Chi-Squared Goodness of Fit and Independence

Inference on Categorical Variables with Multiple Levels

Dr. Gilbert

October 31, 2024

Reminder of Inference and Inferential Tools

We use statistical inference to make or test claims about population parameters which we cannot measure directly

• We make claims by constructing confidence intervals
• We test claims by conducting hypothesis tests

Confidence intervals provide a range of plausible values for a population parameter

• They are centered at the point estimate (sample statistic)
• They open up some “wiggle room” called a margin of error, which is influenced by the critical value and the standard error

\[\left(\begin{array}{c}\text{point}\\ \text{estimate}\end{array}\right) \pm \left(\begin{array}{c}\text{critical}\\ \text{value}\end{array}\right)\left(\begin{array}{c}\text{standard}\\ \text{error}\end{array}\right)\]

Reminder of Inference and Inferential Tools

We use statistical inference to make or test claims about population parameters which we cannot measure directly

• We make claims by constructing confidence intervals
• We test claims by conducting hypothesis tests

Confidence intervals provide a range of plausible values for a population parameter

• They are centered at the point estimate (sample statistic)
• They open up some “wiggle room” called a margin of error, which is influenced by the critical value and the standard error

\[\left(\begin{array}{c}\text{point}\\ \text{estimate}\end{array}\right) \pm \boxed{\left(\begin{array}{c}\text{critical}\\ \text{value}\end{array}\right)\left(\begin{array}{c}\text{standard}\\ \text{error}\end{array}\right)}\]

Reminder of Inference and Inferential Tools

We use statistical inference to make or test claims about population parameters which we cannot measure directly

• We make claims by constructing confidence intervals
• We test claims by conducting hypothesis tests

Confidence intervals provide a range of plausible values for a population parameter

• They are centered at the point estimate (sample statistic)
• They open up some “wiggle room” called a margin of error, which is influenced by the critical value and the standard error

\[\left(\begin{array}{c}\text{point}\\ \text{estimate}\end{array}\right) \pm \left(\begin{array}{c}\text{critical}\\ \text{value}\end{array}\right)\left(\begin{array}{c}\text{standard}\\ \text{error}\end{array}\right)\]

Inferential Tools (Continued)

Hypothesis tests provide a framework for testing claims about a population parameter

1. Assume the claim is false, in reality (null hypothesis)

2. Measure the probability of observing a sample at least as extreme as ours in that reality (this is called the \(p\)-value)

   • If our observed data is “unlikely” (\(p\)-value is lower than the level of significance, \(\alpha\)), then what we’ve observed is incompatible with our assumed reality; we declare evidence that the null hypothesis is false and accept the alternative hypothesis instead
   • If our observed data is not “unlikely” (\(p\)-value at least as large as the level of significance), then our observed data is compatible with our assumed reality – we don’t reject the null hypothesis

Where We Are; Where We’re Going…

Inference On...	Covered
One Binary Categorical Variable	✔
Association Between Two Binary Categorical Variables	✔

Where We Are; Where We’re Going…

Inference On...	Covered
One Binary Categorical Variable	✔
Association Between Two Binary Categorical Variables	✔
One MultiClass Categorical Variable	Today
Associations Between Two MultiClass Categorical Variables	Today

Where We Are; Where We’re Going…

Inference On...	Covered
One Binary Categorical Variable	✔
Association Between Two Binary Categorical Variables	✔
One MultiClass Categorical Variable	Today
Associations Between Two MultiClass Categorical Variables	Today
One Numerical Variable
Association Between a Numerical Variable and a Binary Categorical Variable
Association Between a Numerical Variable and a MultiClass Categorical Variable
Association Between a Numerical Variable and a Single Other Numerical Variable
Association Between a Numerical Variable and Many Other Variables

Where We Are; Where We’re Going…

Inference On...	Covered
One Binary Categorical Variable	✔
Association Between Two Binary Categorical Variables	✔
One MultiClass Categorical Variable	Today
Associations Between Two MultiClass Categorical Variables	Today
One Numerical Variable
Association Between a Numerical Variable and a Binary Categorical Variable
Association Between a Numerical Variable and a MultiClass Categorical Variable
Association Between a Numerical Variable and a Single Other Numerical Variable
Association Between a Numerical Variable and Many Other Variables
Association Between a Categorical Variable and Many Other Variables	✘

Reminder: Inference on a Single Categorical Variable

We’ve been focused on binary (two-class) categorical variables

The single-variable questions we’ve asked are of the form:

• Can we estimate the population proportion?

  • For example, with 95% confidence, what is the proportion of likely voters in New Hampshire who are planning to vote in favor of Amendment 1?

• Is the population proportion greater/less/different than some proposed value?

  • For example, is the proportion of likely voters in New Hampshire who favor Amendment 1 at least 66.67%?

But what if we were interested in categorical variables that have more than just two levels?

Are ideological alignments of voting-aged citizens in the US uniformly distributed across the categories very liberal, liberal, moderate, conservative, and very conservative?

Reminder: Inference on Associations Between Two Categorical Variables

Our multivariable questions have been of the form:

• Can we estimate the difference in population proportions between Group A and Group B?

