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You see one person in line, exactly 18 times. And you see two people in line, exactly eight times. And in your 50 visits, you don't see more than, you never see more than two people in line. I guess this is a very efficient cashier at this frozen yogurt store. So based on this, based on what you have observed, what would be your estimate of the probability of finding no people in line, one people in line, or two people in line, at 4 p.m. on the days after school that you visit the frozen yogurt store? And you'll say you only visited on weekdays where there are school days. So what's the probability of there being no line, a one person line, or a two person line when you visit at 4 p.m. on a school day? | Constructing probability model from observations 7th grade Khan Academy.mp3 |
I guess this is a very efficient cashier at this frozen yogurt store. So based on this, based on what you have observed, what would be your estimate of the probability of finding no people in line, one people in line, or two people in line, at 4 p.m. on the days after school that you visit the frozen yogurt store? And you'll say you only visited on weekdays where there are school days. So what's the probability of there being no line, a one person line, or a two person line when you visit at 4 p.m. on a school day? Well, all you can do is estimate the true probability, the true theoretical probability. We don't know what that is, but you've done 50 observations here, right? This is, and notice this adds up to 50. | Constructing probability model from observations 7th grade Khan Academy.mp3 |
So what's the probability of there being no line, a one person line, or a two person line when you visit at 4 p.m. on a school day? Well, all you can do is estimate the true probability, the true theoretical probability. We don't know what that is, but you've done 50 observations here, right? This is, and notice this adds up to 50. 18 plus eight is 26, 26 plus 24 is 50. So you've done 50 observations here, and so you can figure out, well, what are the relative frequencies of having zero people? What is the relative frequency of one person, or the relative frequency of two people in line? | Constructing probability model from observations 7th grade Khan Academy.mp3 |
This is, and notice this adds up to 50. 18 plus eight is 26, 26 plus 24 is 50. So you've done 50 observations here, and so you can figure out, well, what are the relative frequencies of having zero people? What is the relative frequency of one person, or the relative frequency of two people in line? And then we can use that as the estimates for the probability. So let's do that. So probability estimate. | Constructing probability model from observations 7th grade Khan Academy.mp3 |
What is the relative frequency of one person, or the relative frequency of two people in line? And then we can use that as the estimates for the probability. So let's do that. So probability estimate. I'll do it in the next column. So probability, probability estimate. And once again, we can do that by looking at the relative frequency. | Constructing probability model from observations 7th grade Khan Academy.mp3 |
So probability estimate. I'll do it in the next column. So probability, probability estimate. And once again, we can do that by looking at the relative frequency. The relative frequency of zero, well, we observe that 24 times out of 50, and so 24 out of 50 is the same thing as 0.48, or you could even say that this is 48%. Now, what's the relative frequency of seeing one person in line? Well, you observe that 18 out of the 50 visits, 18 out of the 50 visits, that would be a relative frequency, 18 divided by 50 is 0.36, which is 36% of your visits. | Constructing probability model from observations 7th grade Khan Academy.mp3 |
And once again, we can do that by looking at the relative frequency. The relative frequency of zero, well, we observe that 24 times out of 50, and so 24 out of 50 is the same thing as 0.48, or you could even say that this is 48%. Now, what's the relative frequency of seeing one person in line? Well, you observe that 18 out of the 50 visits, 18 out of the 50 visits, that would be a relative frequency, 18 divided by 50 is 0.36, which is 36% of your visits. And then finally, the relative frequency of seeing a two-person line, that was eight out of the 50 visits, and so that is 0.16, and that is equal to 16% of the visits. And so there's interesting things here. Remember, these are estimates of the probability. | Constructing probability model from observations 7th grade Khan Academy.mp3 |
Well, you observe that 18 out of the 50 visits, 18 out of the 50 visits, that would be a relative frequency, 18 divided by 50 is 0.36, which is 36% of your visits. And then finally, the relative frequency of seeing a two-person line, that was eight out of the 50 visits, and so that is 0.16, and that is equal to 16% of the visits. And so there's interesting things here. Remember, these are estimates of the probability. You're doing this by essentially sampling what the line on 50 different days. You don't know, it's not gonna always be exactly this, but it's a good estimate. You did it 50 times. | Constructing probability model from observations 7th grade Khan Academy.mp3 |
Remember, these are estimates of the probability. You're doing this by essentially sampling what the line on 50 different days. You don't know, it's not gonna always be exactly this, but it's a good estimate. You did it 50 times. And so based on this, you'd say, well, I'd estimate the probability of having a zero-person line is 48%. I'd estimate that the probability of having a one-person line is 36%. I'd estimate that the probability of having a two-person line is 16%, or 0.16. | Constructing probability model from observations 7th grade Khan Academy.mp3 |
You did it 50 times. And so based on this, you'd say, well, I'd estimate the probability of having a zero-person line is 48%. I'd estimate that the probability of having a one-person line is 36%. I'd estimate that the probability of having a two-person line is 16%, or 0.16. And it's important to realize that these are legitimate probabilities. Remember, to be a probability, it has to be between zero and one. It has to be zero and one. | Constructing probability model from observations 7th grade Khan Academy.mp3 |
I'd estimate that the probability of having a two-person line is 16%, or 0.16. And it's important to realize that these are legitimate probabilities. Remember, to be a probability, it has to be between zero and one. It has to be zero and one. And if you look at all of the possible events, it should add up to one, because at least based on your observations, these are the possibilities. Obviously, in a real world, there might be some kind of crazy thing where more people go in line, but at least based on the events that you've seen, these three different events, and these are the only three that you've observed, based on your observations, these three should add, because these are the only three things you've observed, they should add up to one, and they do add up to one. Let's see, 36 plus 16 is 52. | Constructing probability model from observations 7th grade Khan Academy.mp3 |
