6.27 Public Option, Part IlI: Exercise 6.13 presents the results of a poll evaluating support for thehealth care public option plan in 2009. 70% of 819 Democrats and 42% of 783 Independentssupport the public option.(a) Calculate a 95% confidence interval for the difference between (pp - P) and interpret it in thiscontext. We have already checked cond itions for you.%) (please round toThe confidence interval is:the nearest percent) Interpret the confidence interval in context:95% of differences in proportions are contained within this intervalWe can be 95% confident that the difference in sample proportions is contained withinour intervalWe can be 95% confident that the difference in population proportions is containedwithin our interval(b) True or false (Explain): If we had picked a random Democrat and a random Independent at thetime of this poll, it is more likely that the Democrat would support the public option than theIndependent.falsetrue

We have provided following information:Sample 1: DemocratsSample 2: Independents

Answered: 6.27 Public Option, Part IlI: Exercise…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A New York Times/CBS News Poll asked a random sample of U.S. adults the question “Do you favor an amendment to the Constitution that would permit organized prayer in public schools?” Based on this poll, the 95% confidence interval for the population proportion who favor such an amendment is (0.63, 0.69). Based on this poll, a reporter claims that more than two-thirds of U.S. adults favor such an amendment. Use the confidence interval to evaluate the reporter’s claim. Because the value 23=0.66723=0.667 (and values greater than 2323) are in this interval, it is plausible that more than 2323 of the population favor such an amendment. Thus, there is convincing evidence that more than 2323 of U.S. adults favor such an amendment. Because 23=0.66723=0.667 is included in this interval, it is plausible that more than 2323 of U.S. adults favor such an amendment. 95% of the time there will be more than two-thirds of U.S. adults in favor of such an amendment. Because 0.95 > 0.667, the…
4. A poll was conducted in a very small community in country X to see if residents are for or against a new policy that requires them to wear a mask. A total of 500 persons were surveyed. Of the 500 surveyed 200 are against wearing a mask. Compute a 90% confidence interval for the true percent of residents who are against the new policy, and interpret the confidence interval.
According to a report on sleep deprivation by the Centers for Disease Control and Prevention, the proportion of California residents who reported insufficient rest or sleep during each of the preceding 30 days is 8.0%, while this proportion is 8.8% for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents. Calculate a 95% confidence interval for the difference between the proportions of Californians and Oregonians who are sleep deprived and interpret it in context of the data.
A random sample of 62 suspension helmets used by motorcycle riders and automobile race-car drivers was subjected to an impact test, and on 37 of these helmets some damage was observed. Find a 90% two-sided confidence interval on the true proportion (p) of helmets of this type that would show damage from this test. 0.517 <= p <= 0.677 0.597 <= p <= 0.677 0.517 <= p <= 0.699 0.494 <= p <= 0.699 0.494 <= p <= 0.677
A recent survey on people’s opinion as to whether the traditional TV licence should be replaced by a device independent broadcasting fee showed that out of a random sample of 115 individuals from the population, 46 of them stated were in favour, i.e. a sample proportion of 0.40. Which of the following is a 99% confidence interval for the true proportion of the adult Irish population that support the new fee structure? (a) 0.4±0.0895(b) 0.2821 < p < 0.5179 (c) 0.3105 < p < 0.4895 (d) 0.3246 < p < 0.4754 (e) 0.3269 < p < 0.4731 (f) 46±0.1179
1. A recent survey showed that from a sample of 500 packages delivered by a Postal Service, 480were delivered on time. a) Construct a 95% confidence interval for the proportion of all packages that are deliveredon time by the Postal Service.
4) The trade magazine QSR routinely checks the drive-through service times of fast-food restaurants. A 90% confidence interval that results from examing 607 customers in Taco Bell's drive-through has a lower bound of 161.6 seconds and an upper bound of 164.7 seconds. What does this mean?
In a study of academic procrastination, the authors of a paper reported that for a sample of 431 undergraduate students at a midsize public university preparing for a final exam in an introductory psychology course, the mean time spent studying for the exam was 7.54 hours and the standard deviation of study times was 3.90 hours. For purposes of this exercise, assume that it is reasonable to regard this sample as representative of students taking introductory psychology at this university. (a) Construct a 95% confidence interval to estimate ?, the mean time spent studying for the final exam for students taking introductory psychology at this university. (Round your answers to three decimal places.)( , )(b) The paper also gave the following sample statistics for the percentage of study time that occurred in the 24 hours prior to the exam. n = 431 x = 43.88 s = 21.66 Construct a 90% confidence interval for the mean percentage of study time that occurs in the 24 hours prior…
The production manager in an automobile plant is concerned with the number of cars that do not pass the final quality control inspection. In the last two hours he noted that in a sample of 120 cars only 92 were acceptable. The lower and upper limit of a 95% confidence interval for the true percentage of all cars that would not pass the final quality control inspection would be:
A national opinion poll found that 44% of all American adults agree that parents should be given vouchers that can be applied toward their children's education at any public or private school of their choice. The result was based on a small sample. A)How large an SRS is required to obtain a margin of error of at most 0.03 in a 99% confidence interval? B) Answer the question in part A again but this time By using Phat=0.5. By how much do the two sample size differ?
According to a report on sleep deprivation by the Centers for Disease Control and Prevention, the proportion of California residents who reported insufficient rest or sleep during each of the preceding 30 days is 8.0%, while this proportion is 8.8% for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents. Calculate a 95% confidence interval for the difference between the proportions of Californians and Oregonians who are sleep deprived and interpret it in context of the data. Also, conduct a hypothesis, with alpha = 0.05, test to determine if these data provide strong evidence the rate of sleep deprivation is different for the two states. Construct a 95% confidence interval Which of the following correctly displays the standard error? 0.0046 0.009 0.001 0.00002 0.008
In a study of contamination at landfills containing construction and demolition waste, 42 leachate specimens were tested for the presence of several contaminants. Of the 42 specimens, 26 contained detectable levels of lead, and 32 contained detectable levels of chromium. Is it possible, using the methods of this section, to find a 95% confidence interval for the difference between the probability that a specimen will contain a detectable amount of lead and the probability that it will contain a detectable amount of chromium? If so, find the confidence interval. If not, explain why not.

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