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Confidence interval of difference of means

Sal uses a confidence interval to help figure out if a low-fat diet helps obese people lose weight. Created by Sal Khan.

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  • leaf red style avatar for user Destin Andersen
    Well I understand Z Scores and Normal Distribution but i'm having a hard time understanding Confidence Intervals.
    (3 votes)
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    • leaf blue style avatar for user Matthew Daly
      Let's say that I run a sandwich shop. From experience, I know that the number of sandwiches I sell is normally distributed with a mean of 100 and a standard deviation of 10. There are two things that can go wrong with my store -- either I don't have enough customers to meet my labor costs or I run out of the bread and have to turn away all my late customers. So when I use my z-score stuff to realize that there is a 99.75% probability that I will sell between 70 and 130 sandwiches, that's important enough that we want to call [70,130] a 99.75% confidence interval. Then I conclude that if I order 130 rolls and schedule enough workers for a 70 worker day that I should expect only about one "bad day" per year.
      (10 votes)
  • blobby green style avatar for user Elizabeth Topczewski
    @ Andrew M:
    Yes, I have encountered the CLT. Why exactly does it apply here? I thought the CLT said that given a sufficiently large sample size n, the distribution of the means of the samples would be approximately normally distributed. Why should the differences between two distributions of sample means be normally distributed? Those aren't means.

    The only thing I can think of intuitively is that maybe if you add two normal r.v.s together, the resulting distribution is normal? Is this true? Otherwise, what parts of the CLT correspond to the difference of sample means?

    Thanks!
    Beth

    Thanks!
    (5 votes)
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  • leaf green style avatar for user Anshu Dwibhashi
    Okay, so we can make a sample distribution for the people who took the low fat diet and one for those who didn't. But why do we find their difference? My guess is that by finding the difference we get how much more effective is the former than the latter. But then how does the confidence interval part fit into this picture.

    I'm looking for an answer that gives me an intuition more than a technical answer, if possible...
    (3 votes)
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    • male robot johnny style avatar for user Thomas B
      The confidence interval is there to account for the randomness associated with any sort of trial. People lose weight at different rates, some likely stick to the diet better than others, and people just experience general fluctuations in weight that may have effected their starting or ending weights. All these things add "noise" to the measurement and the confidence interval is a way to show how much noise is associated with a given result.

      If the results of a study are well within that noise, it makes the difference much less credible than one where the noise is small relative to the difference in results.
      (4 votes)
  • blobby green style avatar for user Ryan Callahan
    Was this example a real study?
    (5 votes)
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  • blobby green style avatar for user bvkjbourbeau
    Can we use the same calculation if the data is not normally distributed? What about if the sample sizes are not equal?
    (3 votes)
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  • blobby green style avatar for user Danny Weissberger
    Why are you dividing by sample size for variance here? My understanding was that this is only for the standard deviation of a sample distribution of sample means and that does not seem to be the case here. You just have 2 samples of size n/m. Can you please clarify?
    (2 votes)
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  • blobby green style avatar for user Pranavi A V S
    mu x1 is the sample std dev which was given in the qn right, why are we again approximating using population std dev and dividing by 100?
    (2 votes)
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  • mr pink red style avatar for user Eric Waters
    When to use a Z Table?
    I understand the rule for 1 data set; If the sample size is less than 30, use a T-table.
    But for 2 data sets, each data set sample size could be different and you could have a scenario one data set sample size is above 30 and the other is below 30. My guess is that a conservative answer would be to use a T-table if either data set is below 30. Is that a good rule of thumb or are there other factors that come into play?
    (2 votes)
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  • leafers tree style avatar for user Delaine
    At about , why doesn't 4.67/10 + 4.04/10 work?
    (1 vote)
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    • leaf blue style avatar for user Dr C
      The variance of the sum (or difference) of two random variables is additive. So for random variables X and Y:

      V( X + Y ) = V(X) + V(Y)
      V( X - Y ) = V(X) + V(Y)

      This comes from the mathematical theory. I'm not sure where (or if) this is covered on Khan Academy.