  • For example, Find a 90% confidence interval for the difference in the proportion of students who feel a sense of belonging at their university between first-year students and seniors.

• Is the population proportion in Group A greater/less/different than the population proportion in Group B?

  • For example, Is the proportion of students who feel a sense of belonging at their university greater for seniors than for first-year students?

What about associations between categorical variables where at least one has three or more levels?

Is there an association between and individual’s ideology and their perception of the state of their finances (better off, worse off, or about the same) relative to four years ago?

Highlights

• Analysing the form of a test statistic

• The need for a different test statistic

• The need for a new probability distribution

• Chi-Squared Tests for Goodness of Fit (inference on a single, multiclass categorical variable)

  • A Completed Example

• Chi-Squared Tests for Independence (inference on associations between two potentially multiclass categorical variables)

  • A Probability Review
  • A Completed Example

• Additional Examples

A Closer Look at a Test Statistic

So far, I’ve told you that a test statistic takes the form:

\[\begin{array}{c}\text{test}\\ \text{statistic}\end{array} = \frac{\left(\begin{array}{c}\text{point}\\ \text{estimate}\end{array}\right) - \left(\begin{array}{c}\text{null}\\ \text{value}\end{array}\right)}{S_E}\]

• The point estimate comes from our sample data – it is our observed value
• The null value comes from our null hypothesis – it is our expected outcome

Another way to phrase the test statistic formula then is

\[\begin{array}{c}\text{test}\\ \text{statistic}\end{array} = \frac{\text{observed} - \text{expected}}{S_E}\]

For example, if our null hypothesis assumed that 50% of individuals have characteristic “A” and a sample of 200 people included 95 that did, then our observed proportion is 0.475 and our expected proportion was 0.50, so our sample was about 2.5 “percentage points” away from the expected sample.

A New Test Statistic

With binary categorical variables, measuring the proportion associated with just a single category was sufficient – if we know the proportion associated with one outcome, we also know the proportion associated with the other

For example, if we surveyed 100 voters in the US and asked them if they identify more closely with a liberal ideology or a conservative ideology and the results were

Liberal	Conservative
54	?

Then you know what the full table looks like…

Liberal	Conservative
54	46

A New Test Statistic

With binary categorical variables, measuring the proportion associated with just a single category was sufficient – if we know the proportion associated with one outcome, we also know the proportion associated with the other

With multiclass categorical variables, this is not the case

Very Liberal	Liberal	Moderate	Conservative	Very Conservative
?	?	?	?	?

A New Test Statistic

With binary categorical variables, measuring the proportion associated with just a single category was sufficient – if we know the proportion associated with one outcome, we also know the proportion associated with the other

With multiclass categorical variables, this is not the case

Very Liberal	Liberal	Moderate	Conservative	Very Conservative
35	?	?	?	?

A New Test Statistic

With binary categorical variables, measuring the proportion associated with just a single category was sufficient – if we know the proportion associated with one outcome, we also know the proportion associated with the other

With multiclass categorical variables, this is not the case

Very Liberal	Liberal	Moderate	Conservative	Very Conservative
?	?	?	?	?

If we collected data from a sample of 500 citizens and political ideology was uniformly distributed, then we would expect:

Very Liberal	Liberal	Moderate	Conservative	Very Conservative
100	100	100	100	100

But what if we observed:

Very Liberal	Liberal	Moderate	Conservative	Very Conservative
35	105	175	140	45

A New Test Statistic

Expected results from 500 surveyed individuals:

Very Liberal	Liberal	Moderate	Conservative	Very Conservative
100	100	100	100	100

Observed results from 500 surveyed individuals:

Very Liberal	Liberal	Moderate	Conservative	Very Conservative
35	105	175	140	45

Comparing a single observed value to a single expected value is no longer enough to describe our scenario

We could calculate the difference observed - expected for each group though

Very Liberal	Liberal	Moderate	Conservative	Very Conservative
35 - 100 = -65	105 - 100 = 5	175 - 100 = 75	140 - 100 = 40	45 - 100 = -55

A New Test Statistic

Differences (observed - expected) in results from 500 surveyed individuals:

Very Liberal	Liberal	Moderate	Conservative	Very Conservative
-65	5	75	40	-55

We need some way to combine/summarize these deviations.

A New Test Statistic

Differences (observed - expected) in results from 500 surveyed individuals:

Very Liberal	Liberal	Moderate	Conservative	Very Conservative
-65	5	75	40	-55

We need some way to combine/summarize these deviations. If we just add these up, we’ll get 0 because the positive and negative differences will cancel one another out.

“Small” deviations are unsurprising, while “large” deviations should be more rare. To emphasize this, we’ll square the differences, penalizing larger deviations more heavily.

A “large” deviation is relative – a deviation of 50 is very large if the expected count was only 20 to begin with, but a deviation of 50 is quite small if the expected count was 1000. We’ll divide each squared deviation by the expected count to compensate for this.

Our resulting test statistic for this scenario takes the form

\[\sum_{i = 1}^{k}{\frac{\left(\text{observed} - \text{expected}\right)^2}{\text{expected}}}\]

where \(k\) is the number of categories/groups.

