Examples

One-tail test: Weighted die

Your friend gives you a single regular die, and say she is worried that it has been weighted to prefer the outcome of 2. She wants you to test it.

Design a significance test for the data of 20 rolls of the die to determine whether the die is weighted. Use significance level $\alpha =0.05$.

Solution

Let $X$ count the number of 2s that come up.

The Claim: “the die is weighted to prefer 2” The null hypothesis $H_0$: “the die is normal”

Assuming $H_0$ is true, then $X\sim \mathrm{Bin}(\mathrm{20,1}/6)$, and therefore:

$$P_{X|H_0}(k)=\left(\genfrac{}{}{0px}{}{20}{k}\right){(1/6)}^k{(5/6)}^{20-k}$$

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⚠️ Notice that “prefer 2” implies the claim is for more 2s than normal.

Therefore: Choose a one-tail rejection region.

Need $r$ such that:

$$\begin{array}{cc}P[X\ge r|H_0]& =0.05\\ & \\ ⟺P[X<r|H_0]& =0.95\end{array}$$

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Solve for $r$ by computing conditional CDF values:

$k:$	0	1	2	3	4	5	6	7
$F_{X|H_0}(k):$	0.026	0.130	0.329	0.567	0.769	0.898	0.963	0.989

Therefore, choose $r=6$:

$P[X\ge 6|H_0]<0.04$, but $P[X\ge 5|H_0]>0.05$. Final answer:

$$R=\{x|x\ge 6\}$$

Two-tail test: Circuit voltage

A boosted AC circuit is supposed to maintain an average voltage of $130\mathrm{V}$ with a standard deviation of $2.1\mathrm{V}$. Nothing else is known about the voltage distribution.

Design a two-tail test incorporating the data of 40 independent measurements to determine if the expected value of the voltage is truly $130\mathrm{V}$. Use $\alpha =0.02$.

Solution

Use $M_{40}(V)$ as the decision statistic, i.e. the sample mean of 40 measurements of $V$.

The Claim to test: $E[V]\ne 130$

The null hypothesis $H_0$: $E[V]=130$

Rejection region:

$$|M_{40}-130|\ge c$$

where $c$ is chosen so that $P[|M_{40}-130|\ge c]=0.02$

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Assuming $H_0$, we expect that:

$$E[M_{40}]=130,{\sigma }_{M_{40}}^2=\frac{{2.1}^2}{40}\approx 0.110$$

Recall Chebyshev’s inequality:

$$P[|M_{40}-130|\ge c]\le \frac{{\sigma }_{M_{40}}^2}{c^2}\approx \frac{0.110}{c^2}$$

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Now solve:

$$\frac{0.110}{c^2}=0.02\gg \gg c\approx 2.348$$

Therefore the rejection region should be:

$$M_{40}<127.65\cup 132.35<M_{40}$$

One-tail test with a Gaussian: Weight loss drug

Assume that in the background population in a specific demographic, the distribution of a person’s weight $W$ satisfies $W\sim 𝒩({\mathrm{190,24}}^2)$. Suppose that a pharmaceutical company has developed a weight-loss drug and plans to test it on a group of 64 individuals.

Design a test at the $\alpha =0.01$ significance level to determine whether the drug is effective.

Solution

Since the drug is tested on 64 individuals, we use the sample mean $M_{64}(W)$ as the decision statistic.

The Claim: “the drug is effective in reducing weight”

The null hypothesis $H_0$: “no effect: weights on the drug still follow $𝒩({\mathrm{190,24}}^2)$”

Assuming $H_0$ is true, then $W\sim 𝒩({\mathrm{190,24}}^2)$.

⚠️ One-tail test because the drug is expected to reduce weight (unidirectional). Rejection region:

$$M_{64}(W)\le r$$

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Calculate:

$${\sigma }_{M_{64}}^2\gg \gg \frac{{24}^2}{64}\gg \gg 9$$

⚠️ Standardized $M_{64}(W)$ is approximately normal!

(The standardization of $M_{64}(W)$ removes the effect of $\frac{1}{n}$. As if it’s the summation.)

So, standardize and apply CLT:

$$\begin{array}{c}\frac{M_{64}(W)-190}{\sqrt{9}}\sim 𝒩(\mathrm{0,1}),\\ \\ \gg \gg P[M_{64}(W)\le r]\approx P[Z\le \frac{r-190}{3}]=\mathrm{\Phi }\left(\frac{r-190}{3}\right)\end{array}$$

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Solve:

$$\begin{array}{c}P[M_{64}(W)\le r]=0.01\\ \\ \gg \gg \mathrm{\Phi }\left(\frac{r-190}{3}\right)=0.01\\ \\ \gg \gg \mathrm{\Phi }\left(\frac{190-r}{3}\right)=0.99\\ \\ \gg \gg \frac{190-r}{3}=2.33\\ \\ \gg \gg r=183.01\end{array}$$

Therefore, the rejection region:

$$M_{64}(W)\le 183.01$$