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QUESTION 3clinical trial is planned compar C experimenia mcdicouon designed @Wer blood oressure placebo- Suppose that patients Ca recruited from different clinical ...

Question

QUESTION 3clinical trial is planned compar C experimenia mcdicouon designed @Wer blood oressure placebo- Suppose that patients Ca recruited from different clinical sites: Use the following data teste hynertension and clinical site are independent: Run the test at level of significance. Give each of the following receive full credit; 1} the appropriate null and alternative hypotheses; 2) the appropriate test; 3} the decision rule; 4} the calculation of the test_statistic; and your conclusion inc

QUESTION 3 clinical trial is planned compar C experimenia mcdicouon designed @Wer blood oressure placebo- Suppose that patients Ca recruited from different clinical sites: Use the following data teste hynertension and clinical site are independent: Run the test at level of significance. Give each of the following receive full credit; 1} the appropriate null and alternative hypotheses; 2) the appropriate test; 3} the decision rule; 4} the calculation of the test_statistic; and your conclusion including comparison alpha or the critical value. You MUST show vour work Feccive full credit, Partial credit available. Site Hypertensive Not Hypertensive T I Wzpl Patnan Waid:0 (ick WMSu Aut" unld sulnt. ( Tua ollansiltm LV" maC



Answers

Please provide the following information. (a) What is the level of significance? State the null and alternate hypotheses. (b) What sampling distribution will you use? Do you think the sample size is sufficiently large? Explain. Compute the value of the sample test statistic and corresponding $z$ value. (c) Find the $P$ -value of the test statistic. Sketch the sampling distribution and show the area corresponding to the $P$ -value. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level $\alpha ?$ (e) Interpret your conclusion in the context of the application. This problem is based on information taken from The Merck Manual (a reference manual used in most medical and nursing schools). Hypertension is defined as a blood pressure reading over $140 \mathrm{mm}$ Hg systolic and/or over $90 \mathrm{mm}$ Hg diastolic. Hypertension, if not corrected, can cause long-term health problems. In the college-age population (18-24 years), about 9.2\% have hypertension. Suppose that a blood donor program is taking place in a college dormitory this week (final exams week). Before each student gives blood, the nurse takes a blood pressure reading. of 196 donors, it is found that 29 have hypertension. Do these data indicate that the population proportion of students with hypertension during final exams week is higher than $9.2 \% ?$ Use a $5 \%$ level of significance.

The following is a solution for number 27. This looks at the mean systolic blood pressure being less than 100 and 30 for a certain floor for like a nursing student. And first we look to see if the conditions for inference have been met and you're given a normal probability plot. Um Mhm. Uh huh. I did not mean to do that. Sorry, normal probability plot. So uh this is what it kind of looks like in your books. You don't have to do that, but as long as it's linear, then we can assume normality. So that's one condition for inference and then the other one is a box plot and that kind of tells us if there are any outliers or skew nous. So this is what it kind of looks like for you and this definitely does not have any outliers. There are no stars on the outside. And then that really looks pretty darn good to me. That median is basically smack dab in the middle, maybe a little bit left skewed, but for the most part, the conditions for inference of Metz of the normal probability plot, it's linear. So we can assume normality and the box plot, it's free of outliers or really any type of skew as well. So they actually did a mini tab print out where the test statistic is negative 1.41 and a p value of 0.88. So it asked us what are the hypotheses and you don't have to use the context clues here in your original prompt, but h not is mu equals 1 30. And it says less than for the alternative. Some use less than 130 because that's what this nursing patient patient. Uh Thanks. And then it says identify the P value. There it is .088. That's just it's his pee pee. And that's the p value. So .088. And will the nurse rejected? 5? no P value is greater than alpha so failed to reject. Okay. So the conclusion here would be that there is not sufficient evidence to say that The blood pressure the average blood pressure is less than 1:30 on our floor. So not enough evidence to suggest the average. What pressure? All right. Blood pressure on her floor is less than 130.

In this exercise, we're going to be considering a hypothesis test that has been proposed. And for this test, we're going to do a few things number one, we're going to identify the variable. Then we're going to give the two populations population one and two um specify or define the mean for each population. And then also state the non hypothesis and the hypnotic and the alternative hypothesis. And finally uh classified the hypothesis test as two tailed, left tailed or writes tales. No. In this problem, we are considering the Children of diabetic mothers. And the hypothesis test is to decide whether the mean systolic blood pressure of oh D. M. That is diabetic mothers, adolescents exit that of uh oh and M. That is non diabetic mothers, adolescents. Okay, so the variable here is the systolic blood pressure. So that would be the first thing we need to right there for the variable systolic blood pressure. And for the population, we are considering adult lessons that our offspring of diabetic mothers for the first population. So that would be um or D. M. Adolescents. And for the second population population too, we are going to be considering the adolescents that are offspring of a non diabetic mothers. So that's all. And them with their lessons. Next, we shall be defining new one and mewtwo. The population means for the two populations. So a new one uh is going to represent is going to be defined by the the mean systolic blood pressures of ODM adolescents. And mewtwo is the mean systolic blood pressure of who? N. M. Adolescents. Next, the null and alternative hypotheses in the question, the hypothesis is testing to decide whether the means, the systolic blood pressure of OdM adolescents, eggs, seeds that of O. N. M. And we listen. So the null hypothesis that new one is equal to um Youtube. And the alternative hypothesis is me one is greater than mute, too, because of the white exits, so that means greater than and this is since we're using the symbol greater than this is going to be a right tailed test. Mhm.

This problem is called hypertension. We need to identify that the type of design that is being used. Eso for party. This is a This is an observational study. It is an observational study because there are no treatments being imposed. The researchers were simply studying traits that already exist in the subject's. They're not imposing new treatments. Part B. This is considered a ah prospective study because the subjects were identified first, then traits were observed. Part C. The subjects in this study We're roughly the 200 men and women with moderately high blood pressure and normal blood pressure. Um, the study doesn't give us information about the selection method so we can make a comment on that in this election. Method Party. The parameters of interest. Um, that's the difference in memory and reaction time scores between the those with North, the people with normal blood pressure and moderately high blood pressure. And finally, for party, the nature and scope of the conclusion People with moderately high blood pressure did worst on test of memory and reaction time and those with normal blood pressures. And this is very important. Causation cannot be proving using observational study. Okay, so we can say they can't see that. Though blood pressure caused the test memory and reaction time, a lurking variable such as age or overall health might have been the cause boat that's not stated in the study.

Mhm. We're doing a detailed test. So that means our alternative hypothesis is that the two sigma's are not equal. F is going to be S one squared over as two squared which is 1.4321 Our degrees of freedom are gonna be calculated by one list in each of the sample size. For the numerator, we're going to have nine and for the denominator we're going to have 24. Yeah. Now we can go to the back of the book and table eight to figure out what are critical values are. So this is going to be F of 0.25 Since we have a significant level of 5%. Some F 0.25 Okay. And at nine and 24 degrees of freedom, we see that the goal is 2.70 Now, what about F of one minus 10.25? Or as we might say 0.975 This is going to turn out to be 1/3 61 or approximately 0.28 Now that we've figured out those two values we can calculate the are actually just compare the critical value. So we have 1.431 is a critical value and 1431 is between 0.28 and 2.7. Therefore we failed to reject the null hypothesis at these 5% significance level. Is there any other questions, nope?


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