5

Consider drug lesting company tnat provides test for marijuana usage. Among 257 tested subjects, resulls from 27 subjects were wrong (either false positive Or = ial...

Question

Consider drug lesting company tnat provides test for marijuana usage. Among 257 tested subjects, resulls from 27 subjects were wrong (either false positive Or = ialse negative). Use 0.10 significance level t0 test Ihe claim Ihat less Ihan 10 percent of the tesi esuits are wrong Identify the null and altemative hypotheses for this test. Choose the correct answer below:Ho: p=0.1 Hj p<0.1 B. Ho: P<0.1 D=0;- Ho:' p=0.1 H: P>0.1 0 . Ho: P=0.1 H," PF0.1Icentify the test statistic fo

Consider drug lesting company tnat provides test for marijuana usage. Among 257 tested subjects, resulls from 27 subjects were wrong (either false positive Or = ialse negative). Use 0.10 significance level t0 test Ihe claim Ihat less Ihan 10 percent of the tesi esuits are wrong Identify the null and altemative hypotheses for this test. Choose the correct answer below: Ho: p=0.1 Hj p<0.1 B. Ho: P<0.1 D=0;- Ho:' p=0.1 H: P>0.1 0 . Ho: P=0.1 H," PF0.1 Icentify the test statistic for this hypothosis tost. The lest slalisiic lor Ihis nypolhesis lest is (Round IwO decimal places as needed ) Identily Ihe P-value Ior Ihis hypolhests lesL The ~value Ior Inis hypothesis lest (Rcund t0 three decimal places as needed ) Identily the conclusion Ior this hypothesis test Click t0 select your answeris).



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. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section. The company Drug Test Success provides a " 1-Panel-THC" test for marijuana usage. Among 300 tested subjects, results from 27 subjects were wrong (either a false positive or a false negative). Use a 0.05 significance level to test the claim that less than $10 \%$ of the test results are wrong. Does the test appear to be good for most purposes?

Yeah. Has a percentage of 18-25 year olds using marijuana changed since the publication stating it was 13.6%. Well, our first thing to do is look at the randomly selected people done recently, of the 1283 randomly selected 18-25 year olds, 205 of them said that they currently use marijuana, which is about 16%. Now, in terms of our criteria to see if we can use the one property test are we have to have a simple random sample which we have. And the sample size is so big. That's gonna when we multiply the sample size by our hypothesized population proportion, it's definitely going to be bigger than five. So we are going to go on. Those Are null hypothesis is that our population proportion is 13.6%. And then our alternative hypothesis is going to be the population proportion is not 13.6. See if there has been a change from that last publication. So for our Z score, we're going to have our sample proportion minus the hypothesized population proportion. And I'm plugging in my values here for P zero and for N. And when you run this calculation, you get a Z score of 2.51. And when we looked at him up in the table in the back of the book is the score of 2.51 corresponds to .994. But We're going to subtract that from one. This will give us our high end probability, which is About .006 And then we want not equal to so we actually have to double this because this is a two tailed test, which is going to be .012. Now we are looking at a 5% significance level. This .012 is definitely smaller than .05. So we are going to reject H. O. Which tells us essentially, we have enough evidence to claim. The population proportion of 18-25 year olds using marijuana has changed From the 13.6%. It is not that 13.6% anymore.