It has to be zero and one. And if you look at all of the possible events, it should add up to one, because at least based on your observations, these are the possibilities. Obviously, in a real world, there might be some kind of crazy thing where more people go in line, but at least based on the events that you've seen, these three different events, and these are the only three that you've observed, based on your observations, these three should add, because these are the only three things you've observed, they should add up to one, and they do add up to one. Let's see, 36 plus 16 is 52. 52 plus 48, they add up to one. Now, once you do this, you might do something interesting. You might say, okay, you know what? | Constructing probability model from observations 7th grade Khan Academy.mp3 |
Let's see, 36 plus 16 is 52. 52 plus 48, they add up to one. Now, once you do this, you might do something interesting. You might say, okay, you know what? Over the next two years, you plan on visiting 500 times. So visiting 500 times. So based on your estimates of the probability of having no line, of a one-person line, or a two-person line, how many times in your next 500 visits would you expect there to be a two-person line, based on your observations so far? | Constructing probability model from observations 7th grade Khan Academy.mp3 |
You might say, okay, you know what? Over the next two years, you plan on visiting 500 times. So visiting 500 times. So based on your estimates of the probability of having no line, of a one-person line, or a two-person line, how many times in your next 500 visits would you expect there to be a two-person line, based on your observations so far? Well, it's reasonable to say, well, a good estimate of the number of times you'll see a two-person line when you visit 500 times. Well, you say, well, there's gonna be 500 times, and it's a reasonable expectation, based on your estimate of the probability, that 0.16 of the time, you will see a two-person line, or you could say eight out of every 50 times. And so what is this going to be? | Constructing probability model from observations 7th grade Khan Academy.mp3 |
So based on your estimates of the probability of having no line, of a one-person line, or a two-person line, how many times in your next 500 visits would you expect there to be a two-person line, based on your observations so far? Well, it's reasonable to say, well, a good estimate of the number of times you'll see a two-person line when you visit 500 times. Well, you say, well, there's gonna be 500 times, and it's a reasonable expectation, based on your estimate of the probability, that 0.16 of the time, you will see a two-person line, or you could say eight out of every 50 times. And so what is this going to be? Let's see, 500 divided by 50 is just 10. So you would expect, you would expect that 80 out of the 500 times, you would see a two-person line. Now, to be clear, I would be shocked if it's exactly 80 ends up being the case, but this is actually a very good expectation, based on your observations. | Constructing probability model from observations 7th grade Khan Academy.mp3 |
And so what is this going to be? Let's see, 500 divided by 50 is just 10. So you would expect, you would expect that 80 out of the 500 times, you would see a two-person line. Now, to be clear, I would be shocked if it's exactly 80 ends up being the case, but this is actually a very good expectation, based on your observations. It is completely possible, first of all, that your observations were off, that this is just a random chance that you happened to observe this many, or this few times that there were two people in line. So that could be off, but even if these are very good estimates, it's possible that something, that you see a two-person line 85 out of the 500 times, or 65 out of the 500 times. All of those things are possible. | Constructing probability model from observations 7th grade Khan Academy.mp3 |
Nutritionists measured the sugar content in grams for 32 drinks at Starbucks. A cumulative relative frequency graph, let me underline that, a cumulative relative frequency graph for the data is shown below. So they have different, on the horizontal axis, different amounts of sugar in grams, and then we have the cumulative relative frequency. So let's just make sure we understand how to read this. This is saying that zero, or 0%, of the drinks have a sugar content, have no sugar content. This right over here, this data point, this looks like it's at the.5 grams, and then this looks like it's at 0.1. This says that 0.1, or I guess we would say 10%, of the drinks that Starbucks offers has five grams of sugar or less. | Analyzing a cumulative relative frequency graph AP Statistics Khan Academy.mp3 |
So let's just make sure we understand how to read this. This is saying that zero, or 0%, of the drinks have a sugar content, have no sugar content. This right over here, this data point, this looks like it's at the.5 grams, and then this looks like it's at 0.1. This says that 0.1, or I guess we would say 10%, of the drinks that Starbucks offers has five grams of sugar or less. This data point tells us that 100% of the drinks at Starbucks has 50 grams of sugar or less. The cumulative relative frequency, that's why we, for each of these points, we say this is the frequency that has that much sugar or less, and that's why it just keeps on increasing and increasing. As we add more sugar, we're going to see, we're gonna see a larger proportion or a larger relative frequency has that much sugar or less. | Analyzing a cumulative relative frequency graph AP Statistics Khan Academy.mp3 |
This says that 0.1, or I guess we would say 10%, of the drinks that Starbucks offers has five grams of sugar or less. This data point tells us that 100% of the drinks at Starbucks has 50 grams of sugar or less. The cumulative relative frequency, that's why we, for each of these points, we say this is the frequency that has that much sugar or less, and that's why it just keeps on increasing and increasing. As we add more sugar, we're going to see, we're gonna see a larger proportion or a larger relative frequency has that much sugar or less. So let's read the first question. An iced coffee has 15 grams of sugar. Estimate the percentile of this drink to the nearest whole percent. | Analyzing a cumulative relative frequency graph AP Statistics Khan Academy.mp3 |
As we add more sugar, we're going to see, we're gonna see a larger proportion or a larger relative frequency has that much sugar or less. So let's read the first question. An iced coffee has 15 grams of sugar. Estimate the percentile of this drink to the nearest whole percent. So iced coffee has 15 grams of sugar, which would be right over here. And so let's estimate the percentile. So we can see they actually have a data point right over here. | Analyzing a cumulative relative frequency graph AP Statistics Khan Academy.mp3 |
Estimate the percentile of this drink to the nearest whole percent. So iced coffee has 15 grams of sugar, which would be right over here. And so let's estimate the percentile. So we can see they actually have a data point right over here. And we can see that 20%, or 0.2, 20% of the drinks that Starbucks offers has 15 grams of sugar or less. So the percentile of this drink, if I were to estimate it, it looks like it's the relative frequency, 0.2, has that much sugar or less, and so this percentile would be 20%. Once again, another way to think about it, to read this, you could convert these to percentages. | Analyzing a cumulative relative frequency graph AP Statistics Khan Academy.mp3 |
