## 统计代写|回归分析作业代写Regression Analysis代考|How to think about the estimate and its standard error

$\mathrm{Hmmm}$, the estimated slope is shown in the output as $1.6199$, and the standard error is shown in the output as $0.1326$. So the actual slope is most likely in the range $1.6199 \pm 2(0.1316)$, or roughly between $1.6 \pm 0.26$. AHA! The true slope is most likely a positive number! So the $X$ variable has a positive relation to $Y$ !
We used $2.0$ rather than $1.96$ as a multiplier of the standard error because the result is only approximate anyway, so why not? We might as well simplify things by using another approximation, $2.0$ instead of 1.96. It just makes life easier. And it works well in practice, so we generally recommend that you follow the advice given by the above mental conversation.

But there are precise, mathematically exact results that you can use in the case where the data are produced by the classical model. The theory is mathematically deep, but you probably have seen it before, to one degree or another. It involves “Student’s $T$ distribution,” which is ubiquitous in statistics. In a nutshell, the issue revolves around how to deal with the estimate $\hat{\sigma}$ of $\sigma$ in the standard error formula. After all, as shown above, the first interval formula involving $1.96$ and $\sigma$ is exact; the only reason for calling the second interval formula “approximate” is because of the substitution of $\hat{\sigma}$ for $\sigma$. The effect of using $\hat{\sigma}$ rather than $\sigma$ can be precisely, exactly, quantified. A mathematical theorem states that if the classical regression model produces the real data, then the additional variability incurred when you use $\hat{\sigma}$ rather than $\sigma$ is precisely accounted for by using the $T$ (Student’s T) distribution rather than the $Z$ (standard normal) distribution.

Specifically, the critical value $1.96$ is from the $Z$ (standard normal) distribution, the number that puts $95 \%$ probability between $-1.96$ and $1.96$. It is, therefore, the $0.975$ quantile of the standard normal distribution. In $\mathrm{R}$ it is qnorm (.975), which returns the even more precise value $1.959964$.

To account for the error in using the estimate $\hat{\sigma}$ of $\sigma$ in the standard error formula, you need to use the $T$ distribution rather than the $Z$ distribution. The $T$ distribution involves a “degrees of freedom” parameter, which in essence measures the accuracy of $\hat{\sigma}$ as an estimator of $\sigma$. This degrees of freedom quantity is mathematically identical to the divisor used to make the estimated variance an unbiased estimate:
$$d f e=n-\left(# \text { of } \beta^{\prime} s\right)$$
The “e” on “df” refers to “error”: Recall that, $\sigma$, the conditional standard deviation of $Y \mid X=x$, is also the standard deviation of the error term $\varepsilon$. You can think of $d f e$ as the “effective sample size” that is used to estimate the error standard deviation.

There is also a “model degrees of freedom” that we will discuss later, using the symbol $d f m$. The model degrees of freedom means something completely different: It refers to the flexibility (freedom) of the regression model; essentially the number of free parameters $\left(\beta^{\prime}\right.$ s) in the model, excluding the intercept.

To get exact intervals for regression coefficients, you use the quantiles of the $T_{\text {df }}$ distribution, rather than the quantiles of the $Z$ distribution. The mathematics is precise but will not be proved here: It states that, if the data are produced by the classical regression model, then you have the following result.

## 统计代写|回归分析作业代写Regression Analysis代考|Understanding “Exactness” and “Non-exactness” via Simulation

What does “exact” mean in these discussions? It means that the true confidence level is exactly $95 \%$ when you use a $95 \%$ confidence interval. Non-exactness means that the true confidence level is not equal to $95 \%$-it may be higher or lower than $95 \%$. Further, “true confidence level” refers to the true probability that the parameter lies within the prescribed confidence limits.

Here is a simple simulation to illustrate “exactness.” The data are simulated according to the classical model, the $95 \%$ interval for $\beta_1$ is calculated, and we check whether the true $\beta_1$ lies within the interval. Then we repeat that process 100,000 times, finding the proportion of the 100,000 intervals that contain the true $\beta_1$. This proportion should be close to $95 \%$ and will be exactly $95 \%$ with infinitely many (rather than 100,000 ) simulations.

On the other hand, when data are simulated from a model where the assumptions are violated, the proportion will be different from 95\%, even with infinitely many simulations. The simulation code that follows simulates data from the classical model, and also from the model with non-normal conditional distributions used to obtain Figure 1.11.

Thus, in the case where the classical model is true, $94.907 \%$ of the 100,000 samples gave a confidence interval that contained the true $\beta_1=1.5$. According to the mathematical theory, this percentage will be exactly $95 \%$ with infinitely many simulated data sets.

On the other hand, in the simulation where the conditional distributions are non-normal as illustrated in Figure 1.11,96.058\% of the 100,000 samples gave a confidence interval that contained the true $\beta_1=1.5$. The mathematical theory does not state that this percentage will be exactly $95 \%$ with infinitely many simulated data sets. In fact, the true percentage with infinitely many data sets will be more than $95 \%$ in this case.

The non-exactness of the confidence interval is not a huge problem for the given simulation study, because the actual confidence level is close to $95 \%$ in the non-normal case. This study provides an example of our common refrain: You can best understand why and whether violations of assumptions are problematic via simulation.

Violations of assumptions other than normality can cause bigger problems. Figure $3.2$ shows a case where the estimates are biased, and in such cases the intervals will systematically miss the target on the low side, leading to coverage rates close to $0 \%$ in extreme cases. Similarly, heteroscedasticity (non-constant variance) can cause the standard errors to be too small, also leading to coverage rates much lower than $95 \%$, which you can verify by using simulation.

As it turns out, violation of the normality assumption is not usually a major concern for the validity of confidence intervals for the $\beta$ parameters: Even with non-normal conditional distributions $p(y \mid x)$, the Central Limit Theorem dictates that the distribution of the parameter estimates will be approximately normal. Other inferences are not so robust to non-normality: The prediction interval discussed in Section $3.8$ below will behave quite poorly with non-normal processes. Inferences for variance parameters are similarly nonrobust. Further, even when OLS-based inferences are robust in the sense of having confidence levels near $95 \%$ under non-normality, the OLS estimates themselves can be quite inaccurate relative to $\mathrm{ML}$ estimates under non-normality.

# 回归分析代考

## 统计代写|回归分析作业代写回归分析代考|如何思考估计及其标准误差

$\mathrm{Hmmm}$，估计的斜率在输出中表示为 $1.6199$，标准误差在输出中显示为 $0.1326$。所以实际斜率很可能在这个范围内 $1.6199 \pm 2(0.1316)$，或大致介于 $1.6 \pm 0.26$。啊哈!真正的斜率很可能是正数!所以 $X$ 变量与。呈正相关 $Y$

$$d f e=n-\left(# \text { of } \beta^{\prime} s\right)$$
df上的“e”指的是“误差”:回想一下，$\sigma$是$Y \mid X=x$的条件标准差，也是误差项$\varepsilon$的标准差。你可以把$d f e$看作是用来估计误差标准差的“有效样本量”

## 统计代写|回归分析作业代写回归分析代考|通过模拟理解“准确性”和“非准确性”

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