# Point estimation and confidence intervals

04/07/2021

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## Outline

• The following topics will be covered in this lecture:

• General concepts in point estimation
• Bias of estimators
• Variance of estimators
• Standard Error
• An introduction to confidence intervals

## General concepts of point estimation

• Recall, any function of a random sample, i.e., any statistic, is modeled as a random variable.

• If $$h$$ is a general function used to compute some statistic, we thus define

$\hat{\Theta} = h(X_1, \cdots, X_n)$

to be a random variable that will depend on the particular realizations of $$X_1,\cdots, X_n$$.

• We call the probability distribution of a statistic a sampling distribution.

Sampling Distribution
The probability distribution of a statistic is called a sampling distribution.
• The sample mean

$\hat{\Theta} = \overline{X} = h(X_1, \cdots, X_n)= \frac{1}{n}\sum_{i=1}^n X_i$

is now one example for which we have a model of the sampling distribution.

• Specifically, the central limit theorem says that the sampling distribution of the sample mean is $$\overline{X} \sim N\left(\mu, \frac{\sigma^2}{n}\right)$$ when $$X$$ is normal, or if $$n$$ is sufficiently large.

### General concepts of point estimation continued

• Recall, we had a special name for $$\hat{\Theta}$$ in relation to the true parameter value $$\theta$$:
• Point estimators
A point estimate of some population parameter $$\theta$$ is a single numerical value $$\hat{\theta}$$ of a statistic $$\hat{\Theta}$$. This is a particular realization of the random variable $$\hat{\Theta}$$, viewed as a random variable; $$\hat{\Theta}$$ is called the point estimator.
• We want an estimator to be “close” in some sense to the true value of the unknown parameter, but we know that it happens to be a random variable.

• In this way, we need to describe how close this estimator is to the true value in a probabilistic sense.

• As we have seen before, there are important parameters that describe a probability distribution or a data set:

1. the “center” of the data / distribution; and
2. the “spread” of the data / distribution.
• The central limit theorem actually provided both of these (and the sampling distribution) for the sample mean:

1. the “center” of the distribution for $$\hat{\Theta}=\overline{X}$$ was given by $$\mu$$, the true population mean;
2. the “spread” of the distribution for $$\hat{\Theta}=\overline{X}$$ was given by $$\frac{\sigma}{\sqrt{n}}$$, the standard deviation of the population, divided by the square-root of the sample size.
• The two above parameters thus give us a means of describing “how close” the sample mean $$\overline{X}$$ tends to be to the population mean $$\mu$$ in a probabilistic sense.

### Bias of estimators

• The notion of the “center” of the sampling distribution can be useful as a general criteria for estimators.

• Formally, we say that $$\hat{\Theta}$$ is an unbiased estimator of $$\theta$$ if the expected value of $$\hat{\theta}$$ is equal to $$\theta$$.

• This is equivalent to saying that the mean of the probability distribution of $$\hat{\Theta}$$ (or the mean of the sampling distribution of $$\hat{\Theta}$$) is equal to $$\theta$$.

Bias of an Estimator
The point estimator $$\hat{\Theta}$$ is an unbiased estimator for the parameter $$\theta$$ if $\mathbb{E}\left[\hat{\Theta}\right] = \theta$ If the estimator is not unbiased, then the difference $\mathbb{E}\left[\hat{\Theta}\right] - \theta$ is called the bias of the estimator $$\hat{\Theta}$$. When an estimator is unbiased, the bias is zero; that is, \begin{align} \mathbb{E}\left[\hat{\Theta}\right] - \theta &= \theta - \theta \\ &=0 \end{align}
• If we consider the expected value to represent the average value over infinite replications;

• the above says that “over infinite replications of a random sample of size $$n$$, the average value of the point estimator $$\hat{\Theta}$$ will equal the true population parameter $$\theta$$”.
• A particular realization of $$\hat{\Theta}$$ will generally not equal the true value $$\theta$$.

• However, replications of the experiment will give a good approximation of the true value $$\theta$$.