So we can see they actually have a data point right over here. And we can see that 20%, or 0.2, 20% of the drinks that Starbucks offers has 15 grams of sugar or less. So the percentile of this drink, if I were to estimate it, it looks like it's the relative frequency, 0.2, has that much sugar or less, and so this percentile would be 20%. Once again, another way to think about it, to read this, you could convert these to percentages. You could say that 20% has this much sugar or less, 15 grams of sugar or less, so an iced coffee is in the 20th percentile. Let's do another question. So here we are asked to estimate the median of the distribution of drinks. | Analyzing a cumulative relative frequency graph AP Statistics Khan Academy.mp3 |
Once again, another way to think about it, to read this, you could convert these to percentages. You could say that 20% has this much sugar or less, 15 grams of sugar or less, so an iced coffee is in the 20th percentile. Let's do another question. So here we are asked to estimate the median of the distribution of drinks. Hint, think about the 50th percentile. So the median, if you were to line up all of the drinks, you would take the middle drink. And so you could view that as, well, what drink is exactly at the 50th percentile? | Analyzing a cumulative relative frequency graph AP Statistics Khan Academy.mp3 |
So here we are asked to estimate the median of the distribution of drinks. Hint, think about the 50th percentile. So the median, if you were to line up all of the drinks, you would take the middle drink. And so you could view that as, well, what drink is exactly at the 50th percentile? So now let's look at the 50th percentile would be a cumulative relative frequency of 0.5, which would be right over here on our vertical axis. Another way to think about it is 0.5, or 50% of the drinks are going, if we go to this point right over here, what has a cumulative relative frequency of 0.5? We see that we are right at, looks like this is 25 grams. | Analyzing a cumulative relative frequency graph AP Statistics Khan Academy.mp3 |
And so you could view that as, well, what drink is exactly at the 50th percentile? So now let's look at the 50th percentile would be a cumulative relative frequency of 0.5, which would be right over here on our vertical axis. Another way to think about it is 0.5, or 50% of the drinks are going, if we go to this point right over here, what has a cumulative relative frequency of 0.5? We see that we are right at, looks like this is 25 grams. So one way to interpret this is 50% of the drinks have less than, or have 25 grams of sugar or less. So this looks like a pretty good estimate for the median, for the middle data point. So the median is approximately 25 grams, that half of the drinks have 25 grams or less of sugar. | Analyzing a cumulative relative frequency graph AP Statistics Khan Academy.mp3 |
We see that we are right at, looks like this is 25 grams. So one way to interpret this is 50% of the drinks have less than, or have 25 grams of sugar or less. So this looks like a pretty good estimate for the median, for the middle data point. So the median is approximately 25 grams, that half of the drinks have 25 grams or less of sugar. Let's do one more based on the same data set. So here we're asked, what is the best estimate for the interquartile range of the distribution of drinks? So the interquartile range, we wanna figure out, well, what's sitting at the 25th percentile? | Analyzing a cumulative relative frequency graph AP Statistics Khan Academy.mp3 |
So the median is approximately 25 grams, that half of the drinks have 25 grams or less of sugar. Let's do one more based on the same data set. So here we're asked, what is the best estimate for the interquartile range of the distribution of drinks? So the interquartile range, we wanna figure out, well, what's sitting at the 25th percentile? And we wanna think about what's at the 75th percentile. And then we want to take the difference. That's what the interquartile range is. | Analyzing a cumulative relative frequency graph AP Statistics Khan Academy.mp3 |
So the interquartile range, we wanna figure out, well, what's sitting at the 25th percentile? And we wanna think about what's at the 75th percentile. And then we want to take the difference. That's what the interquartile range is. So let's do that. So first, the 25th percentile, we'd wanna look at the cumulative relative frequency. So 25th, this would be 30th, so 25th would be right around here. | Analyzing a cumulative relative frequency graph AP Statistics Khan Academy.mp3 |
That's what the interquartile range is. So let's do that. So first, the 25th percentile, we'd wanna look at the cumulative relative frequency. So 25th, this would be 30th, so 25th would be right around here. And so it looks like the 25th percentile is, that looks like about, I don't know. And we're estimating here, so that looks like it's about, this would be 15, looks like I would say maybe 18 grams. So approximately 18 grams. | Analyzing a cumulative relative frequency graph AP Statistics Khan Academy.mp3 |
So 25th, this would be 30th, so 25th would be right around here. And so it looks like the 25th percentile is, that looks like about, I don't know. And we're estimating here, so that looks like it's about, this would be 15, looks like I would say maybe 18 grams. So approximately 18 grams. Once again, one way to think about it is, 25% of the drinks have 18 grams of sugar or less. And let's look at the 75th percentile. So this is 70th, 75th would be right over there. | Analyzing a cumulative relative frequency graph AP Statistics Khan Academy.mp3 |
So approximately 18 grams. Once again, one way to think about it is, 25% of the drinks have 18 grams of sugar or less. And let's look at the 75th percentile. So this is 70th, 75th would be right over there. Actually, I can draw a straighter line than that. I have a line tool here. So 75th percentile would put me right over there. | Analyzing a cumulative relative frequency graph AP Statistics Khan Academy.mp3 |
So this is 70th, 75th would be right over there. Actually, I can draw a straighter line than that. I have a line tool here. So 75th percentile would put me right over there. I don't know, that looks like, well, I'll go with 39 grams, roughly 39 grams. And so what's the difference between these two? Well, the difference between these two, it looks like it's about 21 grams. | Analyzing a cumulative relative frequency graph AP Statistics Khan Academy.mp3 |
So 75th percentile would put me right over there. I don't know, that looks like, well, I'll go with 39 grams, roughly 39 grams. And so what's the difference between these two? Well, the difference between these two, it looks like it's about 21 grams. So our interquartile range, our estimate of our interquartile range, looking at this cumulative relative frequency distribution, because we're saying, hey, look, it looks like the 25th percentile, it looks like 25% of the drinks have 18 grams or less. 75% of the drinks have 39 grams or less. If we take the difference between these two quartiles, this is the first quartile, this is our third quartile, we're gonna get 21 grams. | Analyzing a cumulative relative frequency graph AP Statistics Khan Academy.mp3 |