The Chi-Squared Distribution

This new test statistic doesn’t follow a normal distribution.

The Chi-Squared Distribution

This new test statistic doesn’t follow a normal distribution. It instead follows a Chi-Squared distribution.

Actually, the Chi-Squared distribution isn’t a single distribution – it is a family of distributions defined by a single parameter…degrees of freedom.

We’ll talk more about degrees of freedom when we get to the actual tests but, for now, here are a few Chi-Squared distributions.

The Chi-Squared Distribution

This new test statistic doesn’t follow a normal distribution. It instead follows a Chi-Squared distribution.

Actually, the Chi-Squared distribution isn’t a single distribution – it is a family of distributions defined by a single parameter…degrees of freedom.

We’ll talk more about degrees of freedom when we get to the actual tests but, for now, here are a few Chi-Squared distributions.

The Chi-Squared distributions are defined over non-negative numbers only, are right-skewed, and [for our course] we’ll only ever be interested in the area in the right tail.

The Chi-Squared Distribution

This new test statistic doesn’t follow a normal distribution. It instead follows a Chi-Squared distribution.

Actually, the Chi-Squared distribution isn’t a single distribution – it is a family of distributions defined by a single parameter…degrees of freedom.

We’ll talk more about degrees of freedom when we get to the actual tests but, for now, here are a few Chi-Squared distributions.

The Chi-Squared distributions are defined over non-negative numbers only, are right-skewed, and [for our course] we’ll only ever be interested in the area in the right tail.

1 - pchisq(q, df)

A Completed Example: Chi-Squared Test for Goodness of Fit

Scenario: We wonder if political ideologies (very liberal, liberal, moderate, conservative, or very conservative) are uniformly distributed. We collect data from a random sample of 500 individuals and observe the following results

Very Liberal	Liberal	Moderate	Conservative	Very Conservative
35	105	175	140	45

Run a test at the \(\alpha = 0.05\) level of significance to determine whether the distribution of political ideologies is uniform.

1. \(\begin{array}{lcl} H_0 & : & \text{The distribution is uniform}\\ H_a & : & \text{The distribution is not uniform}\end{array}\)
2. Set the level of significance at \(\alpha = 0.05\)
3. Calculate the test statistic

A Completed Example: Chi-Squared Test for Goodness of Fit

1. \(\begin{array}{lcl} H_0 & : & \text{The distribution is uniform}\\ H_a & : & \text{The distribution is not uniform}\end{array}\)
2. Set the level of significance at \(\alpha = 0.05\)
3. Calculate the test statistic

Recall the table of differences from earlier

	Very Liberal	Liberal	Moderate	Conservative	Very Conservative
Observed	35	105	175	140	45
Expected	100	100	100	100	100
Difference	-65	5	75	40	-55

\[\begin{array}{c}\text{test}\\ \text{statistic}\end{array} = \sum_{i = 1}^{5}{\frac{\left(\text{observed} - \text{expected}\right)^2}{\text{expected}}}\]

A Completed Example: Chi-Squared Test for Goodness of Fit

1. \(\begin{array}{lcl} H_0 & : & \text{The distribution is uniform}\\ H_a & : & \text{The distribution is not uniform}\end{array}\)
2. Set the level of significance at \(\alpha = 0.05\)
3. Calculate the test statistic

Recall the table of differences from earlier

	Very Liberal	Liberal	Moderate	Conservative	Very Conservative
Observed	35	105	175	140	45
Expected	100	100	100	100	100
Difference	-65	5	75	40	-55

\[\begin{array}{c}\text{test}\\ \text{statistic}\end{array} = \frac{\left(-65\right)^2}{100} + \frac{5^2}{100} + \frac{75^2}{100} + \frac{40^2}{100} + \frac{\left(-55\right)^2}{100}\]

A Completed Example: Chi-Squared Test for Goodness of Fit

1. \(\begin{array}{lcl} H_0 & : & \text{The distribution is uniform}\\ H_a & : & \text{The distribution is not uniform}\end{array}\)
2. Set the level of significance at \(\alpha = 0.05\)
3. Calculate the test statistic

Recall the table of differences from earlier

	Very Liberal	Liberal	Moderate	Conservative	Very Conservative
Observed	35	105	175	140	45
Expected	100	100	100	100	100
Difference	-65	5	75	40	-55

\[\begin{array}{c}\text{test}\\ \text{statistic}\end{array} = 145\]

A Completed Example: Chi-Squared Test for Goodness of Fit

1. \(\begin{array}{lcl} H_0 & : & \text{The distribution is uniform}\\ H_a & : & \text{The distribution is not uniform}\end{array}\)
2. Set the level of significance at \(\alpha = 0.05\)
3. Calculate the test statistic

Alternatively,

observed <- c(35, 105, 175, 140, 45)
expected <- c(100, 100, 100, 100, 100)

sum((observed - expected)^2/expected)

