ECON 3818

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# ECON 3818
## Chapter 20
### Kyle Butts
### 08 February 2022

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## Chapter 20: Inference about a Population Mean

---
# Conditions for Inference about a Mean

We've discussed using confidence intervals and tests of significance for the mean `$\mu$` of a population
 
In general, our analysis relied on two conditions:

- The data is from a .hi[simple random sample] from the population

- Observations from the population have a .hi[normal distribution] with a mean, `$\mu$`, which is generally unknown and a variance `$\sigma^2$`, which we've been assuming is known.

We'll now talk about the situation where `$\mu$` and `$\sigma^2$` are both unknown.

---
# Important Reminder

For a sample mean `$\bar{X}_n$`, the sampling distributino has the following variance and standard deviation:

$$
\text{Variance} = \frac{\sigma^2}{n}
$$

$$
\text{Standard Deviation} = \frac{\sigma}{\sqrt{n}}
$$

---
# When `$\sigma^2$` is Known

Whenever we know the population variance, `$\sigma^2$`, we base confidence intervals and tests for `$\mu$` with the z-statistic:

$$
Z = \frac{\bar{X}-\mu}{\sigma/\sqrt{n}} \implies Z \sim N(0,1)
$$

---

# When `$\sigma^2$` is Unknown

When we don't know the true variance, we must estimate it using the sample variance `$s^2$`. Again, since we're talking about sample means, the standard deviation of a sample mean, based of a .it[population with unknown variance] is:

$$
\frac{\sigma^2}{n} \text{ which is estimated by } \frac{s^2}{n} 
$$

`$\frac{s}{\sqrt{n}}$` is called the .hi.alice[standard error].

---
# When `$\sigma^2$` is Unknown

When we don't know `$\sigma$`, we estimate it by the .it[sample standard deviation], `$s$`. What happens when we standardize?

$$
\frac{\bar{X}-\mu}{s/\sqrt{n}} \sim \quad ?
$$

<br/>
.center.it[This new statistic does not have a normal distribution!]

---
# The `$t$` Distribution

When we standardize based on the sample standard deviation, `$s$`, our statistic has a new distribution called a .hi.daisy[t-distribution]

Consider a simple random sample of size `$n$`, the .hi.daisy[t-statistic]:

$$
\daisy{t} = \frac{\bar{X}-\mu}{s/\sqrt{n}}
$$

has the .daisy[t-distribution] with `$n-1$` degrees of freedom

The .daisy[t-statistic] has the same interpretation as z-statistic: it says how far away `$\bar{X}$` is from its mean `$\mu$`, in standard deviation units.

- The distribution is different because we have one aditional source of .it[uncertainty]: we estimate the standard deviation with `$s$`.

---
# The `$t$`-Distribution

There is a different `$t$` distribution for each sample size, specified by its .hi.coral[degrees of freedom]

- We denote the `$t$` distribution with `$n-1$` degrees of freedom as `$t_{n-1}$`

- The curve has a different shape than the standard normal curve

- It is still symmetric about it's peak at 0
      
  - But, it has more area in the tails.

---
# "More Area in the Tails"?

.ex[Example:] Let's say I want to predict the height of the class by selecting `$n = 3$` people and taking their mean. Sometimes I will pick 3 basketball players or 3 short people.

- If I am using `$n = 20$` people, then it is much more rare to select only tall or only short people (this means that extreme sample means are more rare).

- I might still select the tallest or shortest person in the class, but they're only one of 20 instead of one of 3.

---
# Graphs of `$t$`-Distributions

---
# Graphs of `$t$`-Distributions

Greater degree of dispersion in the `$t$` distribution
      
- It has an additional source of random variability in the sample standard deviation `$s$`
      
The variance from estimating `$s$` decreases when the sample size increases
      
- This means the t-distribution looks more like the z-distribution as sample size grows to infinity

- Once degrees of freedom exceeds 31, the t-distribution is close enough to standard normal to approximate probabilities using the z-distribution

---
# Finding `$t$`-values

---
# Finding `$t$`-values
As you see, once we have more than 30 degrees of freedom, we can use the z-scores

<br/>

---
# Confidence Interval with `$t$`-Distribution

Calculating a confidence interval:

If the variance, `$\sigma^2$`, is known:
      
- Use Z-distribution
      
If the variance, `$\sigma^2$`, is unknown:
      
- 31 or more observations in sample, use Z-distribution

- 30 or fewer observations in sample, use t-distribution

---
# Example
Say you want to construct a 95% confidence interval using 20 observations from population with unknown `$\mu$` and `$\sigma$`, with a sample mean, `$\bar{X}=22.21$` and a sample standard deviation, `$s=1.963$`.

