Homework 8

Solution.

To find best critical regions, we need to:

1. Write down the likelihood ratio $\frac{L\p{\theta_0}}{L\p{\theta_1}}$.

2. Rewrite the inequality $\frac{L\p{\theta_0}}{L\p{\theta_1}} \leq k$ in terms of $u \leq c$ or $u \geq c$, where $u = u\p{X_1, \ldots, X_n}$ is some statistic (for example, $\sum_{i=1}^n X_i$). We typically take logs at this step.

3. If given $\alpha$, you need to solve for $c$ in

   $$\P\p{u \leq c \given H_0} = \alpha \quad\text{or}\quad \P\p{u \geq c \given H_0} = \alpha$$

   using the distribution of $u$. The direction of the inequality is exactly the same as what you got in step 2.

First, let's write down the likelihood ratio.

I will use the notation $y = \sum_{i=1}^5 x_i$ for brevity. Next, we take logs on both sides of $\frac{L\p{\frac{1}{2}}}{L\p{\frac{1}{3}}} \leq k$, which gives

This means that a best critical region is of the form $Y = \sum_{i=1}^5 X_i \leq c$. Recall that a sum of $5$ independent Bernoulli trials (with the same success probability) has distribution $\operatorname{Bin}\p{5, p}$, so we need to solve

for $c$. Using the binomial cdf table, we get

The power is

Solution.

1. The likelihood is

   $$L\p{\theta} = \prod_{i=1}^n \frac{1}{\theta} e^{-\frac{x_i}{\theta}} = \frac{1}{\theta^n} e^{-\frac{1}{\theta} \sum_{i=1}^n x_i}.$$

   Thus,

   $$\frac{L\p{3}}{L\p{5}} = \p{\frac{5}{3}}^n \exp\p{\p{-\frac{1}{3} + \frac{1}{5}} \sum_{i=1}^n x_i}.$$

   We get

   $$\begin{aligned} \p{\frac{5}{3}}^n \exp\p{\p{-\frac{1}{3} + \frac{1}{5}} \sum_{i=1}^n x_i} \leq k &\iff n\ln\p{\frac{5}{3}} + \p{-\frac{1}{3} + \frac{1}{5}} \sum_{i=1}^n x_i \leq \ln k \\ &\iff \p{-\frac{1}{3} + \frac{1}{5}} \sum_{i=1}^n x_i \leq \ln k - n\ln\p{\frac{5}{3}} \\ &\iff \sum_{i=1}^n x_i \geq \frac{\ln k - n\ln\p{\frac{5}{3}}}{-\frac{1}{3} + \frac{1}{5}} = c. \end{aligned}$$

   Note that because $-\frac{1}{3} + \frac{1}{5} < 0$ that the inequality flips.

2. From part 1, we need to solve

   $$\P\p{\sum_{i=1}^n X_i \geq c \given \theta = 3} = 0.1.$$

   Here, we don't know the distribution of $\sum_{i=1}^{12} X_i$, but we do know the distribution of $\frac{2}{\theta} \sum_{i=1}^{12} X_i$, so we write

   $$\begin{gathered} \P\p{\sum_{i=1}^{12} X_i \geq c \given \theta = 3} = \P\p{\frac{2}{3} \sum_{i=1}^{12} X_i \geq \frac{2c}{3} \given \theta = 3} = 0.1 \\ \implies \P\p{\frac{2}{3} \sum_{i=1}^{12} X_i \leq \frac{2c}{3} \given \theta = 3} = 0.9. \end{gathered}$$

   When using the chi-square table, you have to read it carefully. Even though this is equal to $0.9$, we will need to use $\chi^2_{0.1}\p{24} = 33.20$, so

   $$c = \frac{3 \chi^2_{0.1}\p{24}}{2} = 49.8.$$

   Thus, a best critical region is

   $$\boxed{\sum_{i=1}^{12} X_i \geq 49.8}.$$

Solution.

1. The likelihood is

   $$\begin{aligned} L\p{\mu} &= \prod_{i=1}^n \frac{1}{\sqrt{2\pi \sigma^2}} \exp\p{-\frac{1}{2\sigma^2} \p{x_i - \mu}^2} \\ &= \frac{1}{\p{2\pi\sigma^2}^{\frac{n}{2}}} \exp\p{-\frac{1}{2\sigma^2} \sum_{i=1}^n \p{x_i - \mu}^2}. \end{aligned}$$

   To perform a likelihood ratio test, we need to optimize it, so we essentially need to find the MLE. We can optimize the log-likelihood like usual.

   $$\ln L\p{\mu} = -\frac{n}{2} \ln\p{2\pi \sigma^2} - \frac{1}{2\sigma^2} \sum_{i=1}^n \p{x_i - \mu}^2 \\ \implies \frac{\partial}{\partial \mu} \ln L\p{\mu} = \frac{1}{\sigma^2} \sum_{i=1}^n \p{x_i - \mu}.$$

