ECON 3818

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# ECON 3818
## Chapter 16
### Kyle Butts
### 11 October 2021

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## Chapter 16: Confidence Intervals

---
# Statistical Inference

Recall, we're interested in estimating some unknown .cranberry[population] parameter `$\theta$` using the .kelly[sample] `$X_1, ...,X_n$`

We can use some estimator `$\hat{\theta}$`
      
- We can find its bias and its variance

- Can say what it converges to using Law of Large Numbers and the Central Limit Theorem
      
However, we don't have `$n \to \infty$`. We have a .hi[finite] sample.

- Therefore, we want to construct some belief about how good our estimator is.

- For example, if we have a sample mean with 5 individuals, our sampling distribution has a large variance. We want to report that.

---
# Example

Say we collect ACT Scores for 654 students, and calculate `$\bar{X}_{654}=26.8$`. Somehow we also know that the standard deviation of the *population distribution* is `$\sigma$` = 7.5

To visualize:

---
# 95% Rule

From the fact that `$\bar{X}_n \sim N(\mu, \frac{\sigma}{\sqrt{n}})$`, we have:

`\begin{align*}
P(- z_{\alpha/2} \leq \frac{\bar{X}_n - \mu}{\sigma/\sqrt{n}} \leq z_{\alpha/2}) = 1-\alpha,
\end{align*}`

where `$- z_{\alpha/2}$` is the .emph[critical value] with a left-tail probability of 2.5%. With some math, we see:

`\begin{align*}
P(- z_{\alpha/2} \frac{\sigma}{\sqrt{n}} \leq \bar{X}_n - \mu \leq z_{\alpha/2} \frac{\sigma}{\sqrt{n}}) = 1-\alpha \\
\end{align*}`

`\begin{align*}
P(\mu - z_{\alpha/2} \frac{\sigma}{\sqrt{n}} \leq \bar{X}_n \leq \mu + z_{\alpha/2} \frac{\sigma}{\sqrt{n}}) = 1-\alpha \\
\end{align*}`

--
That is, `$1-\alpha$`% of the time, the sample mean falls between `$\mu - z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$` and `$\mu + z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$`.

---
# Central Idea of Confidence Interval

The interval we defined above involves `$\mu$` which we .emph[do not know!] Instead, let's do the math slightly differently:

`\begin{align*}
P(- z_{\alpha/2} \frac{\sigma}{\sqrt{n}} \leq \bar{X}_n - \mu \leq z_{\alpha/2} \frac{\sigma}{\sqrt{n}}) = 1-\alpha \\
\end{align*}`

`\begin{align*}
P(- \bar{X}_n - z_{\alpha/2} \frac{\sigma}{\sqrt{n}} \leq - \mu \leq - \bar{X}_n  + z_{\alpha/2} \frac{\sigma}{\sqrt{n}}) = 1-\alpha \\
\end{align*}`

`\begin{align*}
P(\bar{X}_n - z_{\alpha/2} \frac{\sigma}{\sqrt{n}} \leq \mu \leq \bar{X}_n  + z_{\alpha/2} \frac{\sigma}{\sqrt{n}}) = 1-\alpha \\
\end{align*}`

--
Now we are able to categorize some uncertainty surrounding `$\mu$`. In particular, we have that in repeated sampling of `$\bar{X}_n$` (many samples of size `$n$`), `$\mu$` will be in the interval
`\begin{align*}
	\left[\bar{X}_n - z_{\alpha/2} \frac{\sigma}{\sqrt{n}}, \bar{X}_n + z_{\alpha/2} \frac{\sigma}{\sqrt{n}}\right]
\end{align*}`
`$1 - \alpha$`% of the time.

---
# Confidence Interval

The 95 part of the 68-95-99.7 rule for Normal distributions says that `$\bar{X}_{654}$` is within 2 standard deviations of the mean `$\mu$` in 95% of samples. Therefore `$z_{\alpha/2} \approx 2$`.

Since `$\sigma=7.5$`, the standard deviation of our sampling distribution is `$\frac{7.5}{\sqrt{654}}=0.3$` (from the Central Limit Theorem).

This means for .hi[95% of all samples of size 654], the sample mean `$\bar{X}_{654}$` is within `$0.6$` of the population mean (two std deviations).

<br>

If we estimate that `$\mu$` lies somewhere in the interval from `$\bar{X}_{654}-0.6$` to `$\bar{X}_{654}+0.6$`, we'll be right for 95% of all possible samples.

---
# Confidence Interval

Plugging in all the information we have leads to:

`$$\big[26.8-0.6, 26.8+0.6\big] = \big[ 26.2, 27.4 \big]$$`

This interval is a .hi.ruby[confidence interval], which says that we are .it[95% confident] that the true mean of the BMI, `$\mu$`, is in between 26.2 and 27.4 .sup[*]

.footnote[.sup[*] This is because we got this interval from a method that captures the population mean for 95% of all possible samples]

---
# Confidence Intervals

Let's see what I mean. Let's say the population average ACT Score is 27. I will draw many samples of size 654 from the distribution `$N(27, 7.5)$`.

