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Suppose A = a speeding violation in the last year and B = a cellphone user while driving. If A and B are independent then P(A ∩ B)= P(A)P(B). A ∩ B is the...

Question

Suppose A = a speeding violation in the last year and B = a cellphone user while driving. If A and B are independent then P(A ∩ B)= P(A)P(B). A ∩ B is the event that a driver received a speedingviolation last year and also used a cell phone while driving.Suppose, a study was conducted and the following data weregatheredSpeeding TicketNOT Speeding TicketCell Phone User3134Cell Phone NOT user1639What is the Test Statistics to test the NullHypothesis H0 : Being a cell phone user whiledriving an

Suppose A = a speeding violation in the last year and B = a cell phone user while driving. If A and B are independent then P(A ∩ B) = P(A)P(B). A ∩ B is the event that a driver received a speeding violation last year and also used a cell phone while driving. Suppose, a study was conducted and the following data were gathered Speeding Ticket NOT Speeding Ticket Cell Phone User 31 34 Cell Phone NOT user 16 39 What is the Test Statistics to test the Null Hypothesis H0 : Being a cell phone user while driving and receiving a speeding violation are independent events against the alternative hypothesis Ha : Being a cell phone user while driving and receiving a speeding violation are dependent events?



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.
The Chapter Problem involved passenger cars in Connecticut and passenger cars in New York, but here we consider passenger cars and commercial trucks. Among 2049 Connecticut passenger cars, 239 had only rear license plates. Among 334 Connecticut trucks, 45 had only rear license plates (based on samples collected by the author). A reasonable hypothesis is that passenger car owners violate license plate laws at a higher rate than owners of commercial trucks. Use a 0.05 significance level to test that hypothesis. a. Test the claim using a hypothesis test. b. Test the claim by constructing an appropriate confidence interval.

In this problem, we're going to be testing the effectiveness off seat bells we have to simple random samples off to groups off people. The first group is off. People not wearing seatbelts on the second group is for people who are wearing seatbelts, and we have proportions off people who, um, were killed during a car crash. So P one represents the portion off people who were killed on not wearing seatbelts, and that is that you won out off 2823 and P two. Heart represents the proportion off people who were killed in a car crash. Yet they were wearing seatbelts, and that is 16 out of 7000 765. So in the in this problem, you're going to be testing the the clean that seat belts are effective introducing fatalities, and the first step would be testing the claim using hypothesis test. So, for the hypothesis, the null hypothesis behalf p one equals P two, and for the alternative hypotheses we have P one is greater than P two. This implies that not wearing, uh, does the proportion of people who are killed when they are not wearing seatbelts is much higher than the proportion of people who were killed when they are wearing signals. That means that, uh, the seat belts are effective introducing fatalities. So for the test statistic, we need to substitute the values that you just obtained here into the formula. And when we do that, the value off the calculated value of that is 6.49 Okay. And since this is a one tailed test, the critical value is going to be 1.645 at 0.1 level 0.5 level of significance. Now we can compare these two values off that for the critical value, we can shade from 1.645 on the right and we can see that 6.45 is greater than 6.49 is greater than 1.6 now 45 which means that our calculated value is with being the critical region. And for that reason we conclude that we need to reject the nal hypotheses. And by rejecting the null hypothesis, we conclude that there is sufficient evidence to support the claim that the fatality rate is higher for those not wearing seatbelts. Next, we're going to test the claim by constructing on ah confidence interval and in this case we're going to construct a 90% confidence interval and fast. We need to get the margin of error e by using the formula given. And once we do that, the value off E is 0.33 We then need to substitute it to subtract it to the difference, be one heart and be too hot, and then added the difference to create the range. So in this case, the off the fast part will be p one minutes p two hut minus e and that gives you 0.57 now should be less than P one minus p two. And that is less than than when the some off the difference be one hut on and be may not speak to heart plus e which is 0.123 So we notice that the confidence interval limits do not include zero. And that means that the two fatality rates are not equal. Uh huh. And since the confidence interval limits on Lee include positive values, we can conclude that the fatality rate is higher for those not wearing seat bells. So once again, the claim has Bean supported using that the confidence interval method. Next, we're going to give an explanation to what the results suggest about the effectiveness off seat bells. So as we have seen, the proportions are different, and even when we check the proportions will notice that he won are divided by 202,828 gives us 1.1% and never on the P P. Two hot gives us 0.2%. And as you can see, the fast proportion is much greater. It's significantly larger, 1.1% is significantly larger than 0.2%. So the results suggest that the use of seat belt is associate it with fatalities, um, fatality rates that are lower than those as she waited with no, not using seat belts. So if you're using seatbelt, you're much more secure than when you're not

