5

We are ,95% confident that thettrue difference in proportionsi of smokers in 1995 and 2010 is between Blank 4 and Blanki5Round the bounds of the interval t0 two dec...

Question

We are ,95% confident that thettrue difference in proportionsi of smokers in 1995 and 2010 is between Blank 4 and Blanki5Round the bounds of the interval t0 two decimal places (keep them as proportions; dolnot turn them into percents):Blank #6: Does the interval confirmithal smokingihas decreased from 1995 to 20102INES OR NODiank+Blnla 2Blanlii 3Blanka 4Blany#5Bia

We are ,95% confident that thettrue difference in proportionsi of smokers in 1995 and 2010 is between Blank 4 and Blanki5 Round the bounds of the interval t0 two decimal places (keep them as proportions; dolnot turn them into percents): Blank #6: Does the interval confirmithal smokingihas decreased from 1995 to 20102INES OR NO Diank+ Blnla 2 Blanlii 3 Blanka 4 Blany#5 Bia



Answers

The table gives the results of a survey of $14,000$ college students who were cigarette smokers in a recent year. $$\begin{array}{|l|c|} \hline \begin{array}{l} \text { Number of Cigarettes } \\ \text { per Day } \end{array} & \begin{array}{c} \text { Percent } \\ \text { (as a decimal) } \end{array} \\ \hline \text { Less than } 1 & 0.45 \\ 1 \text { to } 9 & 0.24 \\ 10 \text { to } 19 & 0.20 \\ \text { A pack of } 20 \text { or more } & 0.11 \end{array}$$ Using the percents as probabilinies, approximate the probability that, out of 10 of these shudent smokers selected at random, the following were true. No more than 3 smoked less than 1 cigarette per day.

To solve this example, we're going to need to reference the chart that shows the percent as a decimal of college students who smoke a given number of cigarettes per day. Now, if I take a random group of 10 of these smokers, what is the probability that fewer than two smoked between one and 19 cigarettes per day? Well, there are two different scenarios that match this criteria. I could have one person who smokes between one and 19 cigarettes per day, or I could have no people that fit that criteria. This isn't or I could have either one person or know people that match my criteria. So I'm going to find both of those probabilities and then add them up. That will give me my overall probability that fewer than to match my criteria. So let's take a look at the first case. I have a group of 10 smokers, and I want to know the probability of one of them meeting my criteria. So I'll start with a combination of 10 things taken one at a time. I'll raise the probability that they match my criteria to the first power since I'm looking for one person to meet the criteria and the probability of them not meeting my criteria. I will raise to the ninth Power because I want nine people who do not meet my criteria well, what's the probability that they equal that they are matching my criteria that I'm looking for? Well, I want people who smoke between one and 19 cigarettes that includes the group of 1 to 9 cigarettes, which is 0.24 as a probability. And I could also have smokers who smoke tend to 19 cigarettes a day, and there's a 0.20 probability of that happening. So all together, the probability is somebody smoking 1 to 19 cigarettes is 190.44 So that's what I'm going to raise to the first power. The complement of that is 0.56 and I'll raise that to the ninth Power. What I look at it. No people meeting my criteria. The set up is going to be essentially the same. But instead of a one, I will now have zero cause I want no people meeting my criteria. And instead of nine, I'll have 10 because all 10 of my sample will not meet my criteria So I'm going to put both of those numbers into my calculator, and when I add them up, I'm going to get a value of 0.26864 That is the probability that out of a group of 10 randomly chosen smokers, fewer than two will have smoked between one and 19 cigarettes per day.

So for this problem, we need to reference the chart that's given that shows the percent as a decimal off college students who smoked the given number of cigarettes per day. And I'm curious if I took a group of 10 of these students smokers at random. What is the probability the five of them smoked a pack or more per day? Well, the easiest way to solve this type of problem is used by no male expansion. So I'm taking a group of 10 and I'm looking for five that are going to meet my given criteria. So I'll have 10 things taken five at a time without with order, not mattering. So combination of 10 things taken five at a time. The probability that somebody matches my criteria race to the fifth power because I'm looking for five thumb times. The probability that they don't meet my criteria also raise the fifth power cause out of 10. If five meet my criteria, five won't Well, what is the probability that somebody meets my criteria? Well, I'm looking for people who have smoked a pack or more per day, and according to the chart, there is a 0.11 and probability that somebody is in that category. So the complement of that is 0.89 Those are the That's the percentage of the probability of students that don't meet my criteria. So if you take this and you put it into your calculator, you will get a value of 0.2 to 7. This is the probability that out of a random group of 10 smokers, five of them smoke a pack or more per day.

For this problem, we need to reference the chart that's given that shows the percent as a decimal off the college students who smoke the given number of cigarettes per day, and I want to know what's. The approximate probability is that if I take a group of 10 students selected at random, I will end up with four who smoke fewer than 10 cigarettes per day. Well, the easiest way to solve these is to use binomial expansion. So I have out of a group of 10. I'm looking at four who fit my criteria, so I'll take 10 things taken for at the time. I'm gonna take the probability that they fit my criteria race to the fourth power times the probability that they don't raise to the six power. So what is my criteria? I'm looking for four people who smoke fewer than 10 cigarettes per day. Well, I'm gonna just come over to the side where I've got some room. That means they can smoke less than one cigarette per day, which has a percentage of 0.45 Or they could smoke 1 to 9 cigarettes per day, which is 90.24 if I add those up. I have a probability of 0.69 that they will match my criteria. So that's the number I'm gonna raise to the fourth power. Somewhat four people who meet this criteria, that means the other six will not. So I'm going to raise the complement of 60.69 which is 0.31 I'm raising that to the sixth Power. If I put that into my calculator, I will get a percentage or a decimal of 0.42246 So this is the probability that out of my random group of 10 4 will smoke fewer than 10 cigarettes per day.

Hello. Welcome to the video. And today we're gonna be talking about hypothesis test and specifically, um, determining whether or not the null hypothesis should be rejected. So in this problem, we're told that a study was done to determine, uh, the age at which people start smoking. And we're told that a proposed value for the population mean is 19 and people were doubting this claim. Um, people were doubting that doubting whether or not this claim was true. So they did their own study and found that their sample, they took a sample and found that their mean starting age of the sample was 18.1. So that would be about right here, and that their sample standard deviation was 1.3. So with this information, we can see that, um, our sample mean and the proposed mean are within one standard deviation away from each other. Um, in other words, the Z score off the sample mean of 18.1 is less than one on with this being said, we know that Ah, 68% of the data is, um, contained within one standard deviation off the proposed value off in this case, 19 it might be 64%. It's either 68 or 64. I can't remember right now. But regardless, um, if we test this at the 5% level, um, we would reject it regardless, because we know that it is within the middle 68% and not part of the outer 5%. And therefore because we are certain that, uh, this, um that this observed mean is not statistically significant. We can say that we the correct conclusion would be to fail to reject a civil. And this is because that 18.1 just isn't far away from enough. It just isn't far away from enough from 19. In order for us to sufficiently say that this value should be, um, suspect so once again failed to reject


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