For each sample, I will calculate `$\bar{X}_{654}$` and add and subtract `$0.6$` to form my confidence interval `$[\bar{X}_{654}-0.6, \bar{X}_{654}+0.6]$`.

---
# Confidence Intervals

In that example, the confidence interval was `$\bar{X}_{654} \pm 0.6$`. In general, a confidence interval takes the form

`$$\kelly{\text{estimate}} \pm \daisy{\text{margin of error}}$$`

where the .hi.daisy[margin of error] shows how much variability there is in our estimate

---
# Margins of Error

For a given level of confidence, `$C$` (say 95%), the .daisy[margin of error] for our sample mean as:

$$
\daisy{k} = Z_{\frac{1-C}{2}}} \frac{\sigma}{\sqrt{n}}
$$

<br>

- Let's say `$C = 95\%$`. We want to capture the middle 95%, so `$\frac{1-.95}{2} = 2.5\%$` in each tail.

- `$Z_{0.025} = 1.96$` which is where the `$\sim 2$` standard deviations comes from.

- 90% Confidence Interval:  `$\implies Z_{\frac{1-C}{2}} = Z_{.05} = 1.645$` standard deviations.

---
# Example
Lets determine the .it[exact] margin of error for previous example

`$$k=Z_{\frac{1-C}{2}} \cdot \frac{\sigma}{\sqrt{n}}$$`

If we are calculating 95% confidence interval, where `$\bar{X}_{654}=26.8$` `$\sigma = 7.5$`,  then

`$$k=Z_{0.025} \cdot \frac{7.5}{\sqrt{654}}$$`

We find `$Z_{0.025}$` using the table. `$Z_{0.025}$` is the z-score such that `$P(z>Z_{0.025})=0.025$`

- Look up 0.025 (or 0.975!) and find the corresponding z-score

---
# Example

Using the z-table, we find that `$Z_{0.025} = 1.96$`. This means:

`$$k = 1.96\cdot \frac{7.5}{\sqrt{654}} = 0.57$$`

This means our .it[exact] 95% confidence interval is:

`$$[26.23, 27.37]$$`

---
# Clicker Question

What Z-score will be associated with a 82% confidence interval

---
# Example

Say high-school freshmen are sampled to see how long they spend on social media per day. The sample mean of `$n = 50$` students is `$\bar{X}_{50} = 2.1$` hours. The standard deviation of the population is `$0.5$` hours.

What is the 90% confidence interval for our estimate of the mean number of hours per day?

---
# Clicker Question
A 95% confidence interval for the mean hours freshmen spent on social media per day was calculated to be [2.5 hours, 3.1 hours] based off a sample mean `$\bar{X}_{50} = 2.8$`. The confidence interval was based on a SRS sample of `$n = 50$`.

The standard deviation of the population is:

---
# Margins of Error

So to recap, the margin of error is calculated by:

`$$k=Z_{\frac{1-C}{2}} \cdot \frac{\sigma}{\sqrt{n}}$$`

This means the size of the margin of error is determined by:

- level of confidence

- size of sample (can sometimes control)

- variance (which we can't control)

Discuss with a partner what happens to margin of error when

- increase the level of confidence required

- decrease sample size

- increase variance of population

---
# Margins of Error

Level of confidence
      
- higher the level of confidence we want to have, larger the margin of error
      
Size of sample 
      
- larger the sample, smaller the margin of error
      
- Variance
      
larger the variance, larger the margin of error

---
# Example

Given a sample mean of 8, from a sample of 36 observations, and a variance of 25, construct 90%, 95%, and 99% confidence intervals for the true value, `$\mu$`.

---
# Confidence Intervals

How to .hi[properly] think of a confidence interval:

1. You collect many different samples from the population

2. For each sample, you calculate the mean and the associated confidence interval around that mean

3. You'll have a confidence interval for each sample you collected

4. .hi[95% of the confidence intervals you calculated will include the true mean `$\mu$`]

---
# Confidence Intervals

*Common misconceptions* about confidence intervals

The CI .hi.ruby[does not] tell you:
      
- The true mean is inside the confidence interval

- The probability the true mean is in your CI is C%
      
      
The CI .hi.kelly[does] tell you:
      
- the range of estimates that contain the true mean C% of the time *in repeated sampling*
      
      
      
---
# Confidence Intervals

---
# Clicker Question -- Midterm Example
Veterinary researchers at a major university veterinary hospital calculated a 99% confidence interval for the average age of horses admitted for laminitis (a particular foot disease) as 6.3 to 7.4 years.

Based on this information we conclude that:

<ol type = "a">
	<li> 99% of all horses admitted for laminitis are between 6.3 and 7.4 years old </li>
	
	<li> 99% of the time, the average age of a horse admitted for laminitis will be between 6.3 and 7.4 years </li>
	
	<li> We are 99% confident that the true mean age of horses with laminitis is between 6.3 and 7.4 years old </li>
	
	<li> 99% of all samples of size n=25 will have an average age of horses with laminitis between 6.3 and 7.4 years old. </li>
</ol>