## Variance of estimators

• We use the bias as discussed already to measure the center of a sampling distribution
• An unbiased estimator will have a distribution centered at the true population parameter.
• Yet suppose we have two estimators of the same parameter $$\theta$$, which we will denote $$\hat{\Theta}_1$$ and $$\hat{\Theta}_2$$ respectively.
• It is possible that they are both unbiased (the sampling distributions have the same center), yet they have different spread.
• That is to say, one estimator might tend to vary more than the other. Courtesy of Montgomery & Runger, Applied Statistics and Probability for Engineers, 7th edition

• The spread is a critical measure of how much variation is encountered with respect to resampling.
• We might describe the two concepts with an estimator as follows:
• Accuracy of an estimator - this is represented by the estimator being unbiased, so that we expect it to give an accurate result on average.
• Precision of an estimator - this is represented by the estimator having a small spread, so that the estimates don’t differ wildly from sample to sample.
• It is possible, in general, for an estimator to be either, both or neither of the above.
• We are often interested, thus, in unbiased estimators with a minimum variance as a first choice.
• In some situations biased estimators will actually be preferred, though a general discussion of the tradoffs is beyond our scope.

### Variance of estimators continued

• As a formal definition, we will introduce the following idea:
• Minimum Variance Unbiased Estimator
If we consider all unbiased estimators of $$\theta$$, the one with the smallest variance is called the minimum variance unbiased estimator (MVUE).
• The practical interpretation again is demonsrated by the last figure: Courtesy of Montgomery & Runger, Applied Statistics and Probability for Engineers, 7th edition

• Suppose that $$\hat{\Theta}_1$$ is the MVUE, and $$\hat{\Theta}_2$$ is any other unbiased estimator.
• Then, $\mathrm{var}\left(\hat{\Theta}_1\right) \leq \mathrm{var}\left(\hat{\Theta}_2\right).$
• Practically speaking, the MVUE is the most precise unbiased estimator, as its value changes the least with respect to resampling.
• An important example of a MVUE is actually the sample mean.
• If $$X_1, X_2 , \cdots , X_n$$ is a random sample of size $$n$$ from a normal distribution with mean $$\mu$$ and variance $$\sigma^2$$, the sample mean $$\overline{X}$$ is the MVUE for $$\mu$$.
• Again, other choices exist to estimate $$\mu$$, but among all unbiased estimators, the sample mean is the most precise.
• For non-normal distributions, however, a better choice might be, e.g., a biased estimator.

### Standard error of an estimator

• As noted before, the variance is a “natural” measure of spread mathematically for theoretical reasons, but it is in the units squared of the original units.

• For this reason, when we talk about the spread of an estimator's sampling distribution, we typically discuss the standard error.

The standard error Let $$\hat{\Theta}$$ be an estimator of $$\theta$$. The standard error error of $$\hat{\Theta}$$ is its standard deviation given by $\sigma_\hat{\Theta} = \sqrt{\mathrm{var}\left(\hat{\Theta}\right)}.$ If the standard error involves unknown parameters that can be estimated, substitution of those values into the equation above produces an estimated standard error denoted $$\hat{\sigma}_\hat{\Theta}$$. It is also common to write the standard error as $$\mathrm{SE}\left(\hat{\Theta}\right)$$.
• Q: can anyone recall what the standard error is of the sample mean? That is, what is the standard deviation of the sampling distribution (for a normal sample or $$n$$ large)?

• A: the central limit theorem states that $$\overline{X}$$ follows (exactly for a normal sample or $$n$$ large, approximately) a sampling distribution

$\overline{X}\sim N\left(\mu, \frac{\sigma^2}{n}\right).$

• Therefore, the standard error of the sample mean is precisely,

$\sigma_{\overline{X}} = \frac{\sigma}{\sqrt{n}}.$

### Standard error of an estimator

• As was discussed before, there are times that we may not know all the parameters that describe the standard error.

• For example, suppose we draw $$X_1, \cdots, X_n$$ from a normal population, for which we know neither the mean nor the variance.

• Let the unknown and unobservable theoretical parameters be denoted $$\mu$$ and $$\sigma$$ as usual.

• The sample mean has the sampling distribution,

$\overline{X} \sim N\left( \mu, \frac{\sigma^2}{n}\right),$

and therefore standard error $$\sigma_{\overline{X}} = \frac{\sigma}{\sqrt{n}}$$.

• However, we stated that $$\sigma$$ itself is unknown.