So this first one, Roy's Toys received a shipment of 100,000 rubber duckies from the factory. The factory couldn't promise that all rubber duckies are in perfect form, but they promised that the percentage of defective toys won't exceed 5%. Let me underline that. They promised that the percentage of defective toys won't exceed 5%. Roy wanted to get an estimation of the percentage of defective toys, and since he couldn't go over the entire 100,000 duckies, he took a random sample of 10 duckies. He found that 10% of them were defective. So what's going on here? | Worked example identifying sample study.mp3 |
They promised that the percentage of defective toys won't exceed 5%. Roy wanted to get an estimation of the percentage of defective toys, and since he couldn't go over the entire 100,000 duckies, he took a random sample of 10 duckies. He found that 10% of them were defective. So what's going on here? Roy gets a shipment. There's 100,000 ducks in the shipment. He wants to figure out what percentage of them are defective. | Worked example identifying sample study.mp3 |
So what's going on here? Roy gets a shipment. There's 100,000 ducks in the shipment. He wants to figure out what percentage of them are defective. He can't look at all 100,000 ducks. It's not practical. So he samples 10 of them. | Worked example identifying sample study.mp3 |
He wants to figure out what percentage of them are defective. He can't look at all 100,000 ducks. It's not practical. So he samples 10 of them. One, two, three, four, five, six, seven, eight, nine, 10, and he finds that one out of those 10 are defective, 10% of the 10. So first of all, this is clearly a sample study. This is a sample study. | Worked example identifying sample study.mp3 |
So he samples 10 of them. One, two, three, four, five, six, seven, eight, nine, 10, and he finds that one out of those 10 are defective, 10% of the 10. So first of all, this is clearly a sample study. This is a sample study. How do we know that? Well, he is taking a sample from a broader population in order to estimate a parameter. The parameter is the percentage of those 100,000 duckies that are actually defective. | Worked example identifying sample study.mp3 |
This is a sample study. How do we know that? Well, he is taking a sample from a broader population in order to estimate a parameter. The parameter is the percentage of those 100,000 duckies that are actually defective. Now, the next question is, is what kind of a conclusion can you make? Roy, since he got the shipment and he took a sample and he found that 10% of the sample was defective, he might be all up in arms and say, oh, this toy shipment from the factory, they violated this promise that the percentage of defective toys won't exceed 5% because I sampled 10 toys and 10% of those 10 toys were defective. Well, that isn't a reasonable conclusion because this is a small sample. | Worked example identifying sample study.mp3 |
The parameter is the percentage of those 100,000 duckies that are actually defective. Now, the next question is, is what kind of a conclusion can you make? Roy, since he got the shipment and he took a sample and he found that 10% of the sample was defective, he might be all up in arms and say, oh, this toy shipment from the factory, they violated this promise that the percentage of defective toys won't exceed 5% because I sampled 10 toys and 10% of those 10 toys were defective. Well, that isn't a reasonable conclusion because this is a small sample. This is a small sample. Think about it. He could have sampled five duckies and if he just happened to get one of the defective ones, he would have said, oh, maybe 20% are defective. | Worked example identifying sample study.mp3 |
Well, that isn't a reasonable conclusion because this is a small sample. This is a small sample. Think about it. He could have sampled five duckies and if he just happened to get one of the defective ones, he would have said, oh, maybe 20% are defective. What he's really gotta do is sample, take a larger sample. And once again, whenever you're sampling, there's always a probability that your estimate is going to be not close or definitely not the same as the parameter for the population. But the larger your sample, the higher probability that your estimate is close to the actual parameter for the population. | Worked example identifying sample study.mp3 |
He could have sampled five duckies and if he just happened to get one of the defective ones, he would have said, oh, maybe 20% are defective. What he's really gotta do is sample, take a larger sample. And once again, whenever you're sampling, there's always a probability that your estimate is going to be not close or definitely not the same as the parameter for the population. But the larger your sample, the higher probability that your estimate is close to the actual parameter for the population. And 10 in this is just too low. In future videos, we'll talk about how you can estimate the probability or how you can figure out whether your sample seems sufficient. But for this one, for what Roy did, I don't think 10 duckies is enough. | Worked example identifying sample study.mp3 |
But the larger your sample, the higher probability that your estimate is close to the actual parameter for the population. And 10 in this is just too low. In future videos, we'll talk about how you can estimate the probability or how you can figure out whether your sample seems sufficient. But for this one, for what Roy did, I don't think 10 duckies is enough. If he sampled maybe 100 duckies or more than that and he found that 10% of them were defective, well, that seems less likely to happen just purely due to chance. Let's do a few more of these. Actually, I'll do those in the next videos. | Worked example identifying sample study.mp3 |
They're closed on Sunday. So this is 100% of their customers for a week. If you add that up, you get 100%. I obviously am a little bit suspicious, so I decide to see how good this distribution that he's describing actually fits observed data. So I actually observe the number of customers when they come in during the week, and this is what I get for my observed data. So to figure out whether I want to accept or reject his hypothesis right here, I'm going to do a little bit of a hypothesis test. So I'll make the null hypothesis that the owner's distribution, so that's this thing right here, is correct. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
I obviously am a little bit suspicious, so I decide to see how good this distribution that he's describing actually fits observed data. So I actually observe the number of customers when they come in during the week, and this is what I get for my observed data. So to figure out whether I want to accept or reject his hypothesis right here, I'm going to do a little bit of a hypothesis test. So I'll make the null hypothesis that the owner's distribution, so that's this thing right here, is correct. And then the alternative hypothesis is going to be that it is not correct, that it is not a correct distribution, that I should not feel reasonably okay relying on this. It's not correct. I should reject the owner's distribution. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