[1] 145

Recall the table of differences from earlier

	Very Liberal	Liberal	Moderate	Conservative	Very Conservative
Observed	35	105	175	140	45
Expected	100	100	100	100	100
Difference	-65	5	75	40	-55

\[\begin{array}{c}\text{test}\\ \text{statistic}\end{array} = \frac{\left(-65\right)^2}{100} + \frac{5^2}{100} + \frac{75^2}{100} + \frac{40^2}{100} + \frac{\left(-55\right)^2}{100} = \boxed{~145~}\]

A Completed Example: Chi-Squared Test for Goodness of Fit

1. \(\begin{array}{lcl} H_0 & : & \text{The distribution is uniform}\\ H_a & : & \text{The distribution is not uniform}\end{array}\)
2. Set the level of significance at \(\alpha = 0.05\)
3. Calculate the test statistic

\[\chi^2~\text{test statistic} = 145\]

4. Calculate the \(p\)-value using the Chi-Squared distribution

A Completed Example: Chi-Squared Test for Goodness of Fit

1. \(\begin{array}{lcl} H_0 & : & \text{The distribution is uniform}\\ H_a & : & \text{The distribution is not uniform}\end{array}\)
2. Set the level of significance at \(\alpha = 0.05\)
3. Calculate the test statistic

\[\chi^2~\text{test statistic} = 145\]

4. Calculate the \(p\)-value using the Chi-Squared distribution with degrees of freedom equal to one less than the number of groups (\(k - 1\))

A Completed Example: Chi-Squared Test for Goodness of Fit

1. \(\begin{array}{lcl} H_0 & : & \text{The distribution is uniform}\\ H_a & : & \text{The distribution is not uniform}\end{array}\)
2. Set the level of significance at \(\alpha = 0.05\)
3. Calculate the test statistic

\[\chi^2~\text{test statistic} = 145\]

4. Calculate the \(p\)-value using the Chi-Squared distribution with degrees of freedom equal to one less than the number of groups (\(k - 1\))

1 - pchisq(145, df = 5 - 1) \(\approx\) 0

A Completed Example: Chi-Squared Test for Goodness of Fit

1. \(\begin{array}{lcl} H_0 & : & \text{The distribution is uniform}\\ H_a & : & \text{The distribution is not uniform}\end{array}\)
2. Set the level of significance at \(\alpha = 0.05\)
3. Calculate the test statistic

\[\chi^2~\text{test statistic} = 145\]

Note: Remember that a reported \(p\)-value of “0” means that the \(p\)-value is extremely small and is being rounded to 0 – it is not possible to have a \(p\)-value which is 0 exactly.

4. Calculate the \(p\)-value using the Chi-Squared distribution with degrees of freedom equal to one less than the number of groups (\(k - 1\))

1 - pchisq(145, df = 5 - 1) \(\approx\) 0

A Completed Example: Chi-Squared Test for Goodness of Fit

1. \(\begin{array}{lcl} H_0 & : & \text{The distribution is uniform}\\ H_a & : & \text{The distribution is not uniform}\end{array}\)
2. Set the level of significance at \(\alpha = 0.05\)
3. Calculate the test statistic

\[\chi^2~\text{test statistic} = 145\]

5. The \(p\)-value is less than \(0.05\) (our desired level of significance), so we reject the null hypothesis and accept the alternative

6. The observed data are not compatible with political ideologies being distributed uniformly. The distribution is different.

4. Calculate the \(p\)-value using the Chi-Squared distribution with degrees of freedom equal to one less than the number of groups (\(k - 1\))

1 - pchisq(145, df = 5 - 1) \(\approx\) 0

Recap: Chi-Squared Goodness of Fit

We now have an ability to test whether the levels of a categorical variable over a population follow a particular distribution

Here, we tested for a uniform distribution, but we can test any generic distribution we like – the choice just influences how we calculate our expected counts

Recap: Chi-Squared Goodness of Fit

We now have an ability to test whether the levels of a categorical variable over a population follow a particular distribution

The strategy for testing Goodness of Fit follows the same general structure as the hypothesis tests we’ve seen previously: read the scenario, write the hypotheses, declare the level of significance (\(\alpha\)), calculate the test statistic, convert it to a \(p\)-value, compare the \(p\)-value to \(\alpha\), and interpret the result of the test in context

The main differences were:

1. The null hypothesis assumes the questioned distribution rather than assuming a claim is false.

2. The test statistic is a \(\chi^2\) test statistic, taking the form \(\displaystyle{\sum_{i = 1}^{k}{\frac{\left(\text{observed} - \text{expected}\right)^2}{\text{expected}}}}\)

3. The \(p\)-value is computed using the Chi-Squared distribution and we are always interested in the right tail; we use 1 - pchisq(test_statistic, df)

   • The degrees of freedom (df) is one less than the number of levels/groups under the categorical variable whose distribution we are testing.

Chi-Squared Tests for Independence

Now that we know how to perform statistical inference to test the distribution of a single categorical variable with multiple (more than two) classes, we can use a similar strategy to test for associations between two categorical variables where at least one has more than two classes.