• In this case, we will estimate the standard error as

$\hat{\sigma}_\overline{X} = \frac{s}{\sqrt{n}}$ with the sample standard deviation $$s$$.

• This is what is meant to estimate the standard error.

• This particular example will be extremely important for confidence intervals.

### Standard error of an estimator – example

• An article in the Journal of Heat Transfer (Trans. ASME, Sec. C, 96, 1974, p. 59) described a new method of measuring the thermal conductivity of Armco iron.

• Using a temperature of $$100^\circ$$ F and a power input of 550 watts, the following 10 measurements of thermal conductivity (in Btu/hr-ft-∘ F) were obtained:

$41.60, 41.48, 42.34, 41.95, 41.86, 42.18, 41.72, 42.26, 41.81, 42.04$

• A point estimate of the mean thermal conductivity at $$100^\circ$$ F and 550 watts is the sample mean or

$\overline{x} = 41.924$

• The standard error of the sample mean is $$\sigma_\overline{X}=\frac{\sigma}{\sqrt{n}}$$;

• however, $$\sigma$$ is unknown so that we estimate it by the sample standard deviation $$s = 0.284$$ to obtain

$\hat{\sigma}_\overline{X} = \frac{s}{\sqrt{n}}= \frac{0.284}{\sqrt{10}} \approx 0.0898$

• Notice that the standard error is about 0.2 percent of the sample mean, implying that we have obtained a relatively precise point estimate of thermal conductivity.

### Standard error of an estimator – example

• Assume that thermal conductivity is normally distributed, then two times the standard error is

$2\hat{\sigma}_\overline{X} = 2(0.0898) = 0.1796.$

• The empirical rule says that about 95% of realizations of the sample mean lie within two standard deviations of the true mean $$\mu$$.

• Therefore, we are highly confident that the true mean thermal conductivity is within the interval 41.924 ± 0.1796 or between $$[41.744 , 42.104]$$.

• We will now formalize this logic into confidence intervals.

## Introduction to confidence intervals

• Engineers are often involved in estimating parameters.

• For example, there is an ASTM Standard E23 that defines a technique called the Charpy V-notch method for notched bar impact testing of metallic materials.

• The impact energy is often used to determine whether the material experiences a ductile-to-brittle transition as the temperature decreases.

• Suppose that we have tested a sample of $$10$$ specimens of a particular material with this procedure. We know that we can use the sample average $$\overline{X}$$ to estimate the true mean impact energy $$\mu$$.

• However, we also know that the true mean impact energy is unlikely to be exactly equal to your estimate due to sampling error.

• Reporting the results of your test as a single number is unappealing because $$\overline{X}$$ may be an accurate estimator, but it doesn't tell us how precise the estimate is.

• Rather, measures of the spread of the sampling distribution, e.g., the standard error tell us how precise the estimate is.
• A way to quantify our uncertainty is to report the estimate in terms of a range of plausible values called a confidence interval.

### Introduction to confidence intervals – continued

• Suppose $$\hat{\Theta}$$ is an unbiased estimator for the speed of light $$\theta$$.
• We are assuming that $$\theta$$ is a deterministic but unknown constant.

• For sake of example, also suppose that $$\hat{\Theta}$$ has standard deviation $$\sigma_\hat{\Theta}$$ = 100 km/sec.
• Recall Chebyshev’s inequality,

\begin{align} P\left(\vert \hat{\Theta} - \theta \vert \geq k \sigma_\hat{\Theta}\right) \leq \frac{1}{k^2} \end{align}

• We find

\begin{align} P\left(\vert \hat{\Theta} - \theta \vert < 2 \sigma_\hat{\Theta}\right) > \frac{3}{4} \end{align}

• This tells us that there is a probability of at least 75% that $$\hat{\Theta}$$ is within 200 km/sec of the speed of light $$\theta$$.
• Equivalently, $$\theta \in \left(\hat{\Theta}-200, \hat{\Theta}+200\right)$$ with probability 75%.
• Notice that $$\left(\hat{\Theta}-200, \hat{\Theta}+200\right)$$ is a random interval but we again assume that $$\theta$$ is a fixed constant.

### Introduction to confidence intervals – continued

• Now we will suppose the estimate $$\hat{\Theta}$$ gives us based on some data is $$\hat{\theta}=299852.4$$

• We can say that $$\theta \in (299652.4, 300 052.4)$$ with confidence 75%.