So I'll make the null hypothesis that the owner's distribution, so that's this thing right here, is correct. And then the alternative hypothesis is going to be that it is not correct, that it is not a correct distribution, that I should not feel reasonably okay relying on this. It's not correct. I should reject the owner's distribution. And I want to do this with a significance level of 5%. Or another way of thinking about it, I'm going to calculate a statistic based on this data right here. And it's going to be a chi-square statistic, or another way to view it is that that statistic that I'm going to calculate has approximately a chi-square distribution. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
I should reject the owner's distribution. And I want to do this with a significance level of 5%. Or another way of thinking about it, I'm going to calculate a statistic based on this data right here. And it's going to be a chi-square statistic, or another way to view it is that that statistic that I'm going to calculate has approximately a chi-square distribution. And given that it does have a chi-square distribution with a certain number of degrees of freedom, and we're going to calculate that, what I want to see is the probability of getting this result, or getting a result like this or a result more extreme, less than 5%. If the probability of getting a result like this or something less likely than this is less than 5%, then I'm going to reject the null hypothesis, which is essentially just rejecting the owner's distribution. If I don't get that, if I say, hey, the probability of getting a chi-square statistic that is this extreme or more is greater than my alpha, than my significance level, then I'm not going to reject it. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
And it's going to be a chi-square statistic, or another way to view it is that that statistic that I'm going to calculate has approximately a chi-square distribution. And given that it does have a chi-square distribution with a certain number of degrees of freedom, and we're going to calculate that, what I want to see is the probability of getting this result, or getting a result like this or a result more extreme, less than 5%. If the probability of getting a result like this or something less likely than this is less than 5%, then I'm going to reject the null hypothesis, which is essentially just rejecting the owner's distribution. If I don't get that, if I say, hey, the probability of getting a chi-square statistic that is this extreme or more is greater than my alpha, than my significance level, then I'm not going to reject it. I'm going to say, well, I have no reason to really assume that he's lying. So let's do that. So to calculate the chi-square statistic, what I'm going to do is, so here we're assuming the owner's distribution is correct. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
If I don't get that, if I say, hey, the probability of getting a chi-square statistic that is this extreme or more is greater than my alpha, than my significance level, then I'm not going to reject it. I'm going to say, well, I have no reason to really assume that he's lying. So let's do that. So to calculate the chi-square statistic, what I'm going to do is, so here we're assuming the owner's distribution is correct. So assuming the owner's distribution was correct, what would have been the expected observed? So we have the expected percentage here, but what would have been the expected observed? So let me write this right here, expected. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
So to calculate the chi-square statistic, what I'm going to do is, so here we're assuming the owner's distribution is correct. So assuming the owner's distribution was correct, what would have been the expected observed? So we have the expected percentage here, but what would have been the expected observed? So let me write this right here, expected. I'll add another row. Expected. So we would have expected 10% of the total customers in that week to come in on Monday, 10% of the total customers of that week to come in on Tuesday, 15% to come in on Wednesday. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
So let me write this right here, expected. I'll add another row. Expected. So we would have expected 10% of the total customers in that week to come in on Monday, 10% of the total customers of that week to come in on Tuesday, 15% to come in on Wednesday. Now to figure out what that actual number is, we need to figure out the total number of customers. So let's add up these numbers right here. So we have, let me get the calculator out. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
So we would have expected 10% of the total customers in that week to come in on Monday, 10% of the total customers of that week to come in on Tuesday, 15% to come in on Wednesday. Now to figure out what that actual number is, we need to figure out the total number of customers. So let's add up these numbers right here. So we have, let me get the calculator out. So we have 30 plus 14 plus 34 plus 45 plus 57 plus 20. So there's a total of 200 customers who came into the restaurant that week. So let me write this down. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
So we have, let me get the calculator out. So we have 30 plus 14 plus 34 plus 45 plus 57 plus 20. So there's a total of 200 customers who came into the restaurant that week. So let me write this down. So this is equal to, so I wrote the total over here, total. Ignore this right here. I had 200 customers come in for the week. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
So let me write this down. So this is equal to, so I wrote the total over here, total. Ignore this right here. I had 200 customers come in for the week. So what was the expected number on Monday? Well, on Monday we would have expected 10% of the 200 to come in. So this would have been 20 customers, 10% times 200. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
I had 200 customers come in for the week. So what was the expected number on Monday? Well, on Monday we would have expected 10% of the 200 to come in. So this would have been 20 customers, 10% times 200. On Tuesday, another 10%. So we would have expected 20 customers. Wednesday, 15% of 200. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
So this would have been 20 customers, 10% times 200. On Tuesday, another 10%. So we would have expected 20 customers. Wednesday, 15% of 200. That's 30 customers. On Thursday, we would have expected 20% of 200 customers. So that would have been 40 customers. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
Wednesday, 15% of 200. That's 30 customers. On Thursday, we would have expected 20% of 200 customers. So that would have been 40 customers. Then on Friday, 30%. That would have been 60 customers. And then on Friday, 15% again. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
So that would have been 40 customers. Then on Friday, 30%. That would have been 60 customers. And then on Friday, 15% again. So 15% of 200 would have been 30 customers. So if this distribution is correct, this is the actual number that I would have expected. Now, to calculate our chi-squared statistic, we essentially just take, let me just show it to you. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