The main differences are:

1. The hypotheses will be \(\begin{array}{lcl} H_0 & : & \text{The two variables are independent}\\ H_a & : & \text{An association exists between the two variables}\end{array}\)
2. The tables for our observed and expected counts will be organized in both rows and columns
3. Assuming that the first categorical variable has \(k\) levels and the second categorical variable has \(\ell\) levels, the degrees of freedom for the Chi-Squared distribution in this scenario will be \(\left(k - 1\right)\cdot \left(\ell - 1\right)\).

We’ll quickly review how to work with a “two-way table”, calculating probabilities of independent events, and calculating expected value before heading into an example.

Probability Review

A two-way table is a way to summarize data observed with respect to two categorical variables – its rows correspond to the levels of one of the categorical variables and its columns correspond to levels of the other categorical variable.

Consider the two-way table below which shows species of bird and the type of feeder they were observed feeding from.

	Seed Feeder	Nectar Feeder	Fruit Feeder	Total
Finches	60	15	10	85
Hummingbirds	5	70	5	80
Orioles	20	5	60	85
Woodpeckers	30	0	25	55
Total	115	90	100	305

Probability Review

Consider the two-way table below which shows species of bird and the type of feeder they were observed feeding from.

	Seed Feeder	Nectar Feeder	Fruit Feeder	Total
Finches	60	15	10	85
Hummingbirds	5	70	5	80
Orioles	20	5	60	85
Woodpeckers	30	0	25	55
Total	115	90	100	305

The probability of a randomly selected bird being a Hummingbird is

The probability of a randomly selected bird being found at a nectar feeder is

\[\mathbb{P}\left[\text{Hummingbird}\right] = \frac{80}{305} \approx 0.2262\]

\[\mathbb{P}\left[\text{Nectar Feeder}\right] = \frac{90}{305} \approx 0.2951\]

Probability Review

Consider the two-way table below which shows species of bird and the type of feeder they were observed feeding from.

	Seed Feeder	Nectar Feeder	Fruit Feeder	Total
Finches	60	15	10	85
Hummingbirds	5	70	5	80
Orioles	20	5	60	85
Woodpeckers	30	0	25	55
Total	115	90	100	305

Assuming that species and feeder preference are independent, the probability of a randomly selected bird being a hummingbird and being found at a nectar feeder would be

\[\mathbb{P}\left[\text{Hummingbird and Nectar Feeder}\right] = \mathbb{P}\left[\text{Hummingbird}\right]\cdot\mathbb{P}\left[\text{Nectar Feeder}\right]\]

Probability Review

Consider the two-way table below which shows species of bird and the type of feeder they were observed feeding from.

	Seed Feeder	Nectar Feeder	Fruit Feeder	Total
Finches	60	15	10	85
Hummingbirds	5	70	5	80
Orioles	20	5	60	85
Woodpeckers	30	0	25	55
Total	115	90	100	305

Assuming that species and feeder preference are independent, the probability of a randomly selected bird being a hummingbird and being found at a nectar feeder would be

\[\mathbb{P}\left[\text{Hummingbird and Nectar Feeder}\right] = \left(\frac{80}{305}\right)\cdot\left(\frac{90}{305}\right)\]

Probability Review

Consider the two-way table below which shows species of bird and the type of feeder they were observed feeding from.

	Seed Feeder	Nectar Feeder	Fruit Feeder	Total
Finches	60	15	10	85
Hummingbirds	5	70	5	80
Orioles	20	5	60	85
Woodpeckers	30	0	25	55
Total	115	90	100	305

Assuming that species and feeder preference are independent, the probability of a randomly selected bird being a hummingbird and being found at a nectar feeder would be

\[\mathbb{P}\left[\text{Hummingbird and Nectar Feeder}\right] \approx 0.0774\]

Probability Review

Consider the two-way table below which shows species of bird and the type of feeder they were observed feeding from.

	Seed Feeder	Nectar Feeder	Fruit Feeder	Total
Finches	60	15	10	85
Hummingbirds	5	70	5	80
Orioles	20	5	60	85
Woodpeckers	30	0	25	55
Total	115	90	100	305

Assuming that species and feeder preference are independent, the expected number of hummingbirds observed at the nectar feeder would be:

\[n\cdot\mathbb{P}\left[\text{Hummingbird and Nectar Feeder}\right] \approx 305\cdot 0.0774\]

Probability Review

Consider the two-way table below which shows species of bird and the type of feeder they were observed feeding from.

	Seed Feeder	Nectar Feeder	Fruit Feeder	Total
Finches	60	15	10	85
Hummingbirds	5	70	5	80
Orioles	20	5	60	85
Woodpeckers	30	0	25	55
Total	115	90	100	305

Assuming that species and feeder preference are independent, the expected number of hummingbirds observed at the nectar feeder would be:

\[n\cdot\mathbb{P}\left[\text{Hummingbird and Nectar Feeder}\right] \approx 23.607\]

A Completed Example: Chi-Squared Test for Independence

Scenario: A wildlife researcher is studying whether different bird species have distinct feeding preferences among three types of feeders: Seed Feeder, Nectar Feeder, and Fruit Feeder. The researcher observes four bird species in a particular park: Finches, Hummingbirds, Orioles, and Woodpeckers. Over several days, the researcher records the feeder each bird species prefers in the table below. Test whether species and feeding preference are independent at the 10% level of significance.

