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## Statistics and probability

### Course: Statistics and probability > Unit 13

Lesson 2: Comparing two means- Statistical significance of experiment
- Statistical significance on bus speeds
- Hypothesis testing in experiments
- Difference of sample means distribution
- Confidence interval of difference of means
- Clarification of confidence interval of difference of means
- Hypothesis test for difference of means

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# Hypothesis test for difference of means

Hypothesis Test for Difference of Means. Created by Sal Khan.

## Want to join the conversation?

- Sal, Can we solve the same problem like this:

The mean difference = 1.91, the null hypothesis mean difference is 0. Standard deviation is 0.617. Z = (0-1.91)/0.617 = -3.09. It takes -3.09 standard deviations to get a value 0 in this distribution. which when converted to the probability = normsdist(-3.09) = 0.001 which indicates 0.1% probability which is within our significance level :5%. So we can safely reject the null hypothesis.I probably had built wrong concept in my head.correct me plz.(17 votes)- Yes, you are mostly right, there are two ways to think this problem as you ca try to find out either the:

- "probability of the difference of the samples means being 1.91 or more assuming the difference of the populations means is 0" (this is what Sal does here, except that he only tries to prove that this probability is less than 5%) or the

- "probability of the difference of the samples means being 0 or less (it makes sense because you can theoretically have the people in the first group**gaining**weight and because here we have a distribution centered around 1.91) assuming the difference of the populations means is 1.91"

...and you can see that the probabilities are the same because you get the same Z score (3.09, the sign is meaningless) and corresponding probabilities equal (0.1%, if you looked up the values right, I didn't check your calculations). The math is the same!

So, essentially, you can think of the problem both ways, but Sal's way is standard way used by scientists, so when explaining something to someone else is better to use this form to make sure everybody understands you (also, in other more complicated setups, the variances might not be the same in the context of the null and alternate hypotheses, so going the "standard" way increases the chances of getting things right).

Hope I understood you question good enough to answer it and this helps.

offtopic: Sal almost never answers questions, guess he's too busy making new content for other courses...(8 votes)

**Even with an n>30, I still don't agree**with using the sample's standard deviation as a valid approximation of the population's standard deviation. Say you test your sample the way Sal does it, and realize that the probability of you getting that sample was 1%. Normally, you would reject the null hypothesis. But say the null hypothesis was indeed correct. This means you just happened to choose a lot samples from the far left or far right of the population mean. Since your sample is representative of such a small extreme section, the standard deviation of your sample would have been a lot smaller than the true standard deviation of the total population. Therefore, since Sal used the sample standard deviation as his population standard deviation, he would have**underestimated the population standard deviation**, and consequently**overestimated the z score**of that sample. The degree at which he overestimated? I'm not sure, I think I'd have to solve some recursive function that is really hard to think about right now. Anyways, I'm sure there's something I'm missing, because**I have full faith in Sal Khan**. Please let me know if I have overlooked something.(7 votes)- > "The degree at which he overestimated? I'm not sure, I think I'd have to solve some recursive function that is really hard to think about right now."

I'm not sure about that. There might be some way to formally express the degree of error, but I like jumping to simulations. The Type I error -- rejecting Ho when Ho is actually true -- is slightly inflated. By the time we get to n>30 it's not really by much, but it's a bit above what it should be. This means that using the Z-test when we should use the T-test will conclude a significant result a bit too often.(4 votes)

- If we are assuming that the null hypothesis is true (there is no difference between the two diets), why are we using each of the sample standard deviations separately, as if they are separate populations? If the null hypothesis is true, both samples are taken from the same population. Shouldn't we then take an average of the two standard deviations, or recalculate the standard deviation of the whole n=200 sample?(8 votes)
- I was thinking in the same way so I calculated the the null hypothesis distribution STD from the average of variances and the result was 0.6165 which is so close to the 0.617 that Sal used.(2 votes)

- I thought the hypotheses have to complement each other...? So if one is "equals zero" shouldn't the other be "does not equal zero"?(5 votes)
- I guess the "correct" H(0) should be "u1-u2 <= 0" (I'll call this H(0)_1) instead of H(0): u1-u2 = 0 as Sal writes (I'll call this H(0)_2), bu you can easily see that:

- if we assume H(0)_1 instead of H(0)_2, we would get a Z_1 >= Z_2 (think visually: we "slide" the bell curve to the left ...or if you're math inclined write the inequalities about Z_1, even if you can't really calculate this Z_1)

- next, if Z_1 >= Z_2, then P_1 <= P_2 (where P is the probability of obtaining the given difference of the means)

- next, if P_2 < 0.5% (our significance threshold) and P_1 <= P_2, then P_1 < 0.5%

- so this "correct" H(0)_1 is actually implied by Sal's H(0)_2, so it's practically the same thing.