• Note that $$\theta$$ is an an unknown constant – it is either in the interval or not and there is nothing random about the above statement.
• Therefore, we cannot say that the probability of $$\theta \in (299652.4, 300 052.4)$$ is 75%, but we used information to guarantee that our procedure for estimation will work 75% of the time.
• Confidence intervals give us a systematic procedure as above to guarantee with a level of confidence that our plausible values for the parameter $$\theta$$ include the true value.

• In the remaining course, we will be concerned with dual questions:

1. With what confidence can we estimate a parameter $$\theta$$ as a range of plausible values? And
2. how unlikely would it be for $$\theta$$ to be outside of some range based on our observations?
• These two ideas are known as confidence intervals and hypothesis testing respectively.

### Introduction to confidence intervals – continued

• Suppose that $$X_1 , X_2, \cdots , X_n$$ is a random sample from a normal distribution with unknown mean $$\mu$$ and known variance $$\sigma^2$$.

• By the central limit theorem, we know that the sample mean $$\overline{X}$$ is distributed

$\overline{X} \sim N\left(\mu, \frac{\sigma^2}{n}\right)$

• We may standardize $$\overline{X}$$ by subtracting its mean and dividing by its standard deviation,

\begin{align} Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}. \end{align}

• The random variable $$Z$$ has a standard normal distribution.

• A confidence interval estimate for $$\mu$$ is an interval of the form

$l ≤ \mu ≤ u,$ where the end-points $$l$$ and $$u$$ are computed from the sample data.

• Because different samples will produce different values of $$l$$ and $$u$$, these end-points are values of random variables $$L$$ and $$U$$, respectively.

### Introduction to confidence intervals – continued

• Suppose that we can determine values of $$L$$ and $$U$$ such that the following probability statement is true:

$P\{L \leq \mu \leq U\} = 1 − \alpha$ where $$0 \leq \alpha \leq 1$$.

• There is a probability of $$1 − \alpha$$ of selecting a sample for which the CI will contain the true value of $$\mu$$.

• Once we have selected the sample, so that $$X_1 = x_1 , X_2 = x_2 , \cdots , X_n = x_n$$, and computed $$l$$ and $$u$$, the resulting confidence interval for $$\mu$$ is

$l \leq \mu \leq u.$

• The end-points or bounds $$l$$ and $$u$$ are called the lower- and upper-confidence limits (bounds), respectively, and $$1 − \alpha$$ is called the confidence coefficient (or level).

• Again,

$$l \leq \mu \leq u.$$ is a fixed numerical statement with nothing random about it, this is true or untrue.

• The goal then is to find the procedure that defines the random variables $$L,U$$ for which the procedure will succeed $$(1-\alpha)\times 100\%$$ of the time.

### Introduction to confidence intervals – continued Courtesy of Mario Triola, Essentials of Statistics, 6th edition

• Let’s take an example confidence level of $$95\%$$ – this corresponds to a rate of failure of $$5\%$$ over infinitely many replications.
• Generally, we will write the confidence level as, $(1 - \alpha) \times 100\%$ so that we can associate this confidence level with its rate of failure $$\alpha$$.
• In our problem, because $$Z = \frac{\overline{X} - \mu }{\frac{\sigma}{\sqrt{n}}}$$ has a standard normal distribution, we may write \begin{align} P\left(-z_\frac{\alpha}{2} \leq \frac{\overline{X} - \mu }{\frac{\sigma}{\sqrt{n}}} \leq z_\frac{\alpha}{2}\right) = 1 - \alpha \end{align} where
• we want to find the critical value $$z_\frac{\alpha}{2}$$ for which:
• $$(1-\frac{\alpha}{2})\times 100\%$$ of the area under the normal density lies to the left of $$z_\frac{\alpha}{2}$$; and
• $$(1-\frac{\alpha}{2})\times 100\%$$ of the area under the normal density lies to the right of $$-z_\frac{\alpha}{2}$$.
• Put together, $$(1-\alpha)\times 100\%$$ of values lie within $$[-z_\frac{\alpha}{2},z_\frac{\alpha}{2}]$$ in the standard normal. 