Testing Hypotheses

class: center, middle, inverse, title-slide

.title[
# Testing Hypotheses
]
.subtitle[
## EDP 613
]
.author[
### Week 9
]

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# A Note About The Slides

Currently the equations do not show up properly in Firefox. Other browsers such as Chrome and Safari do work

---

# A Note About Previous Items

We're going to use some introduce some concepts from Chapter 7 here as well

---

# Definitions

The **margin of error** (**MoE**) is

>- *Formally* : the range of values above and below a sample statistic within a confidence interval

>- *In a nutshell* : how many percentage points your results will differ from the real population value

>- NOT the same as a **confidence interval**

(Statistical) **significance** is

>- *Formally* : a measure of the probability of the null hypothesis being true compared to the acceptable level of uncertainty regarding the true answer

>- *In a nutshell* : a result is probably not due to chance so it is likely real

>- NOT the same as a **practical significance**

---

# Interpretation

.pull-left[
<center>
Example result: 
A 95% confidence interval with a 3% margin of error
</center>
]

.pull-right[
<center>
What it means: 
Your statistic will be within 3 percentage points of the real population value 95% of the time
</center>
]

<center>
The MoE is a probability!
</center>

---

# Back to Hypothesis Testing

<center>Recall</center>

.pull-left[
The **null hypothesis `$H_0$`** states

>- *Formally* : that a parameter is equal to a specific value

>- *Informally* : nothing probably happened
]

.pull-right[
The **alternative hypothesis `$H_1$`** states

>- *Formally* : that a parameter differs from the value specified by the null hypothesis

>- *Informally* : something probably happened
]

---

# More about Hypothesis Testing

Say a null hypothesis is `$H_0\colon\mu=50$`. Then three things can occur from a frequentist perspective

- `$H_1 < 50$`: alternative hypothesis states that the parameter is *less* than the value of the null.

>- You know to test for everything to the **left** of `$H_1 = 50$`.

>- Called a **left-tailed test**

- `$H_1 > 50$`: alternative hypothesis states that the parameter is *more* than the value of the null.

>- You know to test for everything to the **right** of `$H_1 = 50$`.

>- Called a **right-tailed test**

- `$H_1 \neq 50$`: alternative hypothesis states that the parameter is *note* the value of the null.

>- You know to test for everything to the **left** and **right** of `$H_1 = 50$`.

>- Called a **two-tailed test**

---

# `$H_1 < H_0$`: Left-Tailed Test

---

# `$H_1 > H_0$`: Right-Tailed Test

---

# `$H_1 \neq H_0$`: Two-Tailed Test

---

# Testing Methods

.pull-left[
<center>

Critical Value (CV )
</center>
]

.pull-right[
<center>

`p` - value
</center>
]

---

# Critical Value

- Uses a **test statistic** - determines how strong the disagreement between a sample mean and a null hypothesis

- Idea: We should reject `$H_0$` if the value of the test statistic is unusual when we assume `$H_0$` to be true

- Process

- We choose a *CV* which forms a boundary between values that are considered unusual and values that are not

- The region containing the unusual values is called the **critical region**
    
--

- If the value of the test statistic is in the critical regions, we *reject* `$H_0$`

---

# Transitioning (Again!)

<center>
Going from a `z`-distribution 
known population variance using `z`-scores
</center>

<center>
a `t`-distribution 
known sample variance using `t`-tests
</center>

---

# Note

We have to assume that for a large enough sample size, the `$t$`-distribution will closely match, or estimate, the a `$z$`-distribution

---

# Things to note

- `$t$`-distribution table can be located in Appendix C

- Assumptions

- *Normality* - Samples are drawn from a population that fits a bell curve 
    
--

- *Independence* - Samples do not share values

- *Random Sampling* - Samples are randomized
  
--

- ***Homogeneity*** (for more than one sample) - Samples have a similar makeup

---

# Steps to Solving

1. *Interpret the Question into Layman's Terms*

2. *Set Acceptable Threshold of Committing a Type I Error*

3. *State the Research Hypothesis*

4. *Calculate the Test Statistic*

5. *Determine the Critical Value*

6. *State the Decision Rule*

7. *Interpret the Results*

---

# One-sample `$t$`-test

- allows us to determine whether the mean of a sample data set is different than a known value

- used when the population variance is not known

- can be used when the sample size is small ~ typically `$N<30$`.