      Then, the 4.67/10 and 4.04/10 are the standard errors (standard deviation) of the two sample means. In order to be able to add them, we first need to convert them into variances by squaring them.
      (3 votes)
  • blobby green style avatar for user marce020
    were we asked to make a 95% C.I.? Or did we choose that? and if we did, why 95% and not any other percentage?
    (2 votes)
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Video transcript

We're trying to test whether a new low-fat diet actually helps obese people lose weight. 100 randomly assigned people are assigned to group one and put on the low-fat diet. Another 100 randomly assigned obese people are assigned to group two and put on a diet of approximately the same amount of food, but not as low in fat. So group two is the control, just the no diet. Group one is the low fat group, to see if it actually works. After four months, the mean weight loss was 9.31 pounds for group one. Let me write this down. Let me make it very clear. So the low fat group, the mean weight loss was 9.31 pounds. So our sample mean for group one is 9.31 pounds, with a sample standard deviation of 4.67. And both of these are obviously very easy to calculate from the actual data. And then for our control group, the sample mean, 7.40 pounds for group two. With a sample standard deviation of 4.04 pounds. And now, if we just look at it superficially, it looks like the low-fat group lost more weight, just based on our samples, than the control group. If we take the difference of them. So if we take the difference between the low-fat group and the control group, we get 9.31 minus 7.40 is equal to, let's get the calculator out, 1.91. So the difference of our samples is 1.91. So just based on what we see, maybe you lose an incremental 1.91 pounds every four months if you are on this diet. And what we want to do in this video is to get a 95% confidence interval around this number. To see that in that 95% confidence interval, maybe, do we always lose weight? Or is there a chance that we can actually go the other way with the low-fat diet? So in this video, 95% confidence interval. In the next video, we'll actually do a hypothesis test using this same data. And now to do a 95% confidence interval, let's think about the distribution that we're thinking about. So let's look at the distribution. Of course we're going to think about the distribution that we're thinking about. We want to think about the distribution of the difference of the means. So it's going to have some true mean here. Which is the mean of the difference of the sample means. Let me write that. It's not a y, it's an x1 and x2. So it's the sample mean of x1 minus the sample mean of x2. And then this distribution right here is going to have some standard deviation. So it's the standard deviation of the distribution of the mean of x1 minus the sample mean of x2. It's going to have some standard deviation here. And we want to make an inference about this. Or I guess, the best way to think about it, we want to get a 95% confidence interval. Based on our sample, we want to create an interval around this, where we're confident that there's a 95% chance that this true mean, the true mean of the differences, lies within that interval. And to do that let's just think of it the other way. How can we construct an interval around this where we are 95% sure that any sample from this distribution, and this is one of those samples, that there is a 95% chance that we will select from this region right over here. So we care about a 95% region right over here. So how many standard deviations do we have to go in each direction? And to do that we just have to look at a Z table. And just remember, if we have 95% in the middle right over here, we're going to have 2.5% over here and we're going to have 2.5% over here. We have to have 5% split between these two symmetric tails. So when we look at a Z table, we want the critical Z value that they give right over here. And we have to be careful here. We're not going to look up 95%, because a Z table gives us the cumulative probability up to that critical Z value. So the Z table is going to be interpreted like this. So there's going to be some Z value right over here where we have 2.5% above it. The probability of getting a more extreme result or Z score above that is 2.5%. And the probability of getting one below that is going to be 97.5%. But if we can find whatever Z value this is right over here, it's going to be the same Z value as that. And instead of thinking about it in terms of a one tail scenario, we're going to think of it in a two tail scenario. So let's look it up for 97.5% on our Z table. Right here. This is 0.975, or 97.5. And this gives us Z value of 1.96. So Z is equal to 1.96. Or 2.5% of the samples from this population are going to be more than 1.96 standard deviations away from the mean. So this critical Z value right here is 1.96 standard deviations. This is 1.96 times the standard deviation of x1 minus x2. And then this right here is going to be negative 1.96 times the same thing. Let me write that. So this right here, it's symmetric. This distance is going to be the same as that distance. So this is negative 1.96 times the standard deviation of this distribution. So let's put it this way, there's a 95% chance that our sample that we got from our distribution-- this is the sample as a difference of these other samples. There's a 95% chance