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# 统计代写|统计推断代考Statistical Inference代写|MS-C1620 Population variance σ2

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## 统计代写|统计推断代考Statistical Inference代写|Population variance σ2

Setting: Suppose $Y_1, Y_2, \ldots, Y_n$ is an iid sample from a $\mathcal{N}\left(\mu, \sigma^2\right)$ population distribution, where both $\mu$ and $\sigma^2$ are unknown. The goal is to construct a level $\alpha$ test for
$$\begin{gathered} H_0: \sigma^2=\sigma_0^2 \ \text { versus } \ H_a: \sigma^2 \neq \sigma_0^2 . \end{gathered}$$

When $H_0$ is true, we know
$$T=\frac{(n-1) S^2}{\sigma_0^2} \sim \chi^2(n-1) .$$
Therefore, a level $\alpha$ test uses the rejection region
$$\mathrm{RR}=\left{t<\chi_{n-1,1-\alpha / 2}^2 \text { or } t>\chi_{n-1, \alpha / 2}^2\right},$$
where
\begin{aligned} \chi_{n-1,1-\alpha / 2}^2 & =\text { lower } \alpha / 2 \text { quantile of } \chi^2(n-1) \ \chi_{n-1, \alpha / 2}^2 & =\text { upper } \alpha / 2 \text { quantile of } \chi^2(n-1) \end{aligned}
see Figure $10.14$ (above).

## 统计代写|统计推断代考Statistical Inference代写|Difference of two population means µ1 − µ2 (independent samples)

Setting: Suppose we have two independent random samples:

$Y_{11}, Y_{12}, \ldots, Y_{1 n_1}$ is an iid sample from a $\mathcal{N}\left(\mu_1, \sigma_1^2\right)$ population distribution

$Y_{21}, Y_{22}, \ldots, Y_{2 n_2}$ is an iid sample from a $\mathcal{N}\left(\mu_2, \sigma_2^2\right)$ population distribution,

where all population parameters are unknown. The goal is to construct a level $\alpha$ test for
$$\begin{gathered} H_0: \mu_1-\mu_2=d_0 \ \text { versus } \ H_a: \mu_1-\mu_2 \neq d_0, \end{gathered}$$
where $d_0$ is a pre-specified difference of the population means. In practice, one often takes $d_0=0$ because the goal is to test whether or not the population means are equal. This inference procedure is called a two-sample $t$ test.

Case 1: $\sigma_1^2=\sigma_2^2=\sigma^2$; i.e., the population variances are equal. When $H_0$ is true, we know
$$T=\frac{\left(\bar{Y}{1+}-\bar{Y}{2+}\right)-d_0}{S_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \sim t\left(n_1+n_2-2\right)$$
where
$$S_p^2=\frac{\left(n_1-1\right) S_1^2+\left(n_2-1\right) S_2^2}{n_1+n_2-2}$$
is the pooled sample variance estimator. Therefore, a level $\alpha$ test uses the rejection region
$$\mathrm{RR}=\left{t<-t_{n_1+n_2-2, \alpha / 2} \text { or } t>t_{n_1+n_2-2, \alpha / 2}\right}=\left{|t|>t_{n_1+n_2-2, \alpha / 2}\right} .$$
Case 2: $\sigma_1^2 \neq \sigma_2^2$; i.e., the population variances are unequal. Under this assumption, we have few options. The reason is that there is no test statistic available for which the exact sampling distribution is known under $H_0$. When $H_0$ is true, it can be shown that
$$T=\frac{\left(\bar{Y}{1+}-\bar{Y}{2+}\right)-d_0}{\sqrt{\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2}}}$$
follows an approximate $t(\nu)$ distribution, where
$$\nu=\frac{\left(\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2}\right)^2}{\frac{S_1^4}{n_1^2\left(n_1-1\right)}+\frac{S_2^4}{n_2^2\left(n_2-1\right)}} .$$

# 统计推断代写

## 统计代写|统计推断代考STATISTICAL INFERENCE代 写|POPULATION VARIANCE $\Sigma 2$

$$H_0: \sigma^2=\sigma_0^2 \text { versus } H_a: \sigma^2 \neq \sigma_0^2 .$$

$$T=\frac{(n-1) S^2}{\sigma_0^2} \sim \chi^2(n-1)$$

$$\chi_{n-1,1-\alpha / 2}^2=\text { lower } \alpha / 2 \text { quantile of } \chi^2(n-1) \chi_{n-1, \alpha / 2}^2=\text { upper } \alpha / 2 \text { quantile of } \chi^2(n-1)$$

## 统计代写|统计推断代考STATISTICAL INFERENCE代 与軍IFFERENCE OF TWO POPULATION MEANS M1 – M2 independentsamples

$Y_{11}, Y_{12}, \ldots, Y_{1 n_1}$ 是来自 $\mathrm{a}$ 的独立同分布样本 $\mathcal{N}\left(\mu_1, \sigma_1^2\right)$ 人口分布
$Y_{21}, Y_{22}, \ldots, Y_{2 n_2}$ 是来自 a 的独立同分布样本 $\mathcal{N}\left(\mu_2, \sigma_2^2\right)$ 人口分布，

$$H_0: \mu_1-\mu_2=d_0 \text { versus } H_a: \mu_1-\mu_2 \neq d_0,$$

$$T=\frac{(\bar{Y} 1+-\bar{Y} 2+)-d_0}{S_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \sim t\left(n_1+n_2-2\right)$$

$$S_p^2=\frac{\left(n_1-1\right) S_1^2+\left(n_2-1\right) S_2^2}{n_1+n_2-2}$$

$$T=\frac{(\bar{Y} 1+-\bar{Y} 2+)-d_0}{\sqrt{\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2}}}$$

$$\nu=\frac{\left(\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2}\right)^2}{\frac{S_1^4}{n_1^2\left(n_1-1\right)}+\frac{S_2^4}{n_2^2\left(n_2-1\right)}}$$

## Matlab代写

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