5

Medical rescarcher wants 0 compare the pulse: rates of smokers and non-smokers- Hc believes that the pulse rate for smokers and non-smoker different and wants t0 le...

Question

Medical rescarcher wants 0 compare the pulse: rates of smokers and non-smokers- Hc believes that the pulse rate for smokers and non-smoker different and wants t0 lest this elaim at the 0.05 level of signilicance; sample of 35 smokers has mean pulse rute of 897 . and sample of 45 non-smokers mean pulse Nie of 83 The population standard deviation of the pulse rates is known to be smnokers and for non-smnokers.Stnte Uic null ard aternative hypotheses for the test;Compute Ihe value of the test stati

medical rescarcher wants 0 compare the pulse: rates of smokers and non-smokers- Hc believes that the pulse rate for smokers and non-smoker different and wants t0 lest this elaim at the 0.05 level of signilicance; sample of 35 smokers has mean pulse rute of 897 . and sample of 45 non-smokers mean pulse Nie of 83 The population standard deviation of the pulse rates is known to be smnokers and for non-smnokers. Stnte Uic null ard aternative hypotheses for the test; Compute Ihe value of the test statistic: Rourd your nnSwET to (WO decinul places. Find the p-value ussociated with the test statistic , Round your answer to four decimal pluces; Make thc decision for the hypothesis test: State the conclusion of the hypothesis test



Answers

Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim.
Among 198 smokers who underwent a "sustained care" program, 51 were no longer smoking after six months. Among 199 smokers who underwent a "standard care" program, 30 were no longer smoking after six months (based on data from "Sustained Care Intervention and Post discharge Smoking Cessation Among Hospitalized Adults," by Rigotti et al., Journal of the American Medical Association, Vol. 312, No. 7). We want to use a 0.01 significance level to test the claim that the rate of success for smoking cessation is greater with the sustained care program. a. Test the claim using a hypothesis test. b. Test the claim by constructing an appropriate confidence interval. c. Does the difference between the two programs have practical significance?

No. Okay, so we know that the number of patients given sustained care is 198. Among that, 82 0.8%. They're no longer smoking after one month. So 198 times 82.8 percent. Well, give us 163.94 Okay, so we can round that 264. Yeah. Andi making, then Right down. Arnel Hypothesis, which is given that P is 1 80 80% since the test claims that 80% of patients stopped smoking after giving state care and our culture bosses will therefore be P is not equal to your right June at a given level of significance. Which is your prince? Your one. Okay. Me? Yeah. So we will get a Z statistic value off 0.99 on the critical value for the normal area. Table movies equals plus or minus +257 eight A p value. 20 above output is just your 200.32 Sure. Yes. On. Since this is greater than our little of significance, we can include that there's not sufficient evidence to support the rejection of the claim that 80% off patients stop smoking when sustained care. What about

In this question. I have been given a table. All right, Andi, I have to calculate. I think the mean is to be calculated by using that table. So if I use the table to find the mean the mean off those 32 these are 32 numbers, right? Yeah. These are ready to farmers focus. So their main turns out to be 14.83 4th expert is 14 0.834 My end is equal to 32. Have I been given the standard Division? Yes. Population standard Division 6.2 Sigma is equal to 6.2. And the claim? What is the claim? It takes smokers too. Quit permanently. Is there enough enough evidence to reject the claim that the meantime, it takes smokers to quit smoking permanently as 15 years? Okay, the mean is 15 years. So immune is equal to 15. This is going to be my null. And my alternative hypothesis will be that Mu is not equal to 15. All right, I need I need to know what I'm going to use. I'm going to use the sad statistic and I have all the information that I'm going to require in order to calculate this, so I'll simply put in the formula 14.83 minus 15 upon 6.2 by route off 32. Now I have to use the calculator to calculate this Just a moment. So this is going to be 14.83 minus 15 divided by 6.2, multiplied by route off 32. This is minus zero point. This is minus 0.155 This is my zed statistic. Okay, Now, if I calculate my people first of all, this is going to be a two tailed test because there's a not equal to sign. And what is my Alfa level? My al 50.5 So if this is a two tailed test night in these two tails, my total areas 0.5 which means in every single tail it will be 0.25 So let me do one thing. Let me find the critical value for 0.25 and this value turns out to be 1.96 which means that since it is symmetrical, this is minus 1.96 and this is plus 1.96 And since my Z value is between this interval, my Z value falls somewhere around here minus 0.155 I will fail to reject minor hypothesis. And if I want to find the p value for this, my P value for this is going to be P Valley for minus for the zero point. Uh, sorry for minus 0.1 point 55 The P value for this is just a moment in a zero point 88 My P value is turning out to be 0.8 ID, 0.88 right. This is my p value. I have used the people you calculate over you, and I can see that my p values greater than Alfa. So from here also, I can say that I will fail to the Jackman and hypothesis. So this was my null hypothesis. I will say that I do not have enough evidence to see or I do not have enough evidence to reject the claim that the meantime it takes the smokers to quit smoking permanently is 15 years. This is going to be my answer