`$\sigma$` is unknown and `$n \leq 31 \implies$` use t-distribution:

$$
CI = \bar{X} \pm t^* \frac{s}{\sqrt{n}}
$$

where `$t^*$` is the critical value:

$$
t^{\frac{1-C}{2}}_{df} = t_{19}^{0.025}
$$

---
# Example

Find `$t^{0.025}_{19}$` using t-distribution table:

.pull-left[
<img src="data:image/png;base64,#t_table3.png" width="100%" style="display: block; margin: auto;" />
]
.pull-right[
`$$t^{0.025}_{19}=2.093$$`
]

---
# Example

`$$CI = \bar{X} \pm t^* \frac{s}{\sqrt{n}}$$`

`$$CI = 22.21 \pm 2.093 \cdot \frac{1.963}{\sqrt{20}}$$`
 
Which means the confidence interval is [21.29, 23.13].

We are 95\% confident that the population mean is between 21.29 and 23.13.

---
# Clicker Question
Which table would you use for the following confidence interval?

99% confidence interval with `$n=1000$` observations

<ol type = "a">
  <li>z-table</li>
  <li>t-table</li>
</ol>

---
# Clicker Question
What critical value `$t^*$` would you use for the following confidence interval?

90% confidence interval with `$n=2$` observations

---
# Clicker Question
What critical value `$t^*$` would you use for the following confidence interval?

95% confidence interval with `$n=16$` observations

</div>

---
# Clicker Question -- Midterm Example
A randomly sampled group of patients at a major U.S. regional hospital became part of a nutrition study on dietary habits. Part of the study consisted of a 50-question survey asking about types of foods consumed. Each question was scored on a scale from one (most unhealthy behavior) to five (most healthy behavior). The answers were summed and averaged. The population of interest is the patients at the regional hospital. A sample of `$n = 15$` patients produced the following statistics `$\bar{X}=3.3$` and `$s=1.2$`. A 99% confidence interval is given by:

---
# Hypothesis Test with `$t$`-Distribution

Like the confidence interval, the `$t$` test is very similar to the `$Z$` test introduced earlier.

If we have  SRS of size `$n$` from population with unknown `$\mu$` and `$\sigma$`. To test the hypothesis: `$H_0: \mu = \mu_0$`, compute the .hi.daisy[t-statistic]:

$$
t=\frac{\bar{X}-\mu_0}{s/\sqrt{n}}
$$

We then use this t-statistic to calculate the p-value

- The p-values are exact if the population does happen to be normally distributed

- Otherwise, they are approximately correct for large enough `$n$`

---
# p-Values in `$t$`-Distribution

---
# Example
Suppose we have a sample of 12 observations and calculate a sample mean `$\bar{X}=113.75$` and a sample standard deviation, `$s=93.90$`, and we want to test the following hypothesis at the `$\alpha=0.10$` level:

$$ 
H_0: \mu=88 
$$

$$ 
H_1: \mu > 88 
$$

The set up is similar to how we've done hypothesis testing before, but we must use the t distribution.

$$ 
P(\bar{X} > 113.75 \ \vert \ \mu = 88) 
$$

---
# Example
When we "standardize", we are going to be using the t-distribution since we don't know the population variance and we have a small sample.

`$$\begin{align*}
P(\bar{X} > 113.75 \ \vert \ \mu=88) &= P(\frac{\bar{X}-\mu_0}{s/\sqrt{n}} > \frac{113.75-88}{93.90/\sqrt{12}}) \\
&= P(t_{11}>0.95)
\end{align*}$$`

---
# Example
Our next step is to find the p-value associated with that t-statistic

We can pin down the p-value between two points by using the t-table used earlier. 
      
- Look at the row, df=11
- See which columns 0.95 lies between, and look at their associated probabilities

---
# Example

We see that along the row with 11 degrees of freedom,  Our t-statistic (0.95) is between 0.25, which has a t-statistic of 0.697, and 0.10, which has a t-statistic of 1.36.

We conclude that `$0.10 \leq p-\text{value} \leq 0.25$`

---
# Example
Our level of significance is `$\alpha=0.10$`, and we concluded that the p-value associated with this hypothesis is test is greater than 0.10.