...so short answer: yes, you're right, but "is greater than 0" (H(0)_2) actually implies "does not equal zero", so proving the former is implied proof for the latter.(3 votes)

- His critical value Z score is 1.65 because of the 95%. I'm a little confused, because my chart says 95% should be 1.96. Am i using a different chart for something unrelated to z scores?(4 votes)
- One sided vs two sided is important. If we're testing for a "difference" then we need to split the type 1 error probability (alpha) into the two tails, so 0.05 translates to a z value of 1.96. If we're only looking for an increase or decrease, then we put all of the alpha probability into one tail, which leads to a z value of 1.65.(9 votes)

- At1:04, I don't understand why the mean of the sampling dist. is the same as the mean of the population dist.

Couldn't your sample mean be quite different? I.e. couldn't you draw a sample that had a higher mean than the overall population ?

Or is he saying that the mean of the distribution of all the samples you draw is going to be the same as the mean of the overall population?(4 votes)- Great question. And yes, your last sentence is the correct explanation! This is why inferential stats can be confusing, and why every single word is critical when we talk about means and samples and distributions. As you say, the mean of ONE SAMPLE could be (in fact, almost always IS) different from the population mean. But the mean value of the distribution of all the sample means (phew!) will be the same as the population mean. Of course, you DON'T HAVE "all the sample means", you only have ONE of them (usually)! So the "distribution of all sample means" is usually something we just imagine as a theoretical abstraction. But it's critical to understand what that distribution represents (we have to imagine doing the same experiment many, many times) in order for hypothesis testing to make sense.(7 votes)

- I'm baffled as to why we keep using our original sample standard deviations as estimates for the population SDs (c.6:50) once we're assuming the null hypothesis. If (and I might be barking up the wrong tree here) the hypothesis is that there's no meaningful difference whatsoever in weight loss effect between the two diets, why should their SDs remain distinct when imagined across the whole population? If the two groups' data are basically identical when viewed globally, shouldn't their SDs be identical too?(5 votes)
- because it would lead to same answer. if you sample twice from the same population then the best variance estimator is ((n1-1)var(x1) + (n2-1)var(x2))/(n1+n2-2) ... i know you understand which symbol means what here .. now calculate for variance of difference of means of two iid samples from this population using the just calculated estimate of variance. It is the same thing as what sal does(2 votes)

- Man, am I confused. I can't understand why we reject the null if there is a low probablility of getting a favorable result. wouldn't that prove the null is correct?(1 vote)
- All of the above is correct but I think Sal said it pretty well himself in an earlier video: We presume the null hypothesis is true -- so if that's the case what are the chances of getting the sample results we got? He just proved that the chances of getting the results we actually got are < 5%, if the null hypothesis is true. The result in this case fell into the rejection region. So the lower the percentage, the less likely it is that we would've gotten that result.(8 votes)

- On 6.43 sal used the same "The standard deviation of the difference of the sample mean" as the video he used before. But on the last video on this chapter (named "Hypothesis test comparing population proportions), he calculated at new Standard Deviation for the Null Hypothesis. Why didn't he do it here? Where in both cases the Null hypothesis was u1-u2=0.(4 votes)
- I think you have a valid observation and unfortunately I don't have an answer for it. I scrolled all the answers in this thread and looks like no one has responded to your query so far. Please let me know if you have found/(or happen to find in future) the reason as to why new SD was not calculated after assuming null hypothesis as true.(3 votes)

- Hello..

Is it ok, if I use general t-test to solve the same problem?(4 votes)- Since you are dealing with means, YES! You very well can. Please flag me or downvote if I am wrong, but upvote if I am correct.(3 votes)