In this problem we're going to be looking at to smoking cessation programs. One of them is the sustained cab, and the other one is the standard camp in the sustained care. We had 198 smokers going through the program on out of those 198 smokers, 51 no longer smoke after six months and for this standard care program among 189 smokers, fatty one no longer smoking after six months. So we're going to use the 0.1 significance level to test the claim that the rate of success for smoking cessation is greater with the sustained care program are compared to them standard care program. So the first part of the question we're going to test the claim using a hypothesis test. After that, we're going to test it using the confidence interval. And then we're going to see whether there's a difference between the two programs, uh, in terms off the proportions. Okay, so let's begin and state the hypothesis on. In this case, the non hypothesis is P one equals P two, which implies that the two proportions have are equal and the alternative hypotheses is P one is greater than P two. So this means that according to the hypothesis that we have a higher proportion off people who no longer smokey the sustained care program compared to the standard care program. So this being, uh, one tales test, the critical value will be 2.33 So we need to substitute the values into the test statistic, and we have the following proportion. For those in the sustained care, we have 51 longest, smoking out off 188. And for the standard care, we have 30 the longest, walking out off the 199. And to get the calculated value of that, we need to substitute these values into the formula. And here p one hot in decimal form is 0.258 and we need to subtract p too hot, which is 0.151 Okay, then the difference. We subtract zero from the difference, which is assumed to be, uh, zero, according to the Nile hypothesis. So then, from there we get the square it off. P one p bar Cuba, developed by N one plus p. Baquba Weber. And to So PBA is given by the sum of all the, uh all the X X one and next to divided by anyone and into So what we need to have here is the sum of 51 and that he divided by the summer of 188 and 199. And when you walk that out together, P bob is 0.204 divided by N one, which is 188 on, we have to multiply the PBA, Cuba and Cuba is one minute 0.204 which is going to be 0.796 So the new Morita is repeated in the next fraction. So it's 0.204 time 0.796 divided by N to an end to is 199. And when we simplify the value off the test statistic that he's 2.6 five now, we can compare the calculated value of that and the critical value off that and you can see. But the critical value is 233 so we can share the right side and the calculated value of that is within the critical region, which is 2.65 And since that is the case, we have to make the conclusion to reject the null hypothesis. No, this means that there is sufficient evidence to support the clean, that the rate of success is for the smoking cessation is greater with the sustained care program compared to them the standard care program. Next, we're going to test the same claim by constructing, uh, confidence interval. And in this case, we're going to construct 98% confidence interval. And for that we need to get the margin of error. E is in the formula given, and when you walk that out you will get e equals 0.9 0.9 35 And when you substitute into this formula for P one heart minus B p one hut minus be too hot minus e, you will obtain the following confidence interval. So to be 0.1 35 it's less than P one main US P two, which is less than zero point 2005 now. In this case, we noticed that the confidence interval limits do not contain zero. And that means that there is significant, a significant difference between the two proportions, as we have had really seen in the test off hypothesis, meaning we could reject the null hypothesis off having the proportions being the same. So because the interval consists off positive numbers only, it appears that the success rate for the sustained care program is greater than the success rate for the standard care program. Lastly, in Passy, we're going to see whether the difference between the two programs have practical significance. So if you check the the percentages, for example, for P one it is 25.8%. And for P two hot, this is 15 0.1%. And based on this sample, the success rates off the programs our 5th 25 points, 8% and 15.1%. And that difference does appear to be substantial. There's a there's a difference. So the difference between the programs does appear to have practical significance, because it's about a difference off 10% between p of P one heart and P to heart

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

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.