---

# What the heck is a *degree of freedom*?

1. Forget statistics

2. Say you only own seven hats and want to wear a different one each day of the week.

3. Process
 - Day 1: Choose from 7
 
 - Day 2: Choose from 6 
 &emsp;&emsp;&emsp;&emsp;. 
 &emsp;&emsp;&emsp;&emsp;.

<ol class="nostyle"> 
 <li class="nostyle">
 <ul> 
 <li class="nostyle">&emsp;&emsp;&emsp;&emsp;.</li>
 <li class="nostyle">&emsp;&emsp;&emsp;&emsp;.</li>
 
 <li>Day 6: Choose from 2</li>
 
 <li>Day 7: Choose from 1</li>
 </ul>
 </li>
</ol>

<ol start=4>
You had 7 - 1 = 6 days of hat freedom!
</ol>

<center>
That's essentially a degree of freedom (written df )
</center>

---

# One Sample Mean

.center2[
.pull-left[
<center>
t-distribution 
</center>

`$$t = \frac{\overline{Y}-\mu}{s/\sqrt{N}}$$`
]

.pull-right[
<center>
with degrees of freedom 
</center>

`$$df = N-1$$`
]
]

---

# Example

.center2[

<table class="table" style="width: auto !important; margin-left: auto; margin-right: auto;">
 <thead>
 <tr>
 <th style="text-align:center;background-color: #212121 !important;"> Region </th>
 <th style="text-align:center;background-color: #212121 !important;"> Median Household Income </th>
 <th style="text-align:center;background-color: #212121 !important;"> Median Male Salary </th>
 <th style="text-align:center;background-color: #212121 !important;"> Median Female Salary </th>
 <th style="text-align:center;background-color: #212121 !important;"> Population </th>
 <th style="text-align:center;background-color: #212121 !important;"> Households </th>
 </tr>
 </thead>
<tbody>
 <tr>
 <td style="text-align:center;width: 10em; "> United States </td>
 <td style="text-align:center;width: 10em; "> $68,703 </td>
 <td style="text-align:center;width: 10em; "> $57,511 </td>
 <td style="text-align:center;width: 10em; "> $43,820 </td>
 <td style="text-align:center;width: 10em; "> 331,449,281 </td>
 <td style="text-align:center;"> 139,684,244 </td>
 </tr>
 <tr>
 <td style="text-align:center;width: 10em; background-color: #212121 !important;"> West Virginia </td>
 <td style="text-align:center;width: 10em; background-color: #212121 !important;"> $48,850 </td>
 <td style="text-align:center;width: 10em; background-color: #212121 !important;"> $57,456 </td>
 <td style="text-align:center;width: 10em; background-color: #212121 !important;"> $47,299 </td>
 <td style="text-align:center;width: 10em; background-color: #212121 !important;"> 1,793,716 </td>
 <td style="text-align:center;background-color: #212121 !important;"> 763,831 </td>
 </tr>
</tbody>
</table>
]

Is the the median household income of West Virginia counties different than the national average?

---

## *Interpret the Question into Layman's Terms*

From Median Household Income, we think WV (48,850 USD) is *practically lower* than the US (68,850 USD), but is it *significantly lower *?

