## 字幕表 動画を再生する

• Hi, I'm Adriene Hill, and welcome back to Crash Course Statistics.

• When you're buying a new car, a new house, or a new pair of jeans, you want to make sure

• you find a good fit.

• Statistics are the same.

• You want to make sure your models or preconceptions are a good fit for the data you have.

• One way to do that is by comparing our observations to our expectations, just like we've done

• in statistical tests over the last couple of episodes.

• Today, we're going to take a break from looking at continuous variables like height

• and IQ and see how we can measure the fit of categorical variables like hair color,

• or academic degree, or tax bracket.

• Turns out a chi-square model fits like that perfect pair of jeans.

• INTRO

• Back in one of our data visualization episodes, we talked about frequency tables, which tell

• you the counts--or frequencies--of different categories.

• Maybe you remember pasta madness?

• When these tables look at two different categorical variables, like favorite pasta and favorite

• pasta sauce, we often call them contingency tables or cross tabulations.

• These tables can help us understand the discrete distribution of one variable, or the relationship

• between two variables.

• Just by looking at the data we can see major differences like most people prefer red sauce....

• Or the distribution of favorite pastas seems to be different for people who like red sauce

• vs. white sauce.

• But sometimes it's hard for us to see smaller differences, or to be sure whether differences

• we see are statistically meaningful.

• Just like when we were comparing means, we need a test.

• A statistical test that can help us extract thesignal from the noise”.

• Statistical tests usually take this form.

• And the test we'll use here -- Chi Square -- is only slightly different from the others

• we've used so far.

• The idea in the numerator--looking at the difference between what we observed and what

• we'd expect if the null is true--is exactly the same.

• The denominator--average variation--is a little different.

• Let's figure out why...in an example.

• A new video game has come out, League of Lemurs, and it has taken the world by storm.

• League of Lemurs has hundreds of unique characters you can play as with four main types: Healers,

• Tanks, Assassins, and Fighters.

• The official League of Lemurs development team says that on average they see 15% of

• players choosing Healers, 20% choosing Tanks, 20% choosing Assassins, and 45% choosing Fighters

• - because who doesn't want to be a fighting lemur.

• But you wonder whether that distribution holds in the top ranks of LOL players.

• The null hypothesis is that the percentages that LOL gave you are correct:

• And the alternative hypothesis is that at least one of these percentages or proportions

• are incorrect.

• So you record 20 games with 10 players each and count the number of Healers, Tanks, Assassins

• and Fighters.

• The data you collected looks like this:

• According to the numbers the LOL developers gave us, we'd have expected 30 Healers,

• 40 Tanks, 40 Assassins, and 90 Fighters.

• So These numbers aren't EXACTLY what we'd expect.

• But we have to ask whether they're different enough for us to consider it statistically significant.

• We need a test statistic.

• Our general formula applies, with the numerator being the count we observed for each category

• minus the count we'd expect.

• But If you tried to add up all these differences: You'll always get zero.

• So we need a better way to measure.

• Using a chi-square, we square them before adding them all up.

• For the denominator instead of a standard error, we just use the expected counts again.

• In this case, the amount that a count deviates from its expected frequency should be scaled

• by the expected frequency.

• A deviation of 1 isn't as big of a deal if the expected count is 2000, but if it's

• 10...that deviation of 1 matters more.

• You might not think it's worth it to go back to the store to demand a refund if you're

• overcharged \$1 for a \$2000 laptop….but it might feel more worth it if you were overcharged

• \$1 for your morning coffee.

• When we add up all these calculations We get a chi-square statistic.

• In this case 3.958.

• Which, like our other test statistics, helps us quantify how well our sample data fits

• a certain distribution.

• Usually the null distribution.

• Like a t-statistic, a chi-square statistic has a distribution that we can use to find a p-value.

• And like t-distributions, chi-square distributions change their shape as degrees of freedom change.

• To find our degrees of freedom we have to think about what kinds of independent information

• we have.

• A frequency table, like the one we just used for our League of Lemurs example, has a certain

• number of cells.

• In this case we had four, one for each character type:

• That means we have four independent pieces of information: each of the four counts.

• But as soon as we know the total counts--in this case the 200 players you recorded--the

• four values aren't ALL independent anymore.

• Because if you know 3 of those values and the total, then with some basic math you can

• figure out the fourth.

• So in this case, our degrees of freedom is the number of categories we have, 4 minus 1.

• In this case, 3.

• Using our chi-square distribution of 3 degrees of freedom, we can now find our p-value!

• Our p-value here is 0.3, so if our cutoff was 0.05 we'd fail to reject the null.

• The sample we took didn't give us any statistically significant evidence that the game developers'

• percentages were wrong.

• All Chi-square tests follow the same formula we just worked through.

