approx 0.091554 .

The value $0.091554$ lies just above $0.09155$ , so it rounds up.

\boxed{\beta\approx 0.0916}

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