   Setting it equal to $0$ and solving gives $\widehat{\mu} = \mean{x}$. Thus,

   $$\begin{aligned} \lambda &= \frac{L\p{230}}{L\p{\mean{x}}} \\ &= \exp\p{-\frac{1}{2\sigma^2} \sum_{i=1}^n \p{\p{x_i - 230}^2 - \p{x_i - \mean{x}}^2}} \\ &= \exp\p{-\frac{1}{2\sigma^2} \sum_{i=1}^n \p{x_i - 230 - \p{x_i - \mean{x}}} \p{x_i - 230 + \p{x_i - \mean{x}}}} \\ &= \exp\p{-\frac{1}{2\sigma^2} \p{\mean{x} - 230} \sum_{i=1}^n \p{2x_i - 230 - \mean{x}}} \\ &= \exp\p{-\frac{1}{2\sigma^2} \p{\mean{x} - 230} \p{2n\mean{x} - 230n - n\mean{x}}} \\ &= \exp\p{-\frac{n}{2\sigma^2} \p{\mean{x} - 230}^2}. \end{aligned}$$

   We get

   $$\begin{aligned} \lambda \leq k &\iff \exp\p{-\frac{n}{2\sigma^2} \p{\mean{x} - 230}^2} \leq k \\ &\iff -\frac{n}{2\sigma^2} \p{\mean{x} - 230}^2 \leq \ln k \\ &\iff \p{\mean{x} - 230}^2 \geq -\frac{2\sigma^2 \ln k}{n} \\ &\iff \abs{\mean{x} - 230} \geq \sqrt{-\frac{2\sigma^2 \ln k}{n}} = c'. \end{aligned}$$

   (I write $c'$ here because we're going to replace the constant one more time.)

   Note that the inequality flips and that $-\ln k \geq 0$, so the negative in the square root looks funny, but isn't actually a problem. This tells us that our critical region has the form

   $$\abs{\mean{X} - 230} \geq c'.$$

   Recall that under $H_0$,

   $$Z = \frac{\mean{X} - 230}{10/\sqrt{16}} \sim \mathcal{N}\p{0, 1}$$

   and that $\abs{\mean{X} - 230} \geq c'$ is equivalent to $\abs{Z} = \frac{\abs{\mean{X} - 230}}{10/\sqrt{16}} \geq \frac{c'}{10/\sqrt{16}} = c$. Thus, to ensure that the test has significance level $\alpha$, we need

   $$\begin{gathered} \P\p{\abs{\mean{X} - 230} \geq c'} = \P\p{\abs{Z} \geq c} = \alpha \\ \implies c = z_{\frac{\alpha}{2}}, \end{gathered}$$

   so the critical region is

   $$\boxed{\abs{Z} \geq z_{\frac{\alpha}{2}}}.$$

2. From the $z$-table, we have $z_{\frac{0.05}{2}} = 1.645$, so we reject if $\abs{z} \geq 1.645$. The observed $z$-statistic is

   $$\abs{z} = \frac{\abs{232.6 - 230}}{10/\sqrt{16}} = 1.04,$$

   so we fail to reject $H_0$ at $\alpha = 0.1$.

Solution.

1. Testing the goodness-of-fit of this model means we need to test the hypothesis

   $$H_0\colon p_i = p_{i0}, \quad p_{i0} = \P\p{\operatorname{Bin}\p{3, \frac{1}{2}} = i}, \quad 0 \leq i \leq 3.$$

   These are

   $$p_{00} = \frac{1}{8}, \quad p_{10} = \frac{3}{8}, \quad p_{20} = \frac{3}{8}, \quad p_{30} = \frac{1}{8}.$$

   There are $4$ probabilities to test, so $k = 4$. The test statistic is

   $$Q_{k-1} = Q_3 = \sum_{i=0}^3 \frac{\p{x_i - np_{i0}}^2}{np_{i0}},$$

   and we reject if $q_3 \geq \chi^2_{0.05}\p{3} = 7.815$.

2. Our observed test statistic is

   $$q_3 = \frac{\p{5 - 52 \cdot \frac{1}{8}}^2}{52 \cdot \frac{1}{8}} + \frac{\p{17 - 52 \cdot \frac{3}{8}}^2}{52 \cdot \frac{3}{8}} + \frac{\p{24 - 52 \cdot \frac{3}{8}}^2}{52 \cdot \frac{3}{8}} + \frac{\p{6 - 52 \cdot \frac{1}{8}}^2}{52 \cdot \frac{1}{8}} = 1.7436,$$

   so we fail to reject $H_0$ at $\alpha = 0.05$.

Solution.

We have $10$ probabilities to test, so $k = 10$. Thus, our test statistic is $Q_{k-1} = Q_9$ and we reject if $q_9 \geq \chi^2_{0.05}\p{9} = 16.92$. From counting the numbers in the list, our observed values are given by the following table.

For each $p_i$, the expected number of observations is $5$, so

Thus, we fail to reject $H_0$ at $\alpha = 0.05$.