	Seed Feeder	Nectar Feeder	Fruit Feeder	Total
Finches	60	15	10	85
Hummingbirds	5	70	5	80
Orioles	20	5	60	85
Woodpeckers	30	0	25	55
Total	115	90	100	305

A Completed Example: Chi-Squared Test for Independence

Scenario: A wildlife researcher is studying whether different bird species have distinct feeding preferences among three types of feeders: Seed Feeder, Nectar Feeder, and Fruit Feeder. The researcher observes four bird species in a particular park: Finches, Hummingbirds, Orioles, and Woodpeckers. Over several days, the researcher records the feeder each bird species prefers in the table below. Test whether species and feeding preference are independent at the 10% level of significance.

1. The hypotheses are \(\begin{array}{lcl} H_0 & : & \text{Species and feeder preference are independent}\\ H_a & : & \text{Species and feeder preference are associated}\end{array}\)
2. The level of significance is \(\alpha = 0.10\)
3. We’ll calculate the test statistic, but we’ll need the observed and expected counts for each cell in the table first…

A Completed Example: Chi-Squared Test for Independence

Observed Counts	Seed Feeder	Nectar Feeder	Fruit Feeder	Total
Finches	60	15	10	85
Hummingbirds	5	70	5	80
Orioles	20	5	60	85
Woodpeckers	30	0	25	55
Total	115	90	100	305

Expected Counts	Seed Feeder	Nectar Feeder	Fruit Feeder
Finches	\(305\cdot\left(\frac{85}{305}\right)\cdot\left(\frac{115}{305}\right)\)	\(305\cdot\left(\frac{85}{305}\right)\cdot\left(\frac{90}{305}\right)\)	\(305\cdot\left(\frac{85}{305}\right)\cdot\left(\frac{100}{305}\right)\)
Hummingbirds	\(305\cdot\left(\frac{80}{305}\right)\cdot\left(\frac{115}{305}\right)\)	\(305\cdot\left(\frac{80}{305}\right)\cdot\left(\frac{90}{305}\right)\)	\(305\cdot\left(\frac{80}{305}\right)\cdot\left(\frac{100}{305}\right)\)
Orioles	\(305\cdot\left(\frac{85}{305}\right)\cdot\left(\frac{115}{305}\right)\)	\(305\cdot\left(\frac{85}{305}\right)\cdot\left(\frac{90}{305}\right)\)	\(305\cdot\left(\frac{85}{305}\right)\cdot\left(\frac{100}{305}\right)\)
Woodpeckers	\(305\cdot\left(\frac{55}{305}\right)\cdot\left(\frac{115}{305}\right)\)	\(305\cdot\left(\frac{55}{305}\right)\cdot\left(\frac{90}{305}\right)\)	\(305\cdot\left(\frac{55}{305}\right)\cdot\left(\frac{100}{305}\right)\)

A Completed Example: Chi-Squared Test for Independence

Observed Counts	Seed Feeder	Nectar Feeder	Fruit Feeder	Total
Finches	60	15	10	85
Hummingbirds	5	70	5	80
Orioles	20	5	60	85
Woodpeckers	30	0	25	55
Total	115	90	100	305

Expected Counts	Seed Feeder	Nectar Feeder	Fruit Feeder
Finches	32.05	25.08	27.87
Hummingbirds	30.16	23.61	26.23
Orioles	32.05	25.08	27.87
Woodpeckers	20.74	16.23	18.03

A Completed Example: Chi-Squared Test for Independence

Observed Counts	Seed Feeder	Nectar Feeder	Fruit Feeder	Total
Finches	60	15	10	85
Hummingbirds	5	70	5	80
Orioles	20	5	60	85
Woodpeckers	30	0	25	55
Total	115	90	100	305

observed <- c(60, 15, 10, 5, 70, 5, 20, 5, 60, 30, 0, 25)

Note: Be sure to exclude the total entries when you store the observed values

A Completed Example: Chi-Squared Test for Independence

Expected Counts	Seed Feeder	Nectar Feeder	Fruit Feeder
Finches	32.05	25.08	27.87
Hummingbirds	30.16	23.61	26.23
Orioles	32.05	25.08	27.87
Woodpeckers	20.74	16.23	18.03

expected <- c(32.05, 25.08, 27.87, 30.16, 23.61, 26.23, 32.05, 25.08, 27.87, 20.74, 16.23, 18.03)

Note: Be sure to input the expected counts in the same order that you used for the observed counts – otherwise you won’t be comparing the right cells to one another.

A Completed Example: Chi-Squared Test for Independence

Scenario: A wildlife researcher is studying whether different bird species have distinct feeding preferences among three types of feeders: Seed Feeder, Nectar Feeder, and Fruit Feeder. The researcher observes four bird species in a particular park: Finches, Hummingbirds, Orioles, and Woodpeckers. Over several days, the researcher records the feeder each bird species prefers in the table below. Test whether species and feeding preference are independent at the 10% level of significance.

