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  • 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.

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カイ二乗検定。クラッシュコース統計学 #29 (Chi-Square Tests: Crash Course Statistics #29)

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    林宜悉 に公開 2021 年 01 月 14 日
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