Properly. 23. Good. Today it's note new. It's smaller battery quickly. 45 and alternative than you. It's bigger than do you mind that having t is equal to X bar minus you note over s over square root off at which is equal to 48 minus 45/5 480.4 over a square root off 25 which is equal to 2.778 for could should be the The value is a number off or interval in the column of title table five containing the T value is in the row off the F equal toe number off size when this one is 20 five minus one, which is 24 so that we value his between oh point over five and a four point or one. So if they devalue smaller than significantly developing, the non hypothesis is rejected and they be value here is smaller than a 0.1. So we reject then on high processes. So we can say that there is sufficient evidence to support Beckley

In this problem, We're going to be looking at the study that was conducted to investigate the association between cell phone use and the hemispheric brain dominance. So we have two samples. The first sample has, uh, 216 subjects who preferred to use their left year on bond, and and yet 166 are right 100. So the first proportion off right 100 people is 166 out off the 216 people who preferred to use their left ear. The second sample was coming from 452 subjects would prefer to use their right here. So this is from the left here. And this is for those who prefer to use their right ear on. For those who prefer to use your there, Right here are 400 36 were right handed out off 452. So we're going to be conducting tests, uh, to test the claim that the writ off right handedness for those who prefer to use their left here, it's less than the rich off right handedness for those who prefer to use their right here. In other words, the fast proportion is much greater than the second proportion. Okay, so let's compare and fast to do a hypothesis test, we will need to right there. Now hypothesis on the alternative hypothesis. So they're not. Hypothesis is p one he calls p two on the second hypothesis, the alternative hypotheses p one iss less than p two. Which is to say that this fraction this proportion is much smaller. That when you're and you're right, when you prefer to use your left here, then fewer people will be, uh, right handed. No for the zed, uh, calculate the value of that. We need to go to substitute the values into the formula. And this gives us the value of that off negative. 7.933 Now we need to get a critical value for this being a one tailed test and the level of significance being 0.1 then the critical value offset would be equal to negative 2.33 Because this time we're saying that p one IHS less than p two. So therefore it's going to be negative and not positive. Now we can compare the two by during the critical region and as you can see this would be negative 2.33 and the the left Month will be the critical region. And with that, we would have to notice that the calculated Valley offset is within the critical region. And for that reason we reject the null hypothesis. Rejecting the null hypotheses means that there is sufficient evidence to support the claim that the rate off right handedness for those who profound to use their left here is less than the rate off right handedness for those who prefer to use your right ear for cell phones. So it's much easier for you to use your your right hand if you're right handed. In other words. So we tested clean using, uh ah confidence interval. And for that, we need to complete the value off the margin of error has shown in this formula, and when you substitute the values you obtained e us zero point 0698 Next you substitute the values into the confidence interval expression and you find the limit will be negative 0.26 six aunt, the other limit his negative 0.1 to 16. So notice that all these values are negative, meaning zero is not contained within the confidence limits. And that means that there is a significant difference between the two proportions. And because the interval consists of negative numbers only it appears that they claim is supported. Therefore, we can see that the difference between the populations does appear to have practical significance.

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


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