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An artide 2019 dcscribcd thc Incidenc cplsodic tension hcadache WoMen SUNvcy was undortatcnand 4598 0 respondents were the age 30 to 39age Eoup The remaining respon...

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An artide 2019 dcscribcd thc Incidenc cplsodic tension hcadache WoMen SUNvcy was undortatcnand 4598 0 respondents were the age 30 to 39age Eoup The remaining respondents were the 18to 29 age gtoup: thc younger EIouP 3593 of women hadsuffered tension headache E their Wies the oldcr agc grouP Mias found that 3773 of women had suffered tension headache in their Iives . Treat ay relevant probabilities from this study population probabilities to help vou answer the questions below. We consider women

An artide 2019 dcscribcd thc Incidenc cplsodic tension hcadache WoMen SUNvcy was undortatcnand 4598 0 respondents were the age 30 to 39age Eoup The remaining respondents were the 18to 29 age gtoup: thc younger EIouP 3593 of women hadsuffered tension headache E their Wies the oldcr agc grouP Mias found that 3773 of women had suffered tension headache in their Iives . Treat ay relevant probabilities from this study population probabilities to help vou answer the questions below. We consider women between 18 and 39 Yeas old: What [ probability rndomly selected woman will have had tension headache? What Is the probj bility randomly selected woman In the younger age group it was kmown tha: she had suffered tension headache? Are age group and headache status independent? Explain using a probability Olculation Suppose you do new study. based on this one, and you randomly sample 20 women the 18 t0 39 Jge group and ask if they have ever suffered tension Radache Let Y be the number of womzn In the smple who sY "ves" Then might be modeled as distnbution with parameters speaific with numencal valucs and notaton for panmctcrs} What is the proba biity that at least 6 out of 20 women inyour sample say they had tension headache? (vou don't have calculate exactly. show howto alculate Suppose vou decided to sample 200 vomen rndomly in the 18-39 age group intead of 20. Whatis pproximate probability that at least 60 them say they have ever suffered tension headache? Include any appropriate checks that allow You t0 compute this probability



Answers

According to the Centers for Disease Control, 15.2\% of American adults experience migraine headaches. Stress is a major contributor to the frequency and intensity of headaches.A massage therapist feels that she has a technique that can reduce the frequency and intensity of migraine headaches. (a) Determine the null and alternative hypotheses that would be used to test the effectiveness of the massage therapist's techniques. (b) A sample of 500 American adults who participated in the massage therapist's program results in data that indicate that the null hypothesis should be rejected. Provide a statement that supports the massage therapist's program. (c) Suppose, in fact, that the percentage of patients in the program who experience migraine headaches is $15.3 \% .$ Was a Type I or Type II error committed?

This is problem # 26 we are given a set of data regarding the number of pixels in an MRI image and the IQ score of a student. So we will begin by finding our Russian equation. So for our some of x, we get eight .76 million For the sum of why we get 173 0.17 Put some of x times y. We get 154 million. For the sum of x squared We get 77.2 times, tend to be 12th 7.7, 2 times 10 to the talk. And for the sum of y squared we get 8695 6. So using these for our regression equation, we get the following, we have white hat equal to 6, 65 times 10 to the mhm minus five X minus 40 point 999. So I would just write this as -41. So we now have our regression line notice that the value of the slope is incredibly small, but this makes sense as the values we are using are very large for inputs and the values of our outputs are fairly small in magnitude relative to them. Next, we will predict Two values firstly, for x equal to one million. We get a predicted Y value equal to 25.53 And for the value of 830,000, we get a predictive value for why hat Equal to 14.22.