## Video transcript

In the last video, we came up
with a 95% confidence interval for the mean weight loss between
the low-fat group and the control group. In this video, I actually want
to do a hypothesis test, really to test if this data
makes us believe that the low-fat diet actually does
anything at all. And to do that let's set up
our null and alternative hypotheses. So our null hypothesis
should be that this low-fat diet does nothing. And if the low-fat diet does
nothing, that means that the population mean on our low-fat
diet minus the population mean on our control should
be equal to zero. And this is a completely
equivalent statement to saying that the mean of the sampling
distribution of our low-fat diet minus the mean of the
sampling distribution of our control should be
equal to zero. And that's because we've seen
this multiple times. The mean of your sampling
distribution is going to be the same thing as your
population mean. So this is the same
thing is that. That is the same
thing is that. Or, another way of saying it is,
if we think about the mean of the distribution of the
difference of the sample means, and we focused on this
in the last video, that that should be equal to zero. Because this thing right over
here is the same thing as that right over there. So that is our null
hypothesis. And our alternative hypothesis, I'll write over here. It's just that it actually
does do something. And let's say that it actually
has an improvement. So that would mean that we
have more weight loss. So if we have the mean of Group
One, the population mean of Group One minus the
population mean of Group Two should be greater then zero. So this is going to be a one
tailed distribution. Or another way we can view it,
is that the mean of the difference of the distributions,
x1 minus x2 is going to be greater then zero. These are equivalent
statements. Because we know that this is the
same thing as this, which is the same thing as this,
which is what I wrote right over here. Now, to do any type of
hypothesis test, we have to decide on a level
of significance. What we're going to do is, we're
going to assume that our null hypothesis is correct. And then with that assumption
that the null hypothesis is correct, we're going to see
what is the probability of getting this sample data
right over here. And if that probability is below
some threshold, we will reject the null hypothesis in
favor of the alternative hypothesis. Now, that probability threshold,
and we've seen this before, is called the
significance level, sometimes called alpha. And here, we're going to decide
for a significance level of 95%. Or another way to think about
it, assuming that the null hypothesis is correct, we want
there to be no more than a 5% chance of getting this
result here. Or no more than a 5% chance of
incorrectly rejecting the null hypothesis when it
is actually true. Or that would be a
type one error. So if there's less than a 5%
probability of this happening, we're going to reject
the null hypothesis. Less than a 5% probability given
the null hypothesis is true, then we're going to reject
the null hypothesis in favor of the alternative. So let's think about this. So we have the null
hypothesis. Let me draw a distribution
over here. The null hypothesis says that
the mean of the differences of the sampling distributions
should be equal to zero. Now, in that situation, what
is going to be our critical region here? Well, we need a result, so
we're going to need some critical value here. Because this isn't a
normalized normal distribution. But there's some critical
value here. The hardest thing is statistics
is getting the wording right. There's some critical value here
that the probability of getting a sample from this
distribution above that value is only 5%. So we just need to figure out
what this critical value is. And if our value is larger than
that critical value, then we can reject the
null hypothesis. Because that means the
probability of getting this is less than 5%. We could reject the null
hypothesis and go with the alternative hypothesis. Remember, once again, we can
use Z-scores, and we can assume this is a normal
distribution because our sample size is large for either
of those samples. We have a sample size of 100. And to figure that out, the
first step, if we just look at a normalized normal distribution
like this, what is your critical Z value? We're getting a result
above that Z value, only has a 5% chance. So this is actually
cumulative. So this whole area right
over here is going to be 95% chance. We can just look
at the Z table. We're looking for 95% percent. We're looking at the
one tailed case. So let's look for 95%. This is the closest thing. We want to err on the side of
being a little bit maybe to the right of this. So let's say 95.05
is pretty good. So that's 1.65. So this critical Z value
is equal to 1.65. Or another way to view it is,
this distance right here is going to be 1.65 standard
deviations. I know my writing
is really small. I'm just saying the standard
deviation of that distribution. So what is the standard
deviation of that distribution? We actually calculated it in
the last video, and I'll recalculate it here. The standard deviation of our
distribution of the difference of the sample means is going to
be equal to the square root of the variance of our
first population. Now, the variance of our first
population, we don't know it. But we could estimate it with
our sample standard deviation. If you take your sample standard
deviation, 4.67 and you square it, you get
your sample variance. And so this is the variance. This is our best estimate
of the variance of the population. And we want to divide that
by the sample size. And then plus our best estimate
of the variance of the population of group two,
which is 4.04 squared. The sample standard deviation
of group two squared. That gives us variance
divided by 100. I did before in the last. Maybe
it's still sitting on my calculator. Yes, it's still sitting
on the calculator. It's this quantity
right up here. 4.67 squared divided
by 100 plus 4.04 squared divided by 100. So it's 0.617. So this right here is
going to be 0.617. So this distance right
here, is going to be 1.65 times 0.617. So let's figure out
what that is. So let's take 0.617
times 1.65. So it's 1.02. This distance right
here is 1.02. So what this tells us is, if
we assume that the diet actually does nothing, there's a
only a 5% chance of having a difference between the means of
these two samples to have a difference of more than 1.02. There's only a 5%
chance of that. Well, the mean that we
actually got is 1.91. So that's sitting out
here someplace. So it definitely falls in
this critical region. The probability of getting this,
assuming that the null hypothesis is correct,
is less than 5%. So it's smaller probability than
our significance level. Actually, let me
be very clear. The significance level,
this alpha right here, needs to be 5%. Not the 95%. I think I might have
said here. But I wrote down the
wrong number there. I subtracted it from
one by accident. Probably in my head. But anyway, the significance
level is 5%. The probability given that the
null hypothesis is true, the probability of getting the
result that we got, the probability of getting that
difference, is less than our significance level. It is less than 5%. So based on the rules that we
set out for ourselves of having a significance level of
5%, we will reject the null hypothesis in favor of the
alternative that the diet actually does make you
lose more weight.