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A classic story involves four carpooling students who missed atest and gave as an excuse a flat tire. On the makeup test,the instructor asked the students to identi...

Question

A classic story involves four carpooling students who missed atest and gave as an excuse a flat tire. On the makeup test,the instructor asked the students to identify the particular tirethat went flat. If they really didn't have a flat tire,would they be able to identify the same tire? A statisticianasked 52 other randomly selected students to identify thetire they would select. The results are listed in the accompanyingtable. Use a 0.05 significance level to testthe statistician's cla

A classic story involves four carpooling students who missed a test and gave as an excuse a flat tire. On the makeup test, the instructor asked the students to identify the particular tire that went flat. If they really didn't have a flat tire, would they be able to identify the same tire? A statistician asked 52 other randomly selected students to identify the tire they would select. The results are listed in the accompanying table. Use a 0.05 significance level to test the statistician's claim that the results fit a uniform distribution. What does the result suggest about the likelihood of four students identifying the same tire when they really didn't have a flat? DATA: Tire Number Selected Left Front 14 Right Front 9 Left Rear 14 Right Rear 15 Test Statistic X2=_____ P value=______



Answers

Conduct the hypothesis test and provide the test statistic and the P-value and/or critical value, and state the conclusion. A classic story involves four carpooling students who missed a test and gave as an excuse a flat tire. On the makeup test, the instructor asked the students to identify the particular tire that went flat. If they really didn't have a flat tire, would they be able to identify the same tire? The author asked 41 other students to identify the tire they would select. The results are listed in the following table (except for one student who selected the spare). Use a 0.05 significance level to test the author's claim that the results fit a uniform distribution. What does the result suggest about the likelihood of four students identifying the same tire when they really didn't have a flat? $$\begin{array}{lc|c|c|c} \hline \text { Tire } & \text { Left Front } & \text { Right Front } & \text { Left Rear } & \text { Right Rear } \\ \hline \text { Number Selected } & 11 & 15 & 8 & 6 \\ \hline \end{array}$$

All right, So we are going to do this problem without having to use based theorem or even using a tree diagram. We just use a visual and then you have all the information. We put it together and get out. And so the visuals are very important in mathematics. So let's begin in the table off seasons text. Okay, then you're told in the four they used means professor of man's textbook. In the winter, they used millions textbook, and in the spring in the spring they used Professor Moz textbook. Okay. And then we also told about satisfaction. We're told that out of the 500 students who used means textbook, only 200 was satisfied, meaning that 300 watt and satisfied. And then out of the 300 who used Professor Medians book handed on 50 was satisfied, meaning that 150 we're not satisfied and 200 used. Professor Moore's book 160 was satisfied, meaning that the remaining 40 were not satisfied. Now these numbers are very important. So 200 simply tells you that these other students who used profits on Maine's book and they want satisfied and this number here tells you that 100 and 50 students used Medians book and they were satisfied. And finally, these number tells you that these are the students who used Professor Moore's book and they were satisfied. Okay, he wants that this food. So then the question is, even student, uh, during one of the semester, the selected at random and admits having been satisfied with the textbook is the student most likely to have used book by men median On what case? Or to figure out that we're going to investigate one at a time. So what is the probability or the chance? Ah, that the student had used means book her means they're using means book. Okay, given that they are satisfied with it. And this will be total number of strands who used to Maine's book and was satisfied, which is 200 divided by. We need to find the total number of students who are satisfied, which was 500 and 491 of satisfied. Okay, so we're going to divide this by 510. This gives use every kind three and nine, and then we go to the next one probability, But this Didn't he used, uh, made Ian's book? Okay, Probability used a made Ian's book. Okay. Okay. Having been satisfied, given that they will satisfy, this will be 100 on 50 divided by 510 which is a little behind three. So zero point took not three little point 2919494 Then finally, what is a chance that the student are the years to mourns? Textbook having being satisfied it And these will be 160 divided by 510 which is zero point 313 So if you look at these results, you realize that, um, it is more likely that the student had to use what he's more likely. No. The student used, uh, professor means textbook. Okay, Because, as you can see, the chance is three continental and the over who is least likely the least likely author is, as you can see, he's professor media least likely. All the is professor movie because you can see the probability of using his book. Having been satisfied with it. He's on the 0.29 for and that's the end of it. Okay. Thank you. Watching

