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3) The table below shows contingency table for sandwich preference at McDonalds for random sample of teens and their parents a) Fill in the expected values to one d...

Question

3) The table below shows contingency table for sandwich preference at McDonalds for random sample of teens and their parents a) Fill in the expected values to one decimab place in the parentheses behind each count:Hamburger_ChickenFishTotalsTeen Parent Totals118 184112b) At the 10% level of significance, is the preference of sandwich dependent on age?

3) The table below shows contingency table for sandwich preference at McDonalds for random sample of teens and their parents a) Fill in the expected values to one decimab place in the parentheses behind each count: Hamburger_ Chicken Fish Totals Teen Parent Totals 118 184 112 b) At the 10% level of significance, is the preference of sandwich dependent on age?



Answers

The following two-way contingency table gives the breakdown of the population in a particular locale according to age and tobacco usage: A person is selected at random. Find the probability of each of the following events. a. The person is a smoker. b. The person is under 30 . c. The person is a smoker who is under 30 .

This question again, were given a contingency table and for education as well as news of supplements, and we are told that an adult is selected at random and to find the probability of the following events. First event is that the person has a high school diploma, so that's here and takes dietary supplements regularly. So that's .6 be the person has an undergraduate degree and takes supplements regularly, so that's point oh nine. See the person takes dietary supplements regularly. So that's basically this column. So that's point two when we add up all of the values here and indeed, the person does not take dietary supplements regularly, so that's 0.5 plus 0.3, which is point okay, so we have all of the probabilities.

Okay, So in this question, we are going to perform a chi Square test to determine of the age. Distribution and location are independent. 0.5 level of significance are Alfa is 0.5 Okay, we have the table with us. We have a judge. Okay. In the age, we have calf, yearling and adult. We have half pure ling and an adult. Okay. And with this, we have the district's This is the Lamar District. This is the nest sports district and be deep. Okay, then there is the fire Whole district. There is f d these abbreviations already funny, but okay. So over here we have the row totals, and at the bottom, we will have the column trees. So let's write this. This is 13. These are all the observed values. 13, 10, 34. Then we have 13, 11, 28. Then we have 15. Too old and 30. Okay, then how about the column rules? This is 57. This is 42 or 52. Okay. Excuse me. That should be 52. This is 52 and then this is 57 then over here we have 41 over here. We have 33 over here we have 92. So what is the grand total? The grand total is 1 66 1 66. All right, now again, in orderto conduct a chi squared as we need the expected values. How do we get the expected values? Expected Value is given by road total multiplied by a column total. Hey, call them total upon the grand total upon the Graham total. Okay, so now again, what I'm going to do is I'm just going to draw the table. There are these three District's these ST is this okay? All right. I'm not writing the labels because labels are going to be the same now, moving on to the calculator, this is going to be 41 multiplied by 57 divided by one. 66. This is 14.8 14.8 This is the expected value for the first one. Then we have 41 in 2 52 divided by 1. 66. This is 12.84 Then we have 41 multiplied by 57. Divide within 66. This is 14 point zero. It again 14.8 We have 33 multiplied by 57 divided by 1. 66. This is 11.33 So this will also be 11.33 Because the column totals are the same in the row, totals are the same. Then we have 33 inches 2, 52 divided by 1 66 the CIS 10.33 10.33 Okay, then we have 92 multiplied by 57 divided by 1. 66. This is 31 point 59 31.59 Then we have 28. Oh, sorry. Then we have 1952 92 multiplied by 52. Developer 1 66 which is 28.81 28.81 And this is going to be 31.59 again. 31.5 nine. Again. Okay, this Let me just write this little more clearly. This is yeah. 10.33 Now we have all the expected values. These are our expected values, X. They did values. Now we need to calculate the Chi square for all of these cells. How do we do that? observed minus expected whole square upon expected. We do this for all the cells, and in the end, we're going toe. Add all the sales up and this is going to give us our chi square statistic. Okay, let's get to work again. We're going to need both of these, uh, both of these tables. Right? So these are going to be our three columns, and these will be our three rows. Okay. Now observed minus expected. Now for the first cell, it is 14.8 and 13. So this is 14.8 minus 13. I square this 1.0, it's square, and I divide this by off. Expected value 14 0.8 This is 0.82 It's let me just write. This is 0.8 Okay, then I have 13 on 12.84 Do you want 16 and divided by 12.8 ft? This is 0.0 2.0 to. Okay, then we have 15 and 14.8 We square this and divide by 14.8 This is 0.6 point 06 Okay, then we have 10 and 11.3. So, 11.33 minus 10. We square this and divide by 11.33 which is 0.15 securities at this 0.16 point 16 Okay, then we have 11. 10.33 Okay, 11 minus 10.33 We square this and divide this by 10.33 which is 0.43 Food 0.43 All right, then. We have 12 and 11.33 12. Minus 11.33 This is 0.67 This is 0.67 in Oh, no. So we have to square this 1st 0.67 and divide by 11 0.33 This is 0.39 600 is right. This is That's it. Is this thing over here? This is going to be 0.4, actually, this is right. This is 0.4 point 04 Okay, then what do we have then? We have 34 31.5 million 34 minus 31.59 We square this and divided by 31.59 Why? What 84 0.1 eat four. Okay, then we have 28 28.81 So this is going to be 0.81 We square this when divided by 28.8 one. This is point 0 to 3 0.2 three. Then we have 30 and 31.5 minutes. This is 1.59 We square this and divided by 31.59 This is 0.8 0.8 All right now we have keister values for all the cells. We just add them up. So this is 0.8 plus 0.2 plus 0.6 plus 0.16 plus point 043 plus 430.4 Last 0.184 Last 0.0 to 3 plus 0.8 This is 0.67 to my chi square value. My chi square statistic is turning out to be 0.672 now. I need the degrees of freedom. Degrees of freedom is going to be okay. Number of rows minus one, multiplied by the number off columns minus one. Okay, so Rose was 33 minus one columns was 33 minus one. This is that the two into two with just four. So my degrees of freedom is four. Now I have an online calculator over here with me. My chi square values 0.672 So this is point. Take 72 on my degrees of freedom is four. What is my Alfa Alfa is 5% level of significance. This is 0.5 and this is 0.95 My fee value is 0.95 My p value is 0.95 What was my Alfa? My Alfa was 0.5 Okay, So as my P value is greater than Alfa, I fail to reject. It's not. I fail to reject. H not now. What was h not h not? Is that these are, in fact independent. These two variables what are these variables? Is distribution on location are independent. So how will actually frame the sentence? Yeah, I will say that I don't have enough statistical evidence. Enough status, tickle evidence, no statistical evidence to suggest that each that is and location that is and location that age and locations are not independent. Mhm are not independent. This is how we go about doing this question

