Homework 7

Solution.

1. Recall that the Wilcoxon signed rank statistic is

   $$W = \sum_{i=1}^n \operatorname{sign}\p{x_i - m} \operatorname{Rank}\p{\abs{x_i - m}}$$

   and we use it with

   $$Z = \frac{W}{\sqrt{n\p{n+1}\p{2n+1}/6}} \approx \mathcal{N}\p{0, 1}.$$

   For a one-sided test, we reject if $z < -z_{0.05} = -1.645$. Computing the signs and ranks gives the following table:

   $$\begin{array}{r|rrrrrrrrrrrrr} x_i & 41195 & 39485 & 38050 & 40890 & 39245 & 31031 & 30906 & 41229 & 36840 & 38345 & 34930 & 40780 & 38050 \\ \abs{x_i - m} & 1195 & 515 & 1950 & 890 & 755 & 8969 & 9094 & 1229 & 3160 & 1655 & 5070 & 780 & 1950 \\ \operatorname{sign}\p{x_i} \operatorname{Rank}\p{\abs{x_i - m}} & 5 & -1 & -8 & 4 & -2 & -12 & -13 & 6 & -10 & -7 & -11 & 3 & -9 \end{array}$$

   Summing the last row, we get $W = -55$, so

   $$z = \frac{-55}{\sqrt{\p{13 \cdot 14 \cdot 27} / 6}} = −1.9219$$

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

2. The sign test statistic is given by

   $$Y = \p{\#\ i \text{'s such that } X_i < m} \sim \operatorname{Bin}\p{n, \frac{1}{2}}.$$

   There are $9$ of these, so our $p$-value is

   $$p\text{-value} = \P\p{Y \geq 9} = \sum_{k=9}^{13} \binom{13}{k} \frac{1}{2^{13}} = 0.1334,$$

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

Python Code

x = np.array([41195, 39485, 38050, 40890, 39245, 31031, 30906, 41229, 36840, 38345, 34930, 40780, 38050])
m = 40000
sign = np.sign(x - m)
rank = np.argsort(np.argsort(np.abs(x - m))) + 1
print(f"w = {np.sum(sign * rank)}") # w = -55
print(f"y = {np.sum(sign < 0)}")    # y = 9

Solution.

1. Recall that

   $$K\p{\mu} = \P\p{\text{reject }H_0 \given \mu}.$$

   In this problem, we reject $H_0$ if and only if $\mean{X}$ lands in the critical region, so

   $$K\p{\mu} = \P\p{\mean{X} \leq 354.05 \mid \mu}.$$

   To calculate this probability, we need to carefully figure out the distribution of $\mean{X}$. Recall that

   $$Z = \frac{\mean{X} - \mu}{\sigma/\sqrt{n}} = \frac{\mean{X} - \mu}{2/\sqrt{12}} \sim \mathcal{N}\p{0, 1},$$

   so

   $$\begin{aligned} \mean{X} \leq 354.05 &\iff \frac{\mean{X} - \mu}{2/\sqrt{12}} \leq \frac{354.05 - \mu}{2/\sqrt{12}}. \end{aligned}$$

   We get

   $$\begin{aligned} K\p{\mu} &= \P\p{\frac{\mean{X} - \mu}{2/\sqrt{12}} \leq \frac{354.05 - \mu}{2/\sqrt{12}} \given \mu} \\ &= \P\p{\mathcal{N}\p{0, 1} \leq \frac{354.05 - \mu}{2/\sqrt{12}}} \\ &= \boxed{\Phi\p{\frac{354.05 - \mu}{2/\sqrt{12}}}}, \end{aligned}$$

   where $\Phi$ is the cdf of a standard normal random variable.

2. We have

   $$\begin{aligned} \alpha &= \P\p{\text{reject }H_0 \given H_0} \\ &= K\p{355} \\ &= \Phi\p{\frac{354.05 - 355}{2/\sqrt{12}}} \\ &= \Phi\p{−1.6454} \\ &\approx -z_{0.05}, \end{aligned}$$

   so the significance level is roughly $\alpha = 0.05$.

3. Since the critical region is given to us, we do not use the $p$-value to make our conclusion. We just check to see if $\mean{x} \in C$ or not.

   $$\mean{x} = 353.8333 \leq 354.05,$$

   so we reject $H_0$.

Solution.

Recall that for one proportion, we use the null hypothesis when computing the standard error:

For our given data, we get

1. The critical region is $z < -z_{0.05} = -1.645$, so we fail to reject $H_0$.

2. The critical region is smaller than in part 1, so we still fail to reject $H_0$.

3. The $p$-value is (taking the average of $\Phi\p{1.12}$ and $\Phi\p{1.13}$ from the $z$-table)

   $$p\text{-value} = \P\p{Z \leq -1.125} = 1 - \P\p{Z \leq 1.125} = 1 - 0.8697 = \boxed{0.1303}.$$

4. The smallest significance level for which we can reject $H_0$ in this test is

   $$\alpha = p\text{-value} = \boxed{0.1303}.$$