This question were asked to identify the population and sample based on different scenarios and tell whether or not they can be used to create a confidence interval. So let's first start off by just defining the variables. So we have P, which is the population proportion and P hot, which is the symptom, the sample proportion. So in taste A our population will be all the cars, whereas the sample size is going to be the cars stopped at the certain checkpoints and like we said, p as the population proportion. So in this case, is all cars with safety problems, and P hot is a sample proportion. So these air the cars that are actually seen with safety problems. So we can further calculate P hat by using the numbers given. And we know that there are 14 of 134 cars stopped have at least one safety problem, so that number ends up being 0.1045 Weaken further transfer that into percentage form and we get that it's 10 points 45% as RP hot volume and were also asked whether these methods can be used to create the confidence interval. So when a sample of data is representative, then it can be used to create a confidence interval and in this case it is because it's sampling all cars. For case be, we are going to find the population and sample once again for the population. We have the general public and for the sample, it's people that are logged into the website. We can further define them as P being the favor. The people in favor of prayer in school where us The sample proportion are the people that voted in this poll who favor prayer in school, we can calculate p hot with the given values were at 488 over 602. We get 0.81 and making that into a percentage value, we get 81% toe. Decide whether or not the sample can be used. We can Onley consider people logging into the website for this case. So in a way it's a bit biased and non random. So you're unable to apply the methods to create the confidence interval in Casey. The population is the parents at school and the sample is the parents expressing opinions through the question here. So the population proportion are all parents who favor the uniforms, whereas the sample proportion P hat are the respondents favoring uniforms. We can calculate the P hot value based on the numbers given 228 over 380 and that gives us a value of 0.6 that can be converted into a percentage of 60%. And since there were 1245 surveys sent home but only 380 returned, there is a complication of non response bias. And so you would use these methods with caution if creating a confidence interval. And the last part D were given a population of students at college, and the sample size is the 16 31,632 College admits the population proportion are all the students who will graduate on time and P hot the sample proportion as the students graduating on time that year. So based on the given values, we have 1388 over 632. Actually, this number is supposed to be 16 32. Sorry, and so based on that we get a value of 0.85 and that could be converted to 85% based on this value, and the sample data for this case was pretty representative. So since it is representative, you can apply these methods to create a confidence interval.

Uh huh. We have a sample of size 1006. And of those 1006 sample members 490 meet a certain criteria. Only. Want to observe our equals 490. We want to use this data to test the claim that the population proportion P is greater than .47 at a confidence level of one or alpha equals 0.01. Now that we've identified the confidence level, we proceeded the following procedural steps to conduct a hypothesis test. A. Is it appropriate to the normal distribution? Yes, it is. Because both N. P and Q. P. R greater than five. What are the hypotheses were testing are known that P 1.47 are alternatives that P is greater than 0.47 This is the one tail test, macro computing P ha. And the test statistic piat is our over and equals 10.487 And the z stat is given by the formula the right, which reduces down to 1.9 Next let's compute Z. Let's use the Z stats computer p value. So we use a Z table from which we see that P equals 1.1379 a Kia. That is the area to the right of Z equals 1.9 on a normal distribution, as has highlighted in yellow on the graph on the right next. We use this P value to reject age. Not no, because he is greater than alpha, And we interpret this to mean that we lack evidence that P is greater than .47.

Okay, so here we want to know if there is sufficient evidence of a difference in cafeteria food satisfaction among the class grade levels and we are doing a chi square test for homogeneity. To answer this question, we should find the P value. Now, a couple steps in finding the P value for this test. I first want to get the degrees of freedom for the test. Woo penguin, wonky. Okay, so the degrees of freedom is typically in the to a table, the number of rows -1 times the number of columns -1. So there were four grade levels. So there's a number of columns minus one and then the there were two opinions about the food so satisfied or not satisfied. And so that's the number of rows there. So two minus one. Okay, so this comes out to be three times one or three degrees of freedom. Now for the p value now that I have the degrees of freedom, the p value is equal to the probability that our chi square value is at least as large as we observed. And what's great about this problem is we have the chi square value already which is 5.998 And so I'll use a calculator to help me figure out this probability there's a chi square C. D. F. Function on the T. I 84 calculator that you can use. You would enter the lower boundary as the chi square value. The upper boundary should technically be infinity. But since you can't actually enter that into the calculator, just use a really big number like 1000. And then you also want to enter the degrees of freedom as three. So entering this into the calculator you get .1117. This is the p value. Now, typically the p value is something that when it's large we don't have good evidence and if it's small we have good evidence. This is large, right? It's larger than 5% it's larger than 10%. So the best answer choice here would be even we just do not have sufficient evidence of um of a difference in the satisfaction among the different class levels. And so answer choice E. Is.


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