Therefore:

$$
\alpha \leq p-\text{value} \implies \text{ Do not reject } H_0
$$

That main difference between solving a hypothesis test with z-statistic versus t-statistic is that we will generally only be able to find a range for the p-value with t-statistic, whereas we found an exact value with a z-statistic.

- You would need 30 seperate `$t$` tables to have matching exactness of Z-table.

---
# Clicker Question

Say we have calculated a `$t$`-statistic of 1.6 from a sample of 24 observations. Do we reject at the `$\alpha=0.10$` level? What about the `$\alpha=0.05$` level?

<ol type = "a">
  <li>Reject at both $ \alpha=0.05 $ and $ \alpha=0.10$</li>
  <li>Do not reject at both $ \alpha=0.05 $ and $ \alpha=0.10$</li>
  <li>Reject at $ \alpha=0.05 $, but do not reject at  $ \alpha=0.10$</li>
  <li>Reject at $ \alpha=0.10 $, but do not reject at  $ \alpha=0.05$</li>
</ol>

---
# Hypothesis Testing with t-Distribution
When we conduct a hypothesis test we follow these steps:

1. Calculate the test statistic
      `$$t=\frac{\bar{X}-\mu_0}{s/\sqrt{n}}$$`

2. Determine range of p-values associated with test statistic using t-table

- Row is determined by degrees of freedom, `$n-1$`

- Find which two columns your t-statistic lies between

3. Similar to hypothesis testing with a Z distribution, if using a two-tailed test, multiply `$p$`-value by 2.

4. Compare range of p-values to `$\alpha$`

---
# Work in Groups

A researcher collected a sample of 12 fish and measured gill length. They calculated `$\bar{X} = 3.12$` and `$s = 0.7$`. Test the following hypothesis at the `$\alpha = 0.05$` level:

$$
H_0: \mu = 3.5
$$

$$
H_1: \mu < 3.5
$$

---
# Example
Observe sample of 27 newborns in a hospital near Chernobyl and record number of fingers. You calculate `$\bar{X} = 9.2$` and `$s = 1.8$`. You want to test to the following hypothesis at the `$\alpha=0.10$` level:
$$
H_0:\mu=10
$$
$$
H_1: \mu \neq 10
$$

---
# Rejection Region from `$t$`-Table

Lookup the critical value `$t^*$`, based off the degrees of freedom and level of significance, `$\alpha$`

Right tailed test:

- Reject `$H_0$` when `$t > t^*$`

Left tailed test:

- Reject `$H_0$` when `$t < -t^*$`

Can work back out the rejection region to find range for `$\bar{X}$`

$$
\daisy{t} = \frac{\bar{X}-\mu}{s/\sqrt{n}} > t^* \implies \bar{X} > t^{*} \frac{s}{\sqrt{n}} + \mu
$$

---
# Example
Suppose you collect a sample of 17 family incomes. You calculate $s = 6,000$. You want to test the following hypothesis:

$$ 
H_0: \mu = \$29,000
$$

$$ 
H_1: \mu > \$29,000
$$

If we test at the `$\alpha = 0.05$` level, what is our rejection region?

---
# Example
Suppose you collect a sample of 25 students and record their test scores. You calculate `$s=4.2$`. If we're testing the following hypothesis:
$$ 
H_0: \mu=82
$$

$$ 
H_1: \mu<82
$$

If we're testing at the `$\alpha=0.01$`, when do we reject the null hypothesis?

---
# Example

Back to our newborn example. Observe sample of 27 newborns in a hospital near Chernobyl and record number of fingers. You calculate `$s = 1.8$`. If we're testing the following hypothesis at the `$\alpha = 0.05$` level, what is the rejection region?

$$ 
H_0:\mu=10
$$

$$ 
H_1: \mu \neq 10
$$

---
# When to Use Which Distribution

If we have SRS, Normal X, and `$\sigma$` known:

- Use Z distribution

If we have SRS, Normal X, and `$\sigma$` unknown:

- Use the t distribution

.purple[What if we don't know that X is normal?]

- `$n<15 \implies$` only use t if X looks very normally distributed.

- `$15 < n < 31 \implies$` use t as long as there are no extreme outliers

- `$n>31$`, probably okay using Z

Recall Law of Large numbers that says as `$n \to \infty$`:
$$ 
\frac{\bar{X}-\mu}{\sigma/\sqrt{n}} \to^d Z\sim(0,1)
$$