Alright, This problem is a binomial probability and we know it's a binomial probability because it meets certain conditions and the conditions it needs to meet are one it needs to have a fixed number of trials. Second thing is each of the trials must be independent of each other, and the third thing that has to be holding true for a binomial probability is that there's only two outcomes, and we define those outcomes as success and failure. Now, when we're doing binomial probabilities, there's a couple different variables involved, and the variables are n which is the number of Trials X, which are all the possible outcomes key, which is the probability of success. And Q, which is the probability of failure. And keep in mind the probability of success and the probability of failure should add up to one. So going through this problem, let's see that we've met these conditions before we solve it. All right. So as we read through this problem, we find that there is a fixed number of trials in this case, and it's six, and we were asking the people whether they felt that the government should fight childhood obesity and we only asked six different people. The values for X could be that zero of the people believed that the government should fight obesity or one of them or two of them, or three or four or five or six. And we found in this survey that 39% felt that, yes, we should be having the government fight childhood obesity. So that means P would be 0.39 and then Q would be the other 61% that said no, they didn't believe that. Now, when we're solving uh, binomial probabilities, the formula that is necessary is that p of X equals the combination, and C x multiplied by p, raised to the X value multiplied que raised to the n minus X value. So the easiest way to handle this is to put our data into the calculator and let the calculator do a lot of the work for us. So before we try to answer Parts A, B and C, let's add the information to our calculator. So I'm gonna bring in my calculator, and I'm going to select the stat feature, and I'm going to edit that we can fill in the information in our list and enlist one. We're going to put all of the X values, so we're gonna put in zero and one to three, four, five or six. So those were the only possible outcomes. If you're talking to six people, zero them could believe it or all the way up to six people can believe. Yes, we need to have the government fight childhood obesity. Now that we have done that, we need to find the probabilities for each one. So we need to put all of these values into the formula. So we need to put the zero in for the exes. We need to put the six in for the end. We need to put the 0.39 in for the P and the 0.61 in for the Q. And we've got to do that for the one and for the two and so forth. So the fast way of approaching that again is to go back to your graphing calculator. So if you sit on top of the l two, we can generate a formula, and instead of us saying, N c X, what we're going to do is end with six and we have stored all our exes. Enlist one. So we're going to type that in. Then we're gonna multiply it by 3.39 And all of our exes were again stored in list one. And then we're going to multiply it by the 0.61 and we're going to raise that to the sixth minus X will again. All of our exes were stored in list one. So when I do that, I'm going to type in six. I'm gonna go and grab the, um, probability, and it doesn't want to let me get that.

