a-survey-by-the-national-institutes-of-health-asked-a-random-sample-of-young-adults-aged-19-to-25-ye

Question

A survey by the National Institutes of Health asked a random sample of young adults (aged 19 to 25 years), “Where do you live now? That is, where do you stay most often?” Here is the full two-way table (omitting a few who refused to answer and one who claimed to be homeless):30
(a) Should we use a chi-square test for homogeneity or a chi-square test of association/independence in this setting? Justify your answer.
(b) State appropriate hypotheses for performing the type of test you chose in part (a). Minitab output from a chi-square test is shown below.
(c) Check that the conditions for carrying out the test are met.
(d) Interpret the P-value in context. What conclusion would you draw?

Caleb Huber · Accepted Answer

we don&#x27;t five. We reject eso his problem. He had this data set. Um, it asks, and only I would write these down as we go. But I made a small it air at the end, so I&#x27;m just going to re read through why? Had already written up here. So we had this data. Let me grab something I can, like, highlight with two point out things. Make this there. You okay? So we have this data of here at this table, and you can total it. And then the lawsuit chi squared tests it as is the chi square test appropriate? Um, we&#x27;re interested in finding whether or not gender and the place of living are statistically significant. And our data is easily organized to a table, so the chi square test makes sense. So that&#x27;s we&#x27;re gonna use the chi square test for that. You have pa theses their policies that there&#x27;s no difference in the distributions of where male and female adults live. The alternative offices is that there is difference industry descriptions of where what? Where male and female adults live. They were. You check conditions. The conditions well, random is a condition, but the data came from random samples, thoughts, good independence. Uh, when I did this totaling here, we see there&#x27;s 4854 total that is less than that number is less than 10% of population and adults swinging. Treat the observations is independent of each other. Expected counts need to be greater than equal to five. Here&#x27;s table of the expected counts. We see that the smallest is still over 100 so we&#x27;re definitely greater than five. So our conditions are met. So all conditions where this milk met for a chi square test. Then we are given the the data. They tell us that the chi square test when they calculate it is 11.13 And of the P value that goes with that data points are won 12. This is all given in the table data in your book. Well, this means What does this mean? It means the probably of finding a difference in distributions. At least as extreme as our sample is 0.8 which is point, which is less than 0.0.5 So it&#x27;s pretty unlikely to find, uh, discretion as extreme as ours. So we&#x27;re going to this reject the no hypothesis and conclude they&#x27;re to be difference in the distribution of where people live. Um, at this with regards to gender since Because if you guys really small, it means that our sample seems like a really extreme case, and so we would assume that something is then wrong with our no hypothesis.

R M · Answer

couple, remember? 51. Ah, here questioning. We say that I have to wait table towns. The count they&#x27;re continually off to get the goodie kille. A very move. One sample. And since we should use the chi square test off a sissy issue in the residents. So here, when you were use one square test Oh, that&#x27;s a deviation. Ah, no, we&#x27;re independence. If there are two samples and then we should use cars perform agility. But here we have one sample. Ah, so we were used Car square. There&#x27;s Golf Association for independence. The question being and additional would work on all hypothesis states that on there is no association between you ready? I want it will turn to that There is association between ready of I mean ah question. See Ah, we know that the necessary requirement o r and that require nous. You run them very well A random sample and here needs satisfied because the sample is given to be trained them second requirement if that will expect it, Holmes or why. And this also requirement is Ah, that&#x27;s fine here because the will expect it counts in doubt, but are more than five and the last requirement in the independency in the buildings. And here we can assume independence before the sample is less than at impression off the population requirement. The ah requirement point number d we know from the other. But the very is equal toe open toe 12 cheese in 1.2%. On this P value is smaller than that significance living open toe time so we can reach back. Another hypothesis s so we can say that there is sufficient evidence and that Berries and association between Grad Eberts

Robin Corrigan · Answer

we&#x27;re asked the state we would use a chi squared goodness of fit test, a chi squared test of homogeneity, a chi squared test of independence or some other statistical test for the following situations. For part A. We are examining whether there is any difference in quality of manufactured cars. Based on the weekday, the car was made using counts of warranty claims filed, so there&#x27;s a somewhat obvious no hypothesis here, and that is that there is no difference in quality from one day to the next. So we&#x27;re testing to see if the data Concorde with this uniform theoretical model. So this is a goodness of fit, test and be. A researcher is testing whether blood cholesterol level is related to heart disease by examining a patient database and looking at cholesterol levels for patients with and without heart disease. So this is a search for an association between two variables. For a single population. However, cholesterol levels are not in counts. It is a continuous variable. Therefore, the chi squared tests are not appropriate in this situation and then see a student is trying to determine if political leaning is associated with choice of major by randomly surveying 500 students. So this is a test on a single population students and looks to see if there is an association between two categorical variables which are choice of major and political leaning, so this is therefore a test of independence.

