---
title: Stat09
tags: Stat
---

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# Chapter 9 Large-sample Tests of Hypotheses

## 9.3 Two population means

### Key ingredients (B1)

- Model: $X_{1i}\stackrel{i.i.d.}{\sim} F(\mu_1,\sigma_1^2)$ for $i=1,\ldots,n_1$ and
$X_{2j}\stackrel{i.i.d.}{\sim} F(\mu_2,\sigma_2^2)$ for $i=1,\ldots,n_2$.
- We are interested in understanding $(\mu_1-\mu_2)$.
- We use the difference between two sample means $(\bar{X}_1-\bar{X}_2)$ to estimate $(\mu_1-\mu_2)$, where
$\bar{X}_1=\frac{1}{n_1}\sum_{i=1}^{n_1}X_{1i}$ and $\bar{X}_2=\frac{1}{n_2}\sum_{j=1}^{n_2}X_{2j}$.
- With the CLT, we have $$\frac{(\bar{X}_1-\bar{X}_2)-(\mu_1-\mu_2)}{\sqrt{ \frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}\stackrel{d}{\rightarrow}Z$$
- When $\sigma_1$ and $\sigma_2$ are unknown, we use the sample variances,
$$s_1^2 = \frac{1}{n_1-1}\sum_{i=1}^{n_1}(X_{1i}-\bar{X}_1)^2, \quad s_2^2 = \frac{1}{n_2-1}\sum_{j=1}^{n_2}(X_{2j}-\bar{X}_2)^2. $$
- With the advanced CLT (Slutsky's theorem), we have $$\frac{(\bar{X}_1-\bar{X}_2)-(\mu_1-\mu_2)}{\sqrt{ \frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\stackrel{d}{\rightarrow}Z.$$

### $H_0: \mu_1-\mu_2 = D_0$

The test statistic and its sampling distribution is
$$Z_{STAT} = \frac{(\bar{X}_1-\bar{X}_2)-D_0}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\stackrel{d}{\rightarrow}Z$$

### Example: $H_0$: $\mu_1-\mu_2=0$ vs $H_a$: $\mu_1-\mu_2\neq 0$.


## 9.4 Population proportion

### Key ingredients (A2)

- Model: $X {\sim} Binomial(n,p)$
- We use sample proportion $\hat{p} = \frac{X}{n}$ to estimate the population proportion $p$.
- With the CLT, we have
$$\frac{\hat{p}-p}{\sqrt{{p}(1-{p})/n}}\stackrel{d}{\rightarrow}Z. $$

### $H_0: p = p_0$

For $H_0: p = p_0$, the test statistic and its sampling distribution is
$$Z_{STAT}=\frac{\hat{p}-p_0}{\sqrt{p_0(1-p_0)/{n}}}\stackrel{d}{\rightarrow}Z$$

### The two-sided test
1. $H_0: p = p_0$ versus $H_a: p \neq p_0$.
2. Set up $\alpha$
3. $Z_{STAT} = \frac{\hat{p}-p_0}{\sqrt{p_0(1-p_0)/n}}\stackrel{d}{\rightarrow}Z$.
4. Calculate the realized statistic $Z^*$ from the data.
5. Find (a) the rejection region = $\{z: z<-z_{\alpha/2}\;,z>z_{\alpha/2}\}$ or (b) the $p$-value= $2*P(Z>|Z^*|)$.
6. Conclude.

### The left-sided test

1. One of the following:
- $H_0:p = p_0$ versus $H_a: p < p_0$
- $H_0:p \geq p_0$ versus $H_a: p < p_0$
- $H_0:p > p_0$ versus $H_a: p \leq p_0$
2. Set up $\alpha$.
3. $Z_{STAT} = \frac{\hat{p}-p_0}{\sqrt{p_0(1-p_0)/n}}\stackrel{d}{\rightarrow}Z$.
4. Calculate the realized statistic $Z^*$ from the data.
5. Find either (a) the rejection region = $\{z: z<-z_{\alpha}\}$ or (b) $p$-value = $P(Z<Z^*)$.
6. Conclude.

### The right-sided test

1. One of the following:
- $H_0: p = p_0$ versus $H_a: p > p_0$
- $H_0:p \leq p_0$ versus $H_a: p > p_0$
- $H_0:p < p_0$ versus $H_a: p \geq p_0$
2. Set up $\alpha$
3. $Z_{STAT} = \frac{\hat{p}-p_0}{\sqrt{p_0(1-p_0)/n}}\stackrel{d}{\rightarrow}Z$.
4. Calculate the realized statistic $Z^*$ from the data.
5. Find (a) the rejection region = $\{z: z>z_{\alpha}\}$ or (b) the $p$-value = $P(Z>Z^*)$.
6. Conclude.

### Example: $H_0$: $p=0.2$ vs $H_a$: $p< 0.2$.

Bernoulli(p=0.20)


## 9.5 Two population proportions

### Key ingredients (B2)

- Model: $X_{1}{\sim} Binomial(n_1,p_1)$ and $X_{2}{\sim} Bernoulli(n_2, p_2)$.
- We are interested in understanding $(p_1-p_2)$.
- Define
$$\hat{p}_1=\frac{X_1}{n_1}, \quad \hat{p}_2=\frac{X_2}{n_2}.$$We use $(\hat{p}_1-\hat{p}_2)$ to estimate $(p_1-p_2)$.
- With the CLT, we have $$\frac{(\hat{p}_1-\hat{p}_2)-(p_1-p_2)}{\sqrt{ \frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2}}}\stackrel{d}{\rightarrow}Z$$
- When $p_1$ and $p_2$ are unknown, assuming $H_0$ is true,i.e., $p_1=p_2=p$, we define $$\hat{p}=\frac{X_1+X_2}{n_1+n_2}$$ as an pool estimate of $p$.
- Apply the advanced CLT (Slutsky's theorem), we have $$\frac{(\hat{p}_1-\hat{p}_2)-(p_1-p_2)}{\sqrt{\hat{p}(1-\hat{p})( \frac{1}{n_1}+\frac{1}{n_2})}}\stackrel{d}{\rightarrow}Z.$$

### $H_0: p_1-p_2 = 0$

For $H_0: p_1-p_2 = 0$, the test statistic and its sampling distribution is
$$Z_{STAT}=\frac{(\hat{p}_1-\hat{p}_2)-0}{\sqrt{\hat{p}(1-\hat{p})( \frac{1}{n_1}+\frac{1}{n_2})}}\stackrel{d}{\rightarrow}Z,$$
where $\hat{p}_1=\frac{X_1}{n_1}$, $\hat{p}_2=\frac{X_2}{n_2}$, and $\hat{p}=\frac{X_1+X_2}{n_1+n_2}$.

### Exemple: $H_0$: $p_1-p_2=0$ vs $H_a$: $p_1-p_2\neq 0$.