Similar Solved Questions

5 answers
Analyzing an RC Circuit In the circuit shown a fixed voltage difference € is applied between points_ A and B After the capacitor €, which starts uncharged, there is a junction that leads to a square section of wire of side length L. The junctions are located on the midpoint of the sides: 2R WJLL R BR
Analyzing an RC Circuit In the circuit shown a fixed voltage difference € is applied between points_ A and B After the capacitor €, which starts uncharged, there is a junction that leads to a square section of wire of side length L. The junctions are located on the midpoint of the sides:...
5 answers
Two consecutive days of the week experience maximum temperature of 520 F at PM and = minimum temperature of 38 Fat AM_ Write cosine equation that would model this behavior. Letx = 0 represent PM on the first day.
Two consecutive days of the week experience maximum temperature of 520 F at PM and = minimum temperature of 38 Fat AM_ Write cosine equation that would model this behavior. Letx = 0 represent PM on the first day....
5 answers
ARC LENGTH AND THE FRENET FRAME For the following curves; determine the arc length parameter the length and the Frenet frame N,B as functions of t). Moreover , compute the curvature in two ways using N then using the tangential and norinal component of the acceleration .7(t) = (2cost,2sint,t2), with t € [0,7/4] 7(t) = (Ga+t",361-+)*2,3 with t € [0,21l 7(t) t, 14, with € [0,20] 7(t) = (6t8 , ~2t8 , -3t3) , with t € [1,2]
ARC LENGTH AND THE FRENET FRAME For the following curves; determine the arc length parameter the length and the Frenet frame N,B as functions of t). Moreover , compute the curvature in two ways using N then using the tangential and norinal component of the acceleration . 7(t) = (2cost,2sint,t2), wit...
5 answers
Simplify and write the trigonometric expression in terms of sine and cosine: tan; g(z) sec? I g(r)Submit Question
Simplify and write the trigonometric expression in terms of sine and cosine: tan; g(z) sec? I g(r) Submit Question...
5 answers
Question 3Find the value of 0 and 4 in the right triangle below:56.31degrees (round to the nearest 2 decimal places)33.69degrees (round to the nearest 2 decimal places)2, Find the value of 0 and $ in the right triangle below:
Question 3 Find the value of 0 and 4 in the right triangle below: 56.31 degrees (round to the nearest 2 decimal places) 33.69 degrees (round to the nearest 2 decimal places) 2, Find the value of 0 and $ in the right triangle below:...
5 answers
Show the reaction of benzaldehyde with one molecule of methanol to form a hemiacetal and then with a second molecule of methanol to form an acetal.
Show the reaction of benzaldehyde with one molecule of methanol to form a hemiacetal and then with a second molecule of methanol to form an acetal....
5 answers
@uestionsCalcuisle Iha go uubldiy 0' Callosk Gonyour roeulez witn !ha firet &d eocord flbeleaWhy should these solubilities b Identical ? ExplainCalculule Kn for 3(IOse using the ueunsolubility .
@uestions Calcuisle Iha go uubldiy 0' Callosk Gonyour roeulez witn !ha firet &d eocord flbelea Why should these solubilities b Identical ? Explain Calculule Kn for 3(IOse using the ueunsolubility ....
5 answers
In the 6/49 lottery game, a player selects six numbers from 1 to 49. What is the probability of picking the six winning numbers?
In the 6/49 lottery game, a player selects six numbers from 1 to 49. What is the probability of picking the six winning numbers?...
5 answers
Kirchhoff laws is applicable to onlyany junction of a circuitlany closed loop and any node of @ circuitany node of a circuitany closed loop of a circuitClear my choice
Kirchhoff laws is applicable to only any junction of a circuit lany closed loop and any node of @ circuit any node of a circuit any closed loop of a circuit Clear my choice...
5 answers
Provide an appropriate response 1) If $25,000 is used to purchase an annuity (4 pts) consisting ' equal payments at the end of each year for the next 8 years and the interest rate is 5% compounded annually, find the amount of each payment:
Provide an appropriate response 1) If $25,000 is used to purchase an annuity (4 pts) consisting ' equal payments at the end of each year for the next 8 years and the interest rate is 5% compounded annually, find the amount of each payment:...
5 answers
Homework: HW LA 5.1,5.2,5.3 Score: 0 of 6 of 10 (9 complete)HW Score: 67.5%, 6.75 of5.3.1Question HelpLet A = PDPand and D as shown below. Compute AD =(Simplify your answers.Enter your answer in the edit fields and then click Check Answer:
Homework: HW LA 5.1,5.2,5.3 Score: 0 of 6 of 10 (9 complete) HW Score: 67.5%, 6.75 of 5.3.1 Question Help Let A = PDP and and D as shown below. Compute A D = (Simplify your answers. Enter your answer in the edit fields and then click Check Answer:...
5 answers
How mass Spectrometer works? (please discuss in completedetails)
How mass Spectrometer works? (please discuss in complete details)...
5 answers
Let Y be random variable that represents the size of network packet in Questions in kilobytes (kB)_ Yis related to the random variable X_ the transmission time of the network packet in ms in Question the following relationship_Y =2+rX (kB),where is the data rate of the network given by;r = 10 Mb/s = 125 kB /msFind the PDF. mean_ and variance of Y
Let Y be random variable that represents the size of network packet in Questions in kilobytes (kB)_ Yis related to the random variable X_ the transmission time of the network packet in ms in Question the following relationship_ Y =2+rX (kB), where is the data rate of the network given by; r = 10 Mb/...
5 answers
Os(der +o Xea din Vetse nd een_Sodlium 0nzs; Nz (q ~ 1 Na 3 Ncs when 5.7 Ofams Sodium fea ( 4s w#h Cx CeSS nkrozin Iab Ae Mc5S 04 Poduc { NJa3 Coledled 15 0.9 9'cs: what (S +o pucentcq_yeld 04 #lu Yec ( Fion ' Ih mola MGSS JG g2. IL Ofcms
os(der +o Xea din Vetse nd een_Sodlium 0nzs; Nz (q ~ 1 Na 3 Ncs when 5.7 Ofams Sodium fea ( 4s w#h Cx CeSS nkrozin Iab Ae Mc5S 04 Poduc { NJa3 Coledled 15 0.9 9'cs: what (S +o pucentcq_yeld 04 #lu Yec ( Fion ' Ih mola MGSS JG g2. IL Ofcms...

-- 0.026163--