This problem is looking at a two way data table about a clinical trial for a headache medicine. So there were a split in the participants here. Some of them got the actual medicine, some of them got a placebo and it measured how long it took for their headaches to go away. Either it went away within 45 minutes or it lingered on past that. And we want to look at some probabilities. So the first one we're going to look at the probability that are randomly selected study participant was given a placebo. So we're looking just at this placebo row really, we want that total of placebo getters recipients. And for probability, remember we always divide by the total in the study, so 100 divided by 250 and we get 0.4. So there's a 40% chance that we randomly select a participant and they actually got the placebo. Then the probability that they're headache went away within 45 minutes. So of all these participants were going to look at this headache went away column, specifically the total. So 188 out of our 250 we get 0.752 or 75.2%. Now the probability that they have a placebo and their headache wanna wait anyway, even though they didn't really get the medicine, that is the single box where those two columns overlap. So we're just going to use that 56. But of course we're finding probabilities. So we divide by our total of 250 and 0.224 Yeah, or 22.4% whose headaches went away, even though they were given the placebo in this last part, we have two different ways. We can do this. Since we have the two way table to find the probability that a randomly selected participant was given the placebo or their headache went away. We can actually look at any box that got highlighted. So the 1 32 the 56 the 44. We can add those up as our total that we are concerned with and divide that by R 250. Alternatively, we can use the addition rule since we've already found the probability of getting the placebo, the probability that the headache went away and we have found the and version placebo and it went away. We already have all the numbers we need for this part. So we could do probability of A. Was your 0.4 plus probability of B 0.752 minus the probability of A. And B. That's what we found in part C. You're a 0.224 So both of these things, Both of these methods should come out to the same exact answer. So if we do the adding up all the boxes way, we get 0.9 to 8% 0.9 to eight or 92.8%. If we use the General edition rule notice we're not going to have to divide anything, but we do get the exact same answer 0.928 or 92.8%.

It's a kind of an interesting question if we we would be assuming that the anxiety level if we are going from easy to difficult is equal to the anxiety level. If the questions were difficult to easy and alternately, we just want to know if there's a difference and easy to difficult and not equal to difficult to easy. And so we're going to assume that that difference is actually zero so that the, I'll just write E G minus D. E. That that difference is zero. And when I put this into my calculator, um we have the lists have different sample sizes. We have that first group of the E. T. D. I have, there are 25 numbers versus the D two E. There are 16 numbers and we could use the degrees of freedom of 15, but I'm going to use the degrees of freedom from the formula. And that degrees of freedom ends up being uh 30 almost 39. So 38 point well, I'm just going to call it 39 it's approximately 39 degrees of freedom. And that test statistic that we're going to get, we need to take the mean, which was 27.1152 minus the mean of the other group, which was 31.7 to 8125 And then divided by the square root of. And the first standard deviation all around that a bit is 6.857 one square, divided by the sample size, which was 25. And then the second standard deviation was 4.26 square divided by the sample size of 16. And when I got that test statistic, the test statistic came out to be, I'll just read it up here negative 2.6 566 And so it came out down here. And since we're doing a two tailed test, we also use this one. And so what's the likelihood of getting a test statistic in this distribution that is less than or equal to that negative 2.6566 That gives us this tale and then we want the other tales to double it. And that p value comes out to be a 0.114 So at a 5% significant level, this is smaller than 5%. So we would have sufficient evidence to reject now. Yeah. And say that the mean anxieties are different so that they're different. The means are different. However, at a 1% significance level we would fail to reject the novel. Mhm. And we'd have to say here that they're actually not. They appeared to be the same. So it does depend on our significance level. How did pick you want to be. But it is an interesting idea in all my years of teaching, I never thought about the arrangement of having them all go easy, too difficult or difficult to easy. So what interesting concept.

We want to conduct a pair differences test at alpha equals 5% significance testing. The claim that the population means X. A and X. B are not equal. The data is given below. With mount shapes and at your distribution on the right. I've already computed D. Bar 2.25 The sample size N equals eight and the standard deviation S. D equals 7.78 We proceeded to five steps listed below to solve first. We'll check the requirements and evaluate hypotheses. So because the distribution shape requires have been met to use the student's T distribution, the degree of freedom is at minus 27. We have no hypothesis. H. And R equals zero alternative mu D. Does not equal zero Alpha equals 00.5 significance. Archer statistic is T equals D. Bar over SD over route and 4.8181 from the tea table R. P value is 0.5 is less than ph 7.25 So we have enough information to conclude our test. Now, since P is greater than Apple, we fail to reject the null hypothesis, which means that we lack evidence that beauty does not equal zero.


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