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## Statistics and probability

### Course: Statistics and probability>Unit 12

Lesson 1: The idea of significance tests

# Simple hypothesis testing

Sal walks through an example about who should do the dishes that gets at the idea behind hypothesis testing. Created by Sal Khan.

## Want to join the conversation?

• Isn't that independent event? Sal
no matter how many times he pick, still got 3/4 chance
(5 votes)
• Yes, the events are independent. If independent events happen n times in a row, then the probability becomes p^n, where p is the probability of the event happening once, and n is the number of times that it happens.
(8 votes)
• how do you find/tell the hypothesis to plan a statistical investigation? like " a psychologist wants to look at the factors that may affect memory. She thinks gender is likely to be a factor, that is, males and females might be different.How could the psychologist test her theory? Write down any factors that you think might make a difference to memory." Do you have any simple methods on understanding this topic??
(4 votes)
• Why can't we solve this problem using P(Bill not picked) = 1 - P(Bill get picked)?
P(Bill get picked) = 1/64. If we use that we get answer as 63/64?
(2 votes)
• That's the probability that he gets picked all three nights. The opposite of "gets picked all three nights" is NOT "avoids dishes all three nights".
(3 votes)
• @2.45, why did you multiply the probability i was thinking it show be add ?.. why can't we add the probability nd why it should only be multiplied?
look: in other way, can the question be re-framed as, what is the "total" probability of bill not picking ? so its "total" so i thought it is addition. but correct me if my intuitions are wrong..
(2 votes)
• We add probabilities in different circumstances than we multiply them. In this case, we multiply because we are finding the odds of several various independent events happening. For example, if we were to toss two coins, the odds of the first coin coming up heads is 1/2. The odds of the second coin coming up heads is also 1/2. How do we find the odds of both coming up heads? We multiply the probabilities, as there is a 50% chance of the first coin coming up heads (in theory, one out of every two times the coin will be heads). Therefore, in the universe of first-coin-flip-heads, 1 out of ever 2 flips there will theoretically come up with a second head. Overall, then, 25% of the time we will get two heads.
I hope that was a little bit helpful.
(2 votes)
• Does anybody know why 3/4 though?
(1 vote)
• 3/4, it consists of the older brothers siblings and he is with his siblings in the denominator. The numerator on the other hand, shows only his younger siblings-the ones drawn from the bowl to wash dishes.
(HOPE THIS MAKES SENSE)
(4 votes)
• Wouldn't the older brothers chance of being selected each night be 1/4 rather than 3/4? There is a 1 in 4 chance he is selected each night. Wouldn't 3 in 4 be the chance that anyone but him be selected? It's possible I missed part of the premise.
(1 vote)
• Remember, were not talking about the older brother being picked, rather, the older brother not being picked, so there's a 3/4 chance that the older brother will not be picked each night, hope this helped,
- Fire Programming
(2 votes)
• I don't manage to see the link between rejecting the hypothesis and the low probability of the observed results.
Using the Alien problem.
A) 20% of the observed sample is rebellious
B) The hypothesis is that 10% are rebellious
Let´s simulate to see how likely is (A) to happen. Simulated results show that it is unlikely (A) to happen therefore i have to reject (B).? that is what I don´t get.
In my mind I have two options, (A) or (B), if (A) doesn´t happen then (B) happens, but here it is like those are linked together and really don't understand where that link is.
(0 votes)
• The two are linked by an assumption that we make. We calculate the probability of A while assuming that B is actually true. That's why we reject the null hypothesis for small probabilities. We're saying, "We made an assumption, and the data we observed are extremely unlikely under that assumption, so they can't both be true. But ... the data are real, and our assumption is, well, and assumption, so we'll believe the data. That makes our assumption wrong."
(6 votes)
• Isn't there a 100% that at least one child hasn't been picked after 3 days?
(1 vote)
• That is correct, however the reason why the probability of Bill not being picked is so low, is because it is very likely that Bill is one of those 3 that have been picked, because he is in a group of 4.
(1 vote)
• I don't understand why Bill's chance of not getting picked 3 nights in a row is 42%, shouldn't it be 1/4 because there are four siblings, and if one of them gets picked randomly every night, then Bill not getting picked is a 3/4 chance, but it's a 1/4 chance because his chance of getting picked on Day 1 is 1/4, his probability of getting picked on Day 2 is 1/4 and his probability of getting picked on Day 3 is 1/4, because he's 1 of 4 siblings, therefore 1/4. I don't understand, can someone help?
(1 vote)
• You are right in that Bill's chance of getting picked any particular day is 1/4. Also that makes his chance of not getting picked any particular day 3/4.
But here, we're not finding the probability that Bill is picked on the fourth day, we're finding the chances that he didn't get picked any of the first three days.
What would you put the chances of someone scoring a basket from half court? If they're good say maybe 10%.
But what would you put their chances of doing it 10 times in a row, or 20? Doesn't it naturally seem like scoring a basket from half court 20 times in a row is much less likely than doing it on a particular attempt?