Now this problem is about asking adults whether they think the government should help fight childhood obesity. And it falls into the binomial probability category because it meets certain conditions. There are going to be a fixed number of trials. The trials are going to be independent of each other, and there's only two outcomes, either success or failure. So as we read through this, it said that 39% of U. S adults think that the government should help fight childhood obesity and you randomly select us six US adults. So there's are fixed number of trials, so N is going to be six. And when you ask one adult, um, the next adult is not going. Their answer is not going to be dependent on the previous one. So the trials are independent. And are there only two outcomes? Yes, either. Yes, they believe the government should help fight childhood obesity or no, they should not. So with binomial probabilities, we have a number of variables, one of them being and the other one is going to be X, which is all the possible outcomes. And if we talked to six different people, zero of them might believe that the government should help fight childhood obesity. Or maybe one of them does. Or two or three or four or five or six. There are two more variables associated with binomial probabilities, and that's going to be the P and the queue and P is the probability of success when you reach out to these people. Um, what's the probability of finding someone that believes the government should help fight the childhood obesity? And since it said 39% of U. S. Adults think that way, then our P is 0.39 Now, cue is the probability of failure. So if 39% of the population believe, yes, we should have the government intervene with fighting childhood obesity, then that means there's 61% that don't believe the government should intervene. And when it comes time to do binomial probabilities, we're going to apply this formula. The probability off an event happening is equal to N. C X multiplied by P to the X power multiplied by Q to the n minus X power. So we're going to take all that information and we're going to apply the formula Thio answer three parts to this question, and in doing so, it's possibly or probably best to set up a whole binomial distribution. So we're going to set up a chart, and then we're gonna answer the questions. So the first column of the chart is going to be all of the different possible outcomes. And then the second part is going to be each of the probabilities. So the first go round to find the probability of zero. We would put a zero in where the X is in three different places and we would put the 0.39 in where the P is on the 0.61 in where the Q is. And we would put a total of six everywhere there's an end, and then we would go back and we would alter that formula with X being one. And then we alter the formula with X being too and so forth. So this is where we are at the moment. So we have six see X multiplied by the P value in this instance is 0.39 raised to the X power multiplied by 61.61 raised to the end minus X power and the end in this case is going to be six minus X power. Now again, the X keeps changing. So the easiest way to fill in the entire probability distribution is to utilize your graphing calculator. So I'm going to bring in the graphing calculator. And if you hit stat and edit and you put in your numbers zero through six and then if you go and sit on top of the l two column and type in six, access your combination under Math Probability and C r. And then all of our exes, we have housed Enlist one. So we're going to tell the calculator toe look, enlist one for every one of those exes. We're then going to multiply that by 0.39 raised to the X power. And again, all of our exes are enlist one. And then we're going to multiply that by 61 raised to the power of six minus the accent again. L one holds all of our X values, and when we hit enter, it's going to fill in the entire chart. So I'm going to come back to our chart and I'm gonna fill in what you see in the graphing calculator. So it's 0.0 515 to 04 and then we see 0.197 6355 Then we see 0.3158 9 to 9. And for three we see 30.2692 857 and for four 0.1 to 9 1247 and the probability of five this 50.0 33 02 to 1. And for six, the probability ISS 60.0 35187 So that's our probability distribution for n being six under these circumstances. So now let's look at the actual problems. Part a part A is asking you to find the probability of exactly two people out of these six believing that the government should get involved with childhood obesity. So we're just gonna look in our chart. We're gonna take the value, So the answer is approximately 3159 for Part B. It's asking us what's the probability that the answer or the number of people is greater than or equal to four. So if that's the case, you're going to take the probabilities for 45 and six and add them together. And when you do that, you should get 0.165 seven ish and for part, see, it's asking you for less than three. So for part C, we want the probability that X is less than three. So in this case, less than three would be 01 and two. So we're going to add those three probabilities together, and you're going to get approximately 0.565 So therefore, our answers to this problem are right here. Probability that there's exactly two out of six is 0.3159 Probability of at least four well, at least four is greater than or equal to four would be 40.1657 and the probability of less than three would be 30.565


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