1. The hypotheses are \(\begin{array}{lcl} H_0 & : & \text{Species and feeder preference are independent}\\ H_a & : & \text{Species and feeder preference are associated}\end{array}\)
2. The level of significance is \(\alpha = 0.10\)
3. We’ll calculate the test statistic, but we’ll need the observed and expected counts for each cell in the table first…

A Completed Example: Chi-Squared Test for Independence

Scenario: A wildlife researcher is studying whether different bird species have distinct feeding preferences among three types of feeders: Seed Feeder, Nectar Feeder, and Fruit Feeder. The researcher observes four bird species in a particular park: Finches, Hummingbirds, Orioles, and Woodpeckers. Over several days, the researcher records the feeder each bird species prefers in the table below. Test whether species and feeding preference are independent at the 10% level of significance.

1. The hypotheses are \(\begin{array}{lcl} H_0 & : & \text{Species and feeder preference are independent}\\ H_a & : & \text{Species and feeder preference are associated}\end{array}\)
2. The level of significance is \(\alpha = 0.10\)
3. We’ll calculate the test statistic, since we have our observed and expected counts

Remember that the Chi-Squared test statistic is \(\displaystyle{\sum{\frac{\left(\text{observed} - \text{expected}\right)^2}{\text{expected}}}}\)

sum((observed - expected)^2/expected)

[1] 249.9124

A Completed Example: Chi-Squared Test for Independence

Scenario: A wildlife researcher is studying whether different bird species have distinct feeding preferences among three types of feeders: Seed Feeder, Nectar Feeder, and Fruit Feeder. The researcher observes four bird species in a particular park: Finches, Hummingbirds, Orioles, and Woodpeckers. Over several days, the researcher records the feeder each bird species prefers in the table below. Test whether species and feeding preference are independent at the 10% level of significance.

1. The hypotheses are \(\begin{array}{lcl} H_0 & : & \text{Species and feeder preference are independent}\\ H_a & : & \text{Species and feeder preference are associated}\end{array}\)

2. The level of significance is \(\alpha = 0.10\)
3. The Chi-Squared test statistic is about 249.91

4. The degrees of freedom for the test is \(\left(k - 1\right)\cdot\left(\ell - 1\right)\)

A Completed Example: Chi-Squared Test for Independence

Scenario: A wildlife researcher is studying whether different bird species have distinct feeding preferences among three types of feeders: Seed Feeder, Nectar Feeder, and Fruit Feeder. The researcher observes four bird species in a particular park: Finches, Hummingbirds, Orioles, and Woodpeckers. Over several days, the researcher records the feeder each bird species prefers in the table below. Test whether species and feeding preference are independent at the 10% level of significance.

1. The hypotheses are \(\begin{array}{lcl} H_0 & : & \text{Species and feeder preference are independent}\\ H_a & : & \text{Species and feeder preference are associated}\end{array}\)

2. The level of significance is \(\alpha = 0.10\)
3. The Chi-Squared test statistic is about 249.91
4. The degrees of freedom for the test is \(\left(4 - 1\right)\cdot\left(3 - 1\right) = 6\)

5. Now we’ll calculate the \(p\)-value

A Completed Example: Chi-Squared Test for Independence

Scenario: A wildlife researcher is studying whether different bird species have distinct feeding preferences among three types of feeders: Seed Feeder, Nectar Feeder, and Fruit Feeder. The researcher observes four bird species in a particular park: Finches, Hummingbirds, Orioles, and Woodpeckers. Over several days, the researcher records the feeder each bird species prefers in the table below. Test whether species and feeding preference are independent at the 10% level of significance.

1. The hypotheses are \(\begin{array}{lcl} H_0 & : & \text{Species and feeder preference are independent}\\ H_a & : & \text{Species and feeder preference are associated}\end{array}\)

2. The level of significance is \(\alpha = 0.10\)
3. The Chi-Squared test statistic is about 249.91
4. The degrees of freedom for the test is \(\left(4 - 1\right)\cdot\left(3 - 1\right) = 6\)
5. Now we’ll calculate the \(p\)-value

A Completed Example: Chi-Squared Test for Independence

Scenario: A wildlife researcher is studying whether different bird species have distinct feeding preferences among three types of feeders: Seed Feeder, Nectar Feeder, and Fruit Feeder. The researcher observes four bird species in a particular park: Finches, Hummingbirds, Orioles, and Woodpeckers. Over several days, the researcher records the feeder each bird species prefers in the table below. Test whether species and feeding preference are independent at the 10% level of significance.

1. The hypotheses are \(\begin{array}{lcl} H_0 & : & \text{Species and feeder preference are independent}\\ H_a & : & \text{Species and feeder preference are associated}\end{array}\)

2. The level of significance is \(\alpha = 0.10\)
3. The Chi-Squared test statistic is about 249.91
4. The degrees of freedom for the test is \(\left(4 - 1\right)\cdot\left(3 - 1\right) = 6\)
5. Now we’ll calculate the \(p\)-value

1 - pchisq(249.91, df = 6) \(\approx\) 0

A Completed Example: Chi-Squared Test for Independence

Scenario: A wildlife researcher is studying whether different bird species have distinct feeding preferences among three types of feeders: Seed Feeder, Nectar Feeder, and Fruit Feeder. The researcher observes four bird species in a particular park: Finches, Hummingbirds, Orioles, and Woodpeckers. Over several days, the researcher records the feeder each bird species prefers in the table below. Test whether species and feeding preference are independent at the 10% level of significance.