And then on Friday, 15% again. So 15% of 200 would have been 30 customers. So if this distribution is correct, this is the actual number that I would have expected. Now, to calculate our chi-squared statistic, we essentially just take, let me just show it to you. And I'll write it, instead of writing chi, I'm going to write a capital X squared. Sometimes someone will write the actual Greek letter chi here. But I'll write the X squared here so that it will be, and let me write it this way. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
Now, to calculate our chi-squared statistic, we essentially just take, let me just show it to you. And I'll write it, instead of writing chi, I'm going to write a capital X squared. Sometimes someone will write the actual Greek letter chi here. But I'll write the X squared here so that it will be, and let me write it this way. This is our chi-squared statistic. But I'm going to write it with a capital X instead of a chi because this is going to have approximately a chi-squared distribution. I can't assume that it's exactly. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
But I'll write the X squared here so that it will be, and let me write it this way. This is our chi-squared statistic. But I'm going to write it with a capital X instead of a chi because this is going to have approximately a chi-squared distribution. I can't assume that it's exactly. So we're dealing with approximations right here. But it's fairly straightforward to calculate. We take, for each of the days, we take the difference between the observed and the expected. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
I can't assume that it's exactly. So we're dealing with approximations right here. But it's fairly straightforward to calculate. We take, for each of the days, we take the difference between the observed and the expected. So it's going to be 30 minus 20. I'll do the first one color-coded. Squared divided by the expected. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
We take, for each of the days, we take the difference between the observed and the expected. So it's going to be 30 minus 20. I'll do the first one color-coded. Squared divided by the expected. So we're essentially taking the square of, almost you could kind of view it, the error between what we observed and expected, or the difference between what we observed and expected. And we're kind of normalizing it by the expected right over here. But we want to take the sum of all of these. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
Squared divided by the expected. So we're essentially taking the square of, almost you could kind of view it, the error between what we observed and expected, or the difference between what we observed and expected. And we're kind of normalizing it by the expected right over here. But we want to take the sum of all of these. I'll just do all of those in yellow. Plus 14 minus 20 squared over 20. Plus 34 minus 30 squared over 30. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
But we want to take the sum of all of these. I'll just do all of those in yellow. Plus 14 minus 20 squared over 20. Plus 34 minus 30 squared over 30. Plus, I'll continue over here, 45 minus 40 squared over 40. Plus 57 minus 60 squared over 60. And then finally, 20 minus 30 squared over 30. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
Plus 34 minus 30 squared over 30. Plus, I'll continue over here, 45 minus 40 squared over 40. Plus 57 minus 60 squared over 60. And then finally, 20 minus 30 squared over 30. I just took the observed minus the expected squared over the expected. I took the sum of it. And this is what gives us our chi-squared statistic. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
And then finally, 20 minus 30 squared over 30. I just took the observed minus the expected squared over the expected. I took the sum of it. And this is what gives us our chi-squared statistic. Now let's just calculate what this number is going to be. So this is going to be equal to, I'll do it over here so we don't run out of space. Let me do this in a new color. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
And this is what gives us our chi-squared statistic. Now let's just calculate what this number is going to be. So this is going to be equal to, I'll do it over here so we don't run out of space. Let me do this in a new color. I'll do it in the orange. This is going to be equal to, this is what? 30 minus 20 is 10 squared, which is 100 divided by 20, which is 5. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
Let me do this in a new color. I'll do it in the orange. This is going to be equal to, this is what? 30 minus 20 is 10 squared, which is 100 divided by 20, which is 5. I might not be able to do all of them in my head like this. Plus, actually let me just write it this way, just so you see what I'm doing. This is going to be 100, this right here is 100 over 20. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
30 minus 20 is 10 squared, which is 100 divided by 20, which is 5. I might not be able to do all of them in my head like this. Plus, actually let me just write it this way, just so you see what I'm doing. This is going to be 100, this right here is 100 over 20. Plus 14 minus 20 is negative 6 squared is positive 36. So plus 36 over 20. Plus 34 minus 30 is 4 squared is 16. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
This is going to be 100, this right here is 100 over 20. Plus 14 minus 20 is negative 6 squared is positive 36. So plus 36 over 20. Plus 34 minus 30 is 4 squared is 16. So plus 16 over 30. Plus 45 minus 40 is 5 squared is 25. So plus 25 over 40. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
Plus 34 minus 30 is 4 squared is 16. So plus 16 over 30. Plus 45 minus 40 is 5 squared is 25. So plus 25 over 40. Plus, the difference here is 3 squared is 9. So it's 9 over 60. Plus, we have a difference of 10 squared is 100 over 30. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
So plus 25 over 40. Plus, the difference here is 3 squared is 9. So it's 9 over 60. Plus, we have a difference of 10 squared is 100 over 30. Plus 100 over 30. And this is equal to, and I'll just get the calculator out for this. This is equal to, this is equal to, we have 100 divided by 20. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
Plus, we have a difference of 10 squared is 100 over 30. Plus 100 over 30. And this is equal to, and I'll just get the calculator out for this. This is equal to, this is equal to, we have 100 divided by 20. Plus 36 divided by 20. Plus 16 divided by 30. Plus 25 divided by 40. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
This is equal to, this is equal to, we have 100 divided by 20. Plus 36 divided by 20. Plus 16 divided by 30. Plus 25 divided by 40. Plus 9 divided by 60. Plus 100 divided by 30. Gives us 11.44. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
Plus 25 divided by 40. Plus 9 divided by 60. Plus 100 divided by 30. Gives us 11.44. So let me write that down. So this right here is going to be 11.44. This is my chi-square statistic, or we could call it a big capital X squared. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