Robin Corrigan · Answer

in this question, we&#x27;re asked to state if we would use the chi squared goodness of fit test, the Chi Square test of homogeneity, the Chi Square test of independence or some other statistical tests for the following situations. For part A. It&#x27;s a brokerage firm wants to find out if customers accounts types affect the types of trades they make, so they collect a random sample of trades made by its customers. So here we are, making inference about a single population, the firm&#x27;s customers, and we&#x27;re trying to see if there is an association between two categorical variables account type and types of trades. So this is a test of independence and be the same firm wants to see if there is an association between the account type and the size of the account. Again, this is a single population, and we&#x27;re looking for an association between two variables. The account type is a categorical variable, for example, silver, gold or platinum, but the account size is a continuous variable. The account size is not presented in counts. It is in dollars and therefore we cannot use the chi squared tests. We could only use some other statistical test so we can say not in counts for part C. A college wants to see if the residential and non residential students select courses according to the same distribution Here. We&#x27;re looking at two populations residential and non residential students, and we&#x27;re sitting if there&#x27;s any difference between them with respect to the distribution of a single categorical variable courses, so this is a test of homogeneity.

Caleb Huber · Answer

all right for this problem. So, you know, I had already written all this up. They made a slight mistake. So it&#x27;s gonna read through what I had already written up. So in this one that we&#x27;re considering people&#x27;s preference, our justification behind buying, um, things through catalog and whether there were looking for this disco difference between it there a Asian or American securities Chi square test? Well, the data represents one sample. The subjects are classified by two things. Their answer and their ethnicity. So a chi square test for independence or association makes sense. The null hypothesis is that ethnicity and reasoning behind shopping via catalogs independent and make the alternate hypothesis just that they&#x27;re not independent in regards to conditions, uh, samples a random sample. So that&#x27;s good. In rights Independence, the sample size is less than 10% of the population being adults. So we treat the sample is being with replacement or, in other words, we can consider as independent observations. So, yeah, that&#x27;s met. Randomness is met and then expected counts. If you look at the data given to you in the book, looks like the minimum expected Countess 5.13 which is less than or equal to five and greater than equal to five So expected counselor as well. So the conditions were met for a Chi Square test. They didn&#x27;t give you a P value of 0.1 This is less than 0.5, that&#x27;s We reject the null hypothesis, and we conclude the ethnicity and the justification behind shopping in a catalogue to not be independent.

R M · Answer

number 52. Ah, Coach in a we can say that the two way table counts the count. Spare Kittinger you Why don&#x27;t get a gherkin? Very bring off to sample and thus we should use the chi Square test of homogeneity. Eso if ah, beer was one example. We will, uh, so if you example and we were in use Pride Square fists, Deputies were in sample. Um, we should use the con spiritist for association or dependency or independency. But here we will you. Because we&#x27;re artists. Uh, former jenny, uh, answer being. We know that they know high processes the event. Very no association green. The variables find good. Good, good. Uh, turn If there is association with me variables a quick trencin. Ah, we know that the necessary requirements necessary requirements is in the central that either in the regular I will expect it Counts are five gate go are greater. Also, it should be in development, depending Ah, requirements are since Friant here. Uh, that&#x27;s why because the sample are giving to be too random variables, Unexpected comments in velvet or more than fine. Soon there are we would expect council edges. Ah, this requirement is a well in here. Um, and it&#x27;s also independent can be assumed and me as soon. And because the sample Yeah, then present off the Richard Serbian. Uh, you know that the value off in the given out is a breakthrough opening one on. We know that the valley is a probability of opening the very off, the just statistic or of any morning to tree. It&#x27;s not our business. If there is no association between the valuable, the probability off getting a similar or more, it&#x27;s three example as the given sample is 0.1 eso We know that the venue in smaller than the significance level. So we reject Obama by processes, as you can say that there is sufficient evidence that it is, and that&#x27;s what&#x27;s the use

Problem

What is the most important reason that students b…

Problem 51 Hard Difficulty

Answer

a. Chi-square test of association/independence
b. $H_{0} :$ There is no association between the variables
$H_{a} :$ There is an association between the variables
c. All conditions are satisfied
d. There is sufficient evidence that there is an association.

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Srikar K.

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a. Chi-square test of association/independenceb. $H_{0} :$ There is no association between the variables$H_{a} :$ There is an association between the variablesc. All conditions are satisfiedd. There is sufficient evidence that there is an association.

The Practice of Statistics for AP*

Srikar K.

Catherine R.

Christine G.

Heather Z.

Data And Variables

Data And Variables - Example 1

Daren S. Starnes, Daniel S. Yates, David S. MooreThe Practice of Statistics for AP*

Srikar K.

Catherine R.

Christine G.

Heather Z.

a. Chi-square test of association/independence
b. $H_{0} :$ There is no association between the variables
$H_{a} :$ There is an association between the variables
c. All conditions are satisfied
d. There is sufficient evidence that there is an association.

Daren S. Starnes, Daniel S. Yates, David S. Moore

The Practice of Statistics for AP*