Okay. So the objective in this question is to test the claim that among smokers try to quit nicotine patch therapy, the majority are smoking year after the treatment. The null hypothesis h not, is that the probability is equal to 2.5 on the alternative hypothesis. H A is that the probability created in 0.5? Okay, you're also given that X is you got 39 and it's equal to 39 plus 32. She'll give us a sample size of 72. On that off level of significance is your 720.5 This just information. We're given the question self. Yeah, now can calculate Z statistic the statistic which is the proportion minus p over the square root off peak you of N v. Q Is one manus Okay, P is our no, not so playing all these values in We'll get value from the calculator, which is 0.83 on the P value. This test from the normal table is 0.203 on since the P value is greater than the level significance Alfa, which is your digital five. The null hypothesis is not rejected now for there is evidence to show that smoke among smokers try to quit with nicotine patches therapy. The majority are still smoking after a year of treatments, majority still smoking. That is what we can conclude.

Okay, now in sustained care. The first case. Let's take off the sustained care Sustained care. Okay, What is going to be my pick up? My peak at is 0.8280 point 828 Okay, Now my end is 1 19 in is 1 98 and I want a 95% confidence level. Which means my Alfa by two is what my Alfa by two is 0.0 to fight. So Z alphabet to will become 1.96 All right. We already know the formula for e the formula for he is going to be Z Alfa by two multiplied by rude over peak cap into one minus p cap or we also I just ask you cap upon in right. Do we have all the values? Yes. So what is goingto be the value of E? If I substitute these, I get my value of 0.5 to 6. My E in this case turns out to be 0105 to 6. Fine. And how do I write the confidence interval it is given by this. Okay, now these are the two formulas which are required here. Okay, So what is my confidence in double for the first case, 77.54 to 88.6 This becomes 77.54% to 88.6%. Okay, this becomes for the sustained care. Now, what about the standard care for standard here for standard game over here. My P cap is 0.6 to 8. 0.6 to 8. This is my peak app. My in is equal to 1 99. My n is equal to 1 99. And my Z Alfa by two is 1.96 Same as before. Z Alfa by two is equal to 1.96 which is the same as before, right? This is 1.96 So again substituting the values to get the answer for E my He turns out to be 0.672 0.672 What am I using? The same formula. And now I will use this formula to find the interval. And what is the interval that I'm getting in this case? The interval that I get over here is 56.8 to 69.52 56.8% to 69.5 2%. Okay, so what can I say? The confidence interval for the sustained care lies completely above the confidence interval for the standard care, which indicates that the sustained care is more effective than the standard care. Sustained care is more effective, sustained care is more effective and this is going to be my answer.