• And there are three main ways that we use chi-square tests.

• The one we just did is called a chi-square Goodness of Fit test, because we tested how

• well certain proportions fit our sample.

• One way to know that you're looking at a Goodness Of Fit Chi-square test is if it only

• has one row.

• We can have many categories, but we're only looking at one variable.

• Like in our case, character class.

• And note: one thing we should always check when doing a Chi-square test is whether the

• expected frequency for every cell is greater than 5.

• If the expected frequency is lower than 5, the results of the chi-square test can be off.

• 5 is arbitrary, just like many of the cutoffs in Statistics, but it's widely accepted.

• But chi-square tests aren't limited to analyzing just ONE categorical variable.

• They can even handle TWO.

• Like with the Test of Independence, the second type of chi-square test.

• Tests of independence look to see whether being a member of one category is independent

• of the other.

• For example, let's look at the annual NerdFighteria Survey -- a survey that Hank and John Green

• request their audience of nerdfighters take every year.

• We wanted to take a look at the two most important questions they asked last year: what Hogwarts

• houses nerdfighters were in AND if they liked pineapple on pizza.

• What we want to know is whether Pineapple on Pizza

• Preference is Independent of Hogwarts House.

• In other words, does liking Pineapple on Pizza affect the probabilities of you identifying

• with each of the houses?

• Luckily, our writer, Chelsea has that data, and she took a random sample of 1000 Nerdfighter's

• responses so we could answer our questions.

• She's a pineapple-loving Ravenclaw, for the record.

• But let's take a look at the data.

• Looks like there's a lot of Ravenclaw Nerdfighters.

• Unlike our Chi Square goodness of fit test, we're not specifying an exact distribution

• for Hogwarts houses and comparing our two groups: Yes pineapple and No pineapple.

• In this situation, we're not too concerned with the exact distribution.

• We just want to know whether it's different for people who like and don't like pineapple

• on pizza.

• A Chi-square test of Independence, can test whether or not one Variable--pineapple preference--is

• independent of another--Hogwarts House.

• And you'll soon see that the calculations we do here are the exact same for the third

• Chi-Square test: the Chi-Square test of Homogeneity.

• Test of homogeneity are looking at whether it's likely that different samples come

• from the same population.

• For example you might want to know whether two samples of water are likely from the same

• lake based on the counts of fish, algae, and bacteria found in them.

• In essence they're testing similar things, and the calculations we're about to do are

• the same for both tests.

• Back to the Nerdfighters.

• To calculate the Chi-Square statistic, we need our observed frequencies which we already

• have, and our expected frequencies, which we need to calculate.

• But it's not quite as straightforward as in the Goodness of fit test.First we'll

• need to calculate some row and column totals:

• We already know that there's 1000 total people, and we can count up all the people

• who don't like pineapple on their pizza to find that there's 479 of them, which

• means there must be 521 people who do like pineapple on their Pizza.

• We have 3 independent pieces of information here, or 3 degrees of freedom.

• In general the formula for degrees of freedom for these chi-square tests is rows minus 1

• times columns minus 1.

• Finally, we can calculate those expected frequencies.

• Remember, the expected counts are what we would expect if the null hypothesis is true.

• In this test, the null hypothesis is that the distribution of Hogwarts House is the

• same for both pineapple lovers and haters.

• But our null hypothesis says nothing about WHAT that distribution is.

• So we can calculate our expected frequencies by taking the total number of Gryffindors

• and dividing it by the total number of people:

• This gives us the expected percentage of people who are Gryffindors.

• Since there's 479 People in our sample who don't like pineapple, we expect 16.1% of

• them or about 77 of them to also be Gryffindors.

• Using this same math, we can calculate the expected frequency for all of our cells.

• Once we have our expected frequency, we just need to use our Chi-square formula on each

• cell, and add them all up to get our Chi-Square Statistic:

• And with our chi-square distribution with 3 degrees of freedom, we can see that our

• p-value of 0.6 is very large compared to our alpha level of 0.05, so we fail to reject

• the null hypothesis that the distribution of Hogwarts Houses is the same regardless of pizza preference.

• If the null were true, we'd expect to see numbers as or more different than ours 60%

• of the time.

• So we don't have evidence that Hogwarts House is dependent on Pineapples on Pizza preference.

• It's often useful to check our assumptions and to see if they're a good fit.

• Whether that's testing whether a population is distributed the way we think it is.

• Are there really the same proportion of Skittles colors in a bag?

• Or whether two variables affect each other, like political party preference and cat and

• dog ownership.

• Since we, as humans, tend to categorize many things, from dog breed to hair color, it can

• be useful to check what we think about how and if those categories interact.

• Thanks for watching. I'll see you next time.

Hi, I'm Adriene Hill, and welcome back to Crash Course Statistics.

B1 中級