Gives us 11.44. So let me write that down. So this right here is going to be 11.44. This is my chi-square statistic, or we could call it a big capital X squared. Sometimes you'll have it written as a chi-square. But this is approximately, this statistic is going to have approximately a chi-square distribution. Anyway, with that said, let's figure out, if we assume that it has roughly a chi-square distribution, what is the probability of getting a result this extreme, or at least this extreme, I guess is another way of thinking about it. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
This is my chi-square statistic, or we could call it a big capital X squared. Sometimes you'll have it written as a chi-square. But this is approximately, this statistic is going to have approximately a chi-square distribution. Anyway, with that said, let's figure out, if we assume that it has roughly a chi-square distribution, what is the probability of getting a result this extreme, or at least this extreme, I guess is another way of thinking about it. Or, another way of saying, is this a more extreme result than the critical chi-square value that there's a 5% chance of getting a result that extreme. So let's do it that way. Let's figure out the critical chi-square value, and if this is more extreme than that, then we will reject our null hypothesis. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
Anyway, with that said, let's figure out, if we assume that it has roughly a chi-square distribution, what is the probability of getting a result this extreme, or at least this extreme, I guess is another way of thinking about it. Or, another way of saying, is this a more extreme result than the critical chi-square value that there's a 5% chance of getting a result that extreme. So let's do it that way. Let's figure out the critical chi-square value, and if this is more extreme than that, then we will reject our null hypothesis. So let's figure out our critical chi-square value. So we have an alpha 5%. And actually, the other thing we have to figure out is the degrees of freedom. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
Let's figure out the critical chi-square value, and if this is more extreme than that, then we will reject our null hypothesis. So let's figure out our critical chi-square value. So we have an alpha 5%. And actually, the other thing we have to figure out is the degrees of freedom. The degrees of freedom here, we're taking 1, 2, 3, 4, 5, 6 sums. So you might be tempted to say the degrees of freedom are 6. But one thing to realize is that if you had all of this information over here, you could actually figure out this last piece of information. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
And actually, the other thing we have to figure out is the degrees of freedom. The degrees of freedom here, we're taking 1, 2, 3, 4, 5, 6 sums. So you might be tempted to say the degrees of freedom are 6. But one thing to realize is that if you had all of this information over here, you could actually figure out this last piece of information. So you actually have 5 degrees of freedom. When you have just kind of n data points like this, and you're measuring kind of the observed versus the expected, your degrees of freedom are going to be n minus 1, because you could figure out that nth data point just based on everything else that you have, all of the other information. So our degrees of freedom here are going to be 5. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
But one thing to realize is that if you had all of this information over here, you could actually figure out this last piece of information. So you actually have 5 degrees of freedom. When you have just kind of n data points like this, and you're measuring kind of the observed versus the expected, your degrees of freedom are going to be n minus 1, because you could figure out that nth data point just based on everything else that you have, all of the other information. So our degrees of freedom here are going to be 5. It's n minus 1. Our significance level is 5%. And our degrees of freedom is also going to be equal to 5. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
So our degrees of freedom here are going to be 5. It's n minus 1. Our significance level is 5%. And our degrees of freedom is also going to be equal to 5. So let's look at our chi-square distribution. We have a degree of freedom of 5. We have a significance level of 5%. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
And our degrees of freedom is also going to be equal to 5. So let's look at our chi-square distribution. We have a degree of freedom of 5. We have a significance level of 5%. We have a significance level of 5%. And so the critical chi-square value is 11.07. So let's go to this chart. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
We have a significance level of 5%. We have a significance level of 5%. And so the critical chi-square value is 11.07. So let's go to this chart. So we have a chi-square distribution with a degree of freedom of 5. So that's this distribution over here in magenta. And we care about a critical value of 11.07. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
So let's go to this chart. So we have a chi-square distribution with a degree of freedom of 5. So that's this distribution over here in magenta. And we care about a critical value of 11.07. So this is right here. You actually even can't see it on this. So if I were to keep drawing this magenta thing all the way over here, if the magenta line just kept the distribution, just kept going over here, you'd have 8. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
And we care about a critical value of 11.07. So this is right here. You actually even can't see it on this. So if I were to keep drawing this magenta thing all the way over here, if the magenta line just kept the distribution, just kept going over here, you'd have 8. Over here you'd have 10. Over here you'd have 12. 11.07 is maybe someplace right over there. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
So if I were to keep drawing this magenta thing all the way over here, if the magenta line just kept the distribution, just kept going over here, you'd have 8. Over here you'd have 10. Over here you'd have 12. 11.07 is maybe someplace right over there. So what it's saying is the probability of getting a result at least as extreme as 11.07 is 5%. Our result, so our critical chi-square value, so we could write even here, our critical chi-square value is equal to, we just saw, 11.07. Let me look at the chart again. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
11.07 is maybe someplace right over there. So what it's saying is the probability of getting a result at least as extreme as 11.07 is 5%. Our result, so our critical chi-square value, so we could write even here, our critical chi-square value is equal to, we just saw, 11.07. Let me look at the chart again. 11.07 is equal to 11.07. The result we got for our statistic is even less likely than that. It's even less likely than that. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