Similar Solved Questions

5 answers
7. Let A, B,and € be subsets of some universal set U.Draw (wo general Venn diagrams for the sets ^ and €. On one. shade the region that represents (B C),and on the other; shade the region Ihat represents Based On the Venn diagrams make conjecture about the relationship between (he sets (k and (A (Are the two sels equal? If not; is one of the sels : subset of the other set?)
7. Let A, B,and € be subsets of some universal set U. Draw (wo general Venn diagrams for the sets ^ and €. On one. shade the region that represents (B C),and on the other; shade the region Ihat represents Based On the Venn diagrams make conjecture about the relationship between (he sets ...
4 answers
Example 8 Let f (€,y) = {396,)+f.9 Where if (€,y) = (0,0) is f continuous? 3r2y Example 9 Let f (T,y) = 279 if (T,y) + (0,0) Where 0 if (,y) = (0,0) is f continuous?
Example 8 Let f (€,y) = {396,)+f.9 Where if (€,y) = (0,0) is f continuous? 3r2y Example 9 Let f (T,y) = 279 if (T,y) + (0,0) Where 0 if (,y) = (0,0) is f continuous?...
5 answers
6) f ktan (tanx)) sec?x dxSelect one:a.COSX +(In | cos ( tan X) | + €cIn Icos(tan x )l + €C. tan / +enone
6) f ktan (tanx)) sec?x dx Select one: a.COSX +( In | cos ( tan X) | + € cIn Icos(tan x )l + € C. tan / + enone...
5 answers
What is the probability of throwing a "six" with a single die?
What is the probability of throwing a "six" with a single die?...
5 answers
Reduce Japisuo) 341 { the equation + 222 ~equation to one 2 + below ofi the standard forms
Reduce Japisuo) 341 { the equation + 222 ~equation to one 2 + below ofi the standard forms...
5 answers
Determine if b is a linear combination of a1, a2, and a3_If yes give the weights-12. a1-2 a2a31118. Let V1V2For what value(s) of h is y in the plane generated by V1 and Vzand
determine if b is a linear combination of a1, a2, and a3_ If yes give the weights- 12. a1 -2 a2 a3 11 18. Let V1 V2 For what value(s) of h is y in the plane generated by V1 and Vz and...
5 answers
Evaluate the following determinate using reduction formula_ -3 ; 139 3 4 -1 ~2 ~1 2 1
Evaluate the following determinate using reduction formula_ -3 ; 139 3 4 -1 ~2 ~1 2 1...
5 answers
Identify the growth rate and the growth factor in the exponential function.$$y=50(1+1)^{t}$$
Identify the growth rate and the growth factor in the exponential function. $$y=50(1+1)^{t}$$...
1 answers
Evaluate the integrals in Exercises $29-50$ $$\int \frac{d x}{1+e^{x}}$$
Evaluate the integrals in Exercises $29-50$ $$\int \frac{d x}{1+e^{x}}$$...
5 answers
If the five numbers summary of a data set are:-14,0,2,3,10 then the =lower outlier limit (lower fence) (abi; 2)10
If the five numbers summary of a data set are:-14,0,2,3,10 then the =lower outlier limit (lower fence) (abi; 2) 10...
5 answers
Lat Ux/y havePAP: Plxsl Lxy) 4 Tf9cyer okx4i} Aeletmine Elu]_{ind Ik 'Pdp
Lat Ux/y have PAP: Plxsl Lxy) 4 Tf9cyer okx4i} Aeletmine Elu]_ {ind Ik 'Pdp...
5 answers
GiicoeUt @ Tocst €r luwss"~het @0c[cna053 70e;1n Fi
GiicoeUt @ Tocst €r luwss"~het @0c[cna 053 70e; 1n Fi...
5 answers
Consider the rational funetion202 +51 2-4Find the domain of the function y = f(r).marks|b) Find the poles and the zeroes of the function y = f(z)marks|Find the limits of the function y f(z) as *Eoo12 marks|
Consider the rational funetion 202 +51 2-4 Find the domain of the function y = f(r). marks| b) Find the poles and the zeroes of the function y = f(z) marks| Find the limits of the function y f(z) as * Eoo 12 marks|...
5 answers
The manager of a bank recorded the amount of tire each customer spent waiting in line during peak business hours one Monday: The frequency table below summarizes the results; (Waiting Number Time of ((minutes) Customers 14 4-7 8-11 14 12-15 16-19 20-23 24-27If we randomly select one of the customers represented in the table; what is the probability that the waiting time is at least 12 minutes or between 8 and 15 minutes?0.3390.66102730712
The manager of a bank recorded the amount of tire each customer spent waiting in line during peak business hours one Monday: The frequency table below summarizes the results; (Waiting Number Time of ((minutes) Customers 14 4-7 8-11 14 12-15 16-19 20-23 24-27 If we randomly select one of the custome...

-- 0.022642--