that 1.91 lies within 1.96 times the standard deviation of that distribution. So you could view it as a standard error of this statistic. So x1 minus x2. Let me finish that sentence. There's a 95% chance that 1.91, which is the sample statistic, or the statistic that we got, is within 1.96 times the standard deviation of this distribution of the true mean of of the distribution. Or we could say it the other way around. There's a 95% chance that the true mean of the distribution is within 1.96 times the standard deviation of the distribution of 1.91. These are equivalent statements. If I say I'm within three feet of you, that's equivalent to saying you're within three feet of me. That's all that's saying. But when we construct it this way, it becomes pretty clear, how do we actually construct the confidence interval? We just have to figure out what this distance right over here is. And to figure out what that distance is, we're going to have to figure out what the standard deviation of this distribution is. Well the standard deviation of the differences of the sample means is going to be equal to, and we saw this in the last video-- in fact, I think I have it right at the bottom here-- it's going to be equal to the square root of the variances of each of those distributions. Let me write it this way. So the variance, I'll kind of re-prove it. The variance of our distribution is going to be equal to the sum of the variances of each of these sampling distributions. And we know that the variance of each of these sampling distributions is equal to the variance of this sampling distribution, is equal to the variance of the population distribution, divided by our sample size. And our sample size in this case is 100. And the variance of this sampling distribution, for our control, is going to be equal to the variance of the population distribution for the control divided by its sample size. And since we don't know what these are, we can approximate them. Especially, because our n is greater than 30 for both circumstances. We can approximate these with our sample variances for each of these distributions. So let me make this clear. Our sample variances for each of these distributions. So this is going to be our sample standard deviation one squared, which is the sample variance for that distribution, over 100. Plus my sample standard deviation for the control squared, which is the sample variance. Standard deviation squared is just the variance divided by 100. And this will give us the variance for this distribution. And if we want the standard deviation, we just take the square roots of both sides. If we want the standard deviation of this distribution right here, this is the variance right now, so we just need to take the square roots. Let's calculate this. We actually know these values. S1, our sample standard deviation for group one is 4.67. We wrote it right here, as well. It's 4.76 and 4.04. The S is 4.67, we're going to have to square it. And the S2 is 4.04, we're going to have to square it. So let's calculate that. So we're going to take the square root of 4.67 squared divided by 100 plus 4.04 squared, divided by 100. And then close the parentheses. And we get 0.617. Let me write it right here. This is going to be equal to 0.617. So if we go back up over here, we calculated the standard deviation of this distribution to be 0.617. So now we can actually calculate our interval. Because this is going to be 0.617. So if you want 1.96 times that, we get 1.96 times that 0.617, I'll just write the answer we just got. So we get 1.21. So the 95% confidence interval is going to be the difference of our means, 1.91, plus or minus this number, 1.21. So what's our confidence interval? So the low end of our confidence interval, and I'm running out of space, 1.91 minus 1.21, that's just 0.7. So the low end is 0.7. And then the high end, 1.91 plus 1.21, that's 2.12. let me just make sure of that. My brain sometimes doesn't work properly when I'm making these videos. 3.12. So just to be clear, there's not a pure 95% chance that the true difference of the true means lies in this. We're just confident that there's a 95% chance. And we always have to put a little confidence there, because remember, we didn't actually know the population standard deviations, or the population variances. We estimated them with our sample. And because of that, we don't know that it's an exact probability. We just have say we're confident that it is a 95% probability. And that's why we just say it's a confidence interval. It's not a pure probability. But it's a pretty neat result. So we're confident that there's a 95% chance that the true difference of these two samples-- and remember, let me make it very clear, the expected value of the sample means is actually the same thing as the expected value of the populations. And so, what this is giving us is actually a confidence interval for the true difference between the populations. If you were to give everyone, every possible person, diet one. And every possible person diet two. This is giving us a confidence interval for the true population means. And so when you look at this, it looks like diet one actually does do something. Because in any case, even at the low end of the confidence interval, you still have a greater weight loss than diet two. Hopefully, that doesn't confuse you too much. In the next video, we're actually going to do a hypothesis test with the same data.