Let me look at the chart again. 11.07 is equal to 11.07. The result we got for our statistic is even less likely than that. It's even less likely than that. The probability is less than our significance level. So then we are going to reject. So the probability of getting that is, let me put it this way, 11.44 is more extreme than our critical chi-square level. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
It's even less likely than that. The probability is less than our significance level. So then we are going to reject. So the probability of getting that is, let me put it this way, 11.44 is more extreme than our critical chi-square level. So it's very unlikely that this distribution is true. So we will reject what he's telling us. We will reject this distribution. | Pearson's chi square test (goodness of fit) Probability and Statistics Khan Academy.mp3 |
She wants to test whether this holds true for teachers in her state. So she is going to take a random sample of these teachers and see what percent of them are members of a union. Let P represent the proportion of teachers in her state that are members of a union. Write an appropriate set of hypotheses for her significance test. So pause this video and see if you can do that. All right, now let's do it together. So what we wanna do for this significance test is set up a null hypothesis and an alternative hypothesis. | Constructing hypotheses for a significance test about a proportion AP Statistics Khan Academy.mp3 |
Write an appropriate set of hypotheses for her significance test. So pause this video and see if you can do that. All right, now let's do it together. So what we wanna do for this significance test is set up a null hypothesis and an alternative hypothesis. Now, your null hypothesis is the hypothesis that, hey, there's no news here. It's what you would expect it to be. And so if you read a report saying that 49% of teachers in the United States were members of labor unions, well, then it would be reasonable to say that the null hypothesis, the no news here, is that the same percentage of teachers in her state are members of a labor union. | Constructing hypotheses for a significance test about a proportion AP Statistics Khan Academy.mp3 |
So what we wanna do for this significance test is set up a null hypothesis and an alternative hypothesis. Now, your null hypothesis is the hypothesis that, hey, there's no news here. It's what you would expect it to be. And so if you read a report saying that 49% of teachers in the United States were members of labor unions, well, then it would be reasonable to say that the null hypothesis, the no news here, is that the same percentage of teachers in her state are members of a labor union. So that percentage, that proportion is P. So this would be the null hypothesis, that the proportion in her state is also 49%. And now what would the alternative be? Well, the alternative is that the proportion in her state is not 49%. | Constructing hypotheses for a significance test about a proportion AP Statistics Khan Academy.mp3 |
And so if you read a report saying that 49% of teachers in the United States were members of labor unions, well, then it would be reasonable to say that the null hypothesis, the no news here, is that the same percentage of teachers in her state are members of a labor union. So that percentage, that proportion is P. So this would be the null hypothesis, that the proportion in her state is also 49%. And now what would the alternative be? Well, the alternative is that the proportion in her state is not 49%. This is the thing that, hey, there would be news here. There'd be something interesting to report. There's something different about her state. | Constructing hypotheses for a significance test about a proportion AP Statistics Khan Academy.mp3 |
Well, the alternative is that the proportion in her state is not 49%. This is the thing that, hey, there would be news here. There'd be something interesting to report. There's something different about her state. And how would she use this? Well, she would take a sample of teachers in her state, figure out the sample proportion, figure out the probability of getting that sample proportion if we were to assume that the null hypothesis is true. If that probability is lower than a threshold, which she should have said ahead of time, her significance level, then she would reject the null hypothesis, which would suggest the alternative. | Constructing hypotheses for a significance test about a proportion AP Statistics Khan Academy.mp3 |
There's something different about her state. And how would she use this? Well, she would take a sample of teachers in her state, figure out the sample proportion, figure out the probability of getting that sample proportion if we were to assume that the null hypothesis is true. If that probability is lower than a threshold, which she should have said ahead of time, her significance level, then she would reject the null hypothesis, which would suggest the alternative. Let's do another example here. According to a very large poll in 2015, about 90% of homes in California had access to the internet. Market researchers want to test if that proportion is now higher. | Constructing hypotheses for a significance test about a proportion AP Statistics Khan Academy.mp3 |
If that probability is lower than a threshold, which she should have said ahead of time, her significance level, then she would reject the null hypothesis, which would suggest the alternative. Let's do another example here. According to a very large poll in 2015, about 90% of homes in California had access to the internet. Market researchers want to test if that proportion is now higher. So they take a random sample of 1,000 homes in California and find that 920, or 92% of homes sampled, have access to the internet. Let P represent the proportion of homes in California that have access to the internet. Write an appropriate set of hypotheses for their significance test. | Constructing hypotheses for a significance test about a proportion AP Statistics Khan Academy.mp3 |
Market researchers want to test if that proportion is now higher. So they take a random sample of 1,000 homes in California and find that 920, or 92% of homes sampled, have access to the internet. Let P represent the proportion of homes in California that have access to the internet. Write an appropriate set of hypotheses for their significance test. So once again, pause this video and see if you can figure it out. So once again, we want to have a null hypothesis and we want to have an alternative hypothesis. The null hypothesis is the, hey, there's no news here. | Constructing hypotheses for a significance test about a proportion AP Statistics Khan Academy.mp3 |
Write an appropriate set of hypotheses for their significance test. So once again, pause this video and see if you can figure it out. So once again, we want to have a null hypothesis and we want to have an alternative hypothesis. The null hypothesis is the, hey, there's no news here. And so that would say that, you know, it's kind of the status quo, that the proportion of people who have internet is still the same as the last study, is still the same at 90%. Or I could write 90%, or I could write 0.9 right over here. Now some of you might have been tempted to put 92% there. | Constructing hypotheses for a significance test about a proportion AP Statistics Khan Academy.mp3 |
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