---

## *Set Acceptable Threshold of Committing a Type I Error*

`$\alpha =0.5$`

---

## *State the Research Hypothesis*

.pull-left[
`$$H_0$$` 
The Median Household Income of West Virginia counties is NOT significantly less than the national average
]

.pull-right[
`$$H_1$$` The Median Household Income of West Virginia counties is significantly less than the national average
]

---

## *Calculate the Test Statistic* (1/2)

.pull-left[
<center>Population</center> 
$$\mu = \$68,703$$
]

.pull-right[
<center>Sample</center> 
`$$N=61$$` $$s = \$5075.28$$ $$\overline{Y} = \$45732.20$$
]

--
 
 
 
<center>df</center> 
`$$61-1 = 60$$`

---

## *Calculate the Test Statistic* (2/2)

`\begin{aligned}
t &= \dfrac{45732.20-68703.00}{5075.28/\sqrt{61}} \\\\
 &\approx -0.579
\end{aligned}`

---

## *Determine the Critical Value*

In Appendix C, we see that for `$df = 60$` at `$\alpha =0.05$`, we have `$t_{crit} = 1.671$`

---

## *State the Decision Rule*

Since `$-0.579 < 1.671$` we reject `$H_0$`

---

## *Interpret the Results*

*The Median Household Income is significantly less than the national average!*
 
---

# Two-sample Mean

Used to compare one sample mean to another.

- We use two different test:

- Equal variances
  
  - Unequal variances (assumed)

- **Homoscedasticity** – the assumption of equal variances.

---

# Difference Between Two Independent Means

>- Observations in each sample are not related

>- Need to compare differences between the sample means

.pull-left[
<center>Estimated Standard Error</center> 
`$$S_{\overline{Y_1}-\overline{Y_1}} = \sqrt{\dfrac{(N_1-1)\cdot s_1^2+(N_2-1)\cdot s_2^2}{(N_1+N_2)-2}}\cdot\sqrt{\dfrac{N_1+N_2}{N_1\cdot N_2}}$$`
]

.pull-right[
<center>difference between means t-statistic</center> 
`$$t = \dfrac{\overline{Y_1}-\overline{Y_2}}{S_{\overline{Y_1}-\overline{Y_1}}}$$`
]

--
 
 
 
<center>df</center> 
`$$df = (N_1+N_2)-2$$`

---

# Difference Between Proportions

.pull-left[
<center>Estimated Standard Error</center> 
`$$S_{p_1 - p_2} = \sqrt{\dfrac{p_1(1-p_1)}{N_1} + \dfrac{p_2(1-p_2)}{N_2}}$$`

.pull-right[
<center>difference between means z-statistic</center> 
`$$z = \dfrac{p_1-p_2}{S_{p_1-p_2}}$$`
]

---

# Example

As part of the Pew Internet and American Life Project, researchers conducted two surveys in late 2009. The first survey asked a random sample of 800 U.S. teens about their use of social media and the Internet. A second survey posed similar questions to a random sample of 2253 U.S. adults. In these two studies, 73% of teens and 47% of adults said that they use social-networking sites. Use these results to construct and interpret a 95% confidence interval for the difference between the proportion of all U.S. teens and adults who use social-networking sites.

---

## *Interpret the Question into Layman's Terms*

Is there a difference between the percent of teens and adults who use social networking sites?

---

## *Set Acceptable Threshold of Committing a Type I Error*

`$\alpha = 0.05$`

---

## *State the Research Hypothesis*

.pull-left[
`$$H_0$$` 
There is no difference between the proportion of teens and adults who use social media
]

.pull-right[
`$$H_1$$` 
There is a difference between the proportion of teens and adults who use social media
]

---

## *Calculate the Test Statistic*

`\begin{align}
S_{p_1 - p_2} &= \sqrt{\dfrac{0.73(1-0.73)}{800} + \dfrac{0.47(1-0.47)}{2253}}\\\\
&\approx 0.0189
\end{align}`
 
 
`\begin{align}
z &= \dfrac{0.73-0.47}{0.0189}\\
&\approx 13.76
\end{align}`

---

## *State the Decision Rule*

`$z=13.76$` is greater than `$p$` value implying that we reject `$H_0$`

---

## *Interpret the Results*

We are 95% confident that in late 2009 more teens than adults in the United States engaged in social media

---

## That's it. Take a break before our R session!