1. The hypotheses are \(\begin{array}{lcl} H_0 & : & \text{Species and feeder preference are independent}\\ H_a & : & \text{Species and feeder preference are associated}\end{array}\)
2. The level of significance is \(\alpha = 0.10\)
3. The Chi-Squared test statistic is about 249.91
4. The degrees of freedom for the test is \(\left(4 - 1\right)\cdot\left(3 - 1\right) = 6\)
5. Again, the \(p\)-value is so small that it rounds to 0

A Completed Example: Chi-Squared Test for Independence

Scenario: A wildlife researcher is studying whether different bird species have distinct feeding preferences among three types of feeders: Seed Feeder, Nectar Feeder, and Fruit Feeder. The researcher observes four bird species in a particular park: Finches, Hummingbirds, Orioles, and Woodpeckers. Over several days, the researcher records the feeder each bird species prefers in the table below. Test whether species and feeding preference are independent at the 10% level of significance.

1. The hypotheses are \(\begin{array}{lcl} H_0 & : & \text{Species and feeder preference are independent}\\ H_a & : & \text{Species and feeder preference are associated}\end{array}\)
2. The level of significance is \(\alpha = 0.10\)
3. The Chi-Squared test statistic is about 249.91
4. The degrees of freedom for the test is \(\left(4 - 1\right)\cdot\left(3 - 1\right) = 6\)
5. Again, the \(p\)-value is so small that it rounds to 0
6. The \(p\)-value is less than the level of significance, so we reject \(H_0\) and accept \(H_a\)

A Completed Example: Chi-Squared Test for Independence

Scenario: A wildlife researcher is studying whether different bird species have distinct feeding preferences among three types of feeders: Seed Feeder, Nectar Feeder, and Fruit Feeder. The researcher observes four bird species in a particular park: Finches, Hummingbirds, Orioles, and Woodpeckers. Over several days, the researcher records the feeder each bird species prefers in the table below. Test whether species and feeding preference are independent at the 10% level of significance.

1. The hypotheses are \(\begin{array}{lcl} H_0 & : & \text{Species and feeder preference are independent}\\ H_a & : & \text{Species and feeder preference are associated}\end{array}\)
2. The level of significance is \(\alpha = 0.10\)
3. The Chi-Squared test statistic is about 249.91
4. The degrees of freedom for the test is \(\left(4 - 1\right)\cdot\left(3 - 1\right) = 6\)
5. Again, the \(p\)-value is so small that it rounds to 0
6. The \(p\)-value is less than the level of significance, so we reject \(H_0\) and accept \(H_a\)
7. This sample provides evidence to suggest that bird species and feeding preference are associated

Examples to Try: Streaming Service Popularity

Scenario: A media company expects certain subscription preferences among viewers for various streaming services: Netflix, Hulu, Amazon Prime, and Disney+, with expected preferences of 40%, 25%, 20%, and 15%, respectively. A sample of 500 viewers reveals the following counts:

Streaming Service	Observed Count
Netflix	210
Hulu	130
Amazon Prime	100
Disney+	60

Conduct a test at the 5% level of significance to determine whether the sample provides evidence against this expected distribution.

Examples to Try: Department and Work Arrangement

Scenario: A company wants to investigate if employees’ work arrangement preferences (In-office, Hybrid, or Remote) vary by department (IT, Marketing, HR). A survey of 300 employees yields the following results:

	In-office	Hybrid	Remote	Total
IT	30	60	40	130
Marketing	25	35	30	90
HR	20	25	35	80
Total	75	120	105	300

Conduct a test at the 1% level of significance to determine whether this data provides evidence to suggest that work arrangements are different across departments.

Examples to Try: Student Major versus Preferred Social Media Platform

Scenario: A university wants to see if there’s a link between students’ major category (STEM, Arts, Business, Humanities) and their preferred social media platform (Instagram, TikTok, Twitter, LinkedIn). They survey 200 students and gather the following results back:

	Instagram	TikTok	Twitter	LinkedIn	Total
STEM	20	15	30	35	100
Arts	15	25	20	5	65
Business	10	5	15	25	55
Humanities	5	5	10	5	25
Total	50	50	75	70	200

Examples to Try: Voting Method Preferences

Scenario: An electoral commission wants to verify if voter preferences for different voting methods in a city (In-person, Mail-in, Drop-off, Online) match their pre-election forecast based on trends in similar areas. They projected that 50% would prefer in-person voting, 20% mail-in, 15% drop-off, and 15% online. A random sample of 1000 likely voters revealed the following preferences.

Voting Method	Observed Count	Expected Percentage
In-person	520	50%
Mail-in	200	20%
Drop-off	150	15%
Online	130	15%
Total	1,000

Does the sample provide evidence against their projections?

Next Time…

Inference on Numerical Variables