Same thing here. 3/4 is the probability that Bill isn't picked any particular day. But when he isn't picked the first day, not the second day, not the third day? That becomes (3/4)*(3/4)*(3/4)=27/64=42% approx.
He doesn't get picked 12 times? (3/4)^12.

Sorry if the basketball reference I used didn't have the right terms. But I hope that clears up some stuff.

Cheers!
(2 votes)
• Acasa insurance company offers plans for home insurance..plan A has a \$700 yearly prem. With a \$3000 deductible. Plan B has a \$1200 yearly premium with a \$1000 deductible ... water damage ( from broken pipes.overflow etc.,) result in approximately 12% of claims overall.with an average claim being \$5200. What is the expected value ( loss or gain) to acasa ..A - 1060..B 436 .. C -360 D-100
(1 vote)

## Video transcript

Let's say that we have four siblings right over here. They're trying to decide how to pick who should do the dishes each night. The oldest sibling right over here he decides, "I'll just put all of our names into a bowl and then I'll "just randomly pick one of our names out of the bowl "each night and then that person is going to be -," so this is the bowl right over here and I'm going to put four sheets of paper in there. Each of them is going to have one of their names. He's just going to randomly pick it out each night and that's the person who's going to do their dishes. They all say, "That seems like a reasonably fair thing to do," so they start that process. Let's say that after the first three nights that he, the oldest brother here. Let's call him Bill. Let's say after three nights Bill has not had to do the dishes. At that point the rest of the siblings are starting to think maybe, just maybe something fishy is happening. What I want to think about is what is the probability of that happening. What's the probability of three nights in a row Bill does not get picked? If we assume that we were randomly taking, if Bill was truly randomly taking these things out of the bowl and not cheating in some way. What's the probability that that would happen? That three nights in a row Bill would not be picked. I encourage you to pause the video and think about that. Let's think about the probability that Bill's not picked on a given night. If it's truly random, so we're going to assume that Bill is not cheating. Assume truly random and that each of the sheets of paper have a one in four chance of being picked, what's the probability that Bill does not get picked? The probability that, I guess let me write this, Bill not picked on a night. Well, there's four equally likely outcomes and three of them result in Bill not getting picked, so there's a three fourths probability that Bill is not picked on a given night. What's the probability that Bill's not picked three nights in a row? Let me write that down. The probability Bill not picked three nights in a row. Well that's the probability he's not picked on the first night, times the probability that he's not picked on the third night. Times the probability that he's not picked on the third night. That's going to be three to the third power, or three times three times three, that's 27 over four to the third power. Four times four times four is 64 and if we want to express that as a decimal. Let me get my calculator out. That is 27 divided by 64 is equal to, and I'll just round to the nearest hundredth here, 0.42. That is equal to 0.42. This doesn't seem that unlikely. It's a little less likely than even odds but you wouldn't question someone's credibility. There's a 42 percent, roughly a 42 percent chance that three nights in a row Bill would not be picked. This seems like if you're assuming truly random it's a reasonable, your hypothesis that it's truly random, there's a good chance that you're right. There's a 42 percent chance you would have the outcome you saw if your assumption is true. Let's say you keep doing this. You trust your older brother, why would he want to cheat out his younger siblings. Let's say that Bill's not picked 12 nights in a row. Then everyone's starting to get a little bit suspicious with Bill right over here. They say, "We're going to give him the benefit of the doubt." Assuming that he's being completely honest, that this a completely random process, what is the probability that he would not be picked 12 nights in a row? I'll just write that down. The probability Bill, it's really the same stuff that I wrote up here. I'll just say, Bill not picked 12 nights in a row. That's going to be, you're going to take 12 three fourths and multiply them together. It's going to be three fourths to the 12th power. What is this going to be equal to? I'll just write three divided four which is going to be 0.75 to the 12th power. This is a much smaller, this is now, this is going to be 0.3, I guess we could go to one more decimal place, 0.32 or we could say- 0.032 I should say, which is approximately equal to, let me write that, which is equal to 3.2 percent. Now you have every right to start thinking that something is getting fishy. You could say, "If there was --," this is what statisticians actually do, they often define a threshold. "If the probability of this happening purely by "chance is more than five percent then I'll say, "'Maybe it was happening by chance,' but if the "probability of this happening purely by chance," and the threshold that statisticians often use is five percent but that's somewhat arbitrarily defined. This is a fairly low probably that it would happen fairly by chance, so you might be tempted to reject the hypothesis that it was truly random, that Bill is cheating in some way. And you could imagine if it wasn't 12 in a row, if it was 20 in a row then this probability becomes really, really, really, really, really small, so your hypothesis that it's truly random starts really coming to doubt.