# 统计代写|应用线性模型代写Applied Linear Models代考|STAT6620

## 统计代写|应用线性模型代写Applied Linear Models代考|Example (continued)

Confidence intervals on $b_1$ will be calculated for the example used earlier, a non-symmetric interval from (89) and a symmetric interval from (91). For both we use
$$\hat{b}1=50 / 24=2.08$$ from (28), $$\hat{\sigma}=\sqrt{5.75}=2.40 \quad \text { and } \quad N-r=2$$ from Table $3.5$, and $$a^{11}=20 / 144=0.139$$ from (86) and (44). Then in (89) a non-symmetric confidence interval for $b_1$ is \begin{aligned} 2.08-2.40 t{2, \alpha, U} \sqrt{0.139} \text { to } & 2.08+2.40 t_{2, \alpha, L} \sqrt{0.139} \ &=2.08-0.89 t_{2, \alpha, U} \text { to } 2.08+0.89 t_{2, \alpha, L} \quad \text { (97) } \end{aligned}
From tabulated values of the $t_2$-distribution, [e.g., Vogler and Norton (1957)] we find that
$$\operatorname{Pr}(t \leq-3.6)=0.04 \quad \text { and } \quad \operatorname{Pr}(t \geq 7.1)=0.01 \text {, }$$
so that by (88), for $\alpha=0.05$,
$$t_{2,05, L}=-3.6 \quad \text { and } \quad t_{2,05, U}=7.1$$
and so in (97) the confidence interval becomes
$$2.08-0.89(7.1) \text { to } 2.08-0.89(-3.6)=(-4.23,5.08) \text {. }$$
It is questionable, of course, as to what kind of situation would reasonably lead to needing a non-symmetric confidence interval with the $t$-distribution. The example illustrates, however, how such intervals can be calculated and doing so emphasizes the important fact that there are many such intervalsbecause there are many values $t_{N-r, \alpha, L}$ and $t_{N-r, \alpha, U}$ that satisfy (88). In contrast, there is only one symmetric confidence interval, the interval which has the optimal property that for given $N-r$ and $\alpha$ it is the interval of shortest length. This is the interval given in (91) for which, for the example, $(90)$ is
$$\operatorname{Pr}{t \geq 4.30}=0.025$$
for $t \sim t_2$. Hence the symmetric interval on $b_1$ is, from (91),
\begin{aligned} 2.08 \pm 2.40 t_{2, \frac{1}{2} \alpha} \sqrt{0.139} &=2.08 \pm 0.89 t_{2,0.025} \ &=2.08 \pm 0.89(4.30) \ &=(-1.75,5.91) \end{aligned}

## 统计代写|应用线性模型代写Applied Linear Models代考|Testing linear hypotheses

The literature of linear models abounds with discussions of different kinds of hypotheses that can be of interest in widely differing fields of application. Four hypotheses of particular interest are: (i) $H: \quad \mathbf{b}=\mathbf{0}$, the hypothesis that all elements of $\mathbf{b}$ are zero. (ii) $H: \quad \mathbf{b}=\mathbf{b}0$, the hypothesis that $b_i=b{i 0}$ for $i=0,1,2, \ldots, k$, i.e., that each $b_i$ is equal to some specified value $b_{i 0}$. (iii) $H: \quad \boldsymbol{\lambda}^{\prime} \mathbf{b}=m$, that some linear combination of the elements of $\mathbf{b}$ equals a specified constant. (iv) $H: \quad \mathbf{b}_q=\mathbf{0}$, that some of the $b_i$ ‘s, $q$ of them where $q<k$, are zero. Although the calculations for the $F$-statistic for these hypotheses and variants of them appear, on the surface, to differ markedly from one kind of hypothesis to another, we will show that all linear hypotheses can be handled by one universal procedure. Specific hypotheses such as those listed above are then just special cases of the general procedure.
The general hypothesis we consider is
$$H: \quad \mathbf{K}^{\prime} \mathbf{b}=\mathbf{m}$$
where $\mathbf{b}$, of course, is the $(k+1)$-order vector of parameters of the model; $\mathbf{K}^{\prime}$ is any matrix of $s$ rows and $k+1$ columns; and $\mathbf{m}$ is a vector, of order $s$, of specified constants. There is only one limitation on $\mathbf{K}^{\prime}$ : that it have full row rank, i.e., $r\left(\mathbf{K}^{\prime}\right)=s$. This simply means that the linear functions of $\mathbf{b}$ which form the hypothesis must be linearly independent; that is, the hypothesis must be made up of linearly independent functions of $\mathbf{b}$ and must contain no functions which are linear combinations of others therein. This is quite reasonable because it means, for example, that if the hypothesis relates to $b_1-b_2$ and $b_2-b_3$ then there is no point in having it also relate, explicitly, to $b_1-b_3$. Clearly, this condition on $\mathbf{K}^{\prime}$ is not at all restrictive in limiting the application of the hypothesis $H: \quad \mathbf{K}^{\prime} \mathbf{b}=\mathbf{m}$ to real problems. Furthermore, although it might seem necessary to also require that $\mathbf{m}$ be such that the equations $\mathbf{K}^{\prime} \mathbf{b}=\mathbf{m}$ be consistent, this is automatically achieved by demanding that $\mathbf{K}^{\prime}$ have full row rank, for the equations $\mathbf{K}^{\prime} \mathbf{b}=\mathbf{m}$ are then consistent for any vector $\mathbf{m}$.

## 统计代写|应用线性模型代写Applied Linear Models代考|Example (continued)

$$\hat{b} 1=50 / 24=2.08$$

$$\hat{\sigma}=\sqrt{5.75}=2.40 \quad \text { and } \quad N-r=2$$

$$a^{11}=20 / 144=0.139$$

$$2.08-2.40 t 2, \alpha, U \sqrt{0.139} \text { to } 2.08+2.40 t_{2, \alpha, L} \sqrt{0.139}=2.08-0.89 t_{2, \alpha, U} \text { to } 2.08+0.89 t_{2, \alpha, L}$$

$$\operatorname{Pr}(t \leq-3.6)=0.04 \text { and } \operatorname{Pr}(t \geq 7.1)=0.01 \text {, }$$

$$t_{2,05, L}=-3.6 \quad \text { and } \quad t_{2,05, U}=7.1$$

$$2.08-0.89(7.1) \text { to } 2.08-0.89(-3.6)=(-4.23,5.08) .$$

$$\operatorname{Pr} t \geq 4.30=0.025$$

$$2.08 \pm 2.40 t_{2, \frac{1}{2} \alpha} \sqrt{0.139}=2.08 \pm 0.89 t_{2,0.025} \quad=2.08 \pm 0.89(4.30)=(-1.75,5.91)$$

## 统计代写|应用线性模型代写Applied Linear Models代考|Testing linear hypotheses

$$H: \quad \mathbf{K}^{\prime} \mathbf{b}=\mathbf{m}$$

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