Suppose an insurance company randomly surveyed 633 drivers and found that 601 claimed they always buckled up. Construct a 94% confidence interval for the fraction of all drivers who buckle up. p' = 8/2 N/D = = Margin of Error: E = We are 94% confident that the fraction of drivers who always buckle up is between and
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- In a random sample of 392 people that were tested for the Norcovirus, it was found that 314 did not have the virus. A 97% confidence interval was constructed to estimate the proportion of the population that does not have the Norcovirus. This confidence interval came out to be (0.76,0.84).A sample of size 25 was taken from a very large population. Sample statistics, x¯ and s, were used to find that the 95% confidence interval was (138.616, 163.384). x¯= s=A random sample of 500 males included 215 that approved of the performance of a President. A random sample of 500 females included 185 that approved of the performance of a President. Determine each of the following parts that will lead you to construct a 95% confidence interval for the overall difference in the percentage of males and females that approve of the President's performance. group one males, gorup 2 females Calculate E, the margin of error, for a 95% level of confidence, Create the final confidence interval for the difference in percentages
- Alfredo wants a 85% confidence interval for the true average gas mileage for a Gremlin after randomly sampling 9 Gremlins. It is known that the true sigma for Gremlin gas mileage is 4 mpg, and the average of the 9 Gremlins in Alfredo's study was 24 mpg. Find the confidence interval assuming the distribution is normal.A researcher surveys a group of college students to determine the negative life events that they experienced in the past 5 years and their current feeling of well-being. For N = 18 participants with 2 or fewer negative experiences, the average well-being score was X = 42 with SS = 398. For the N = 16 participants with 5 to 10 negative experiences the average score is X = 48.6 with SS = 370. Construct a 95% confidence interval for the difference between the population means for the two groups.We wish to estimate what percent of adult residents in a certain county are parents. Out of 100 adult residents sampled, 48 had kids. Based on this, construct a 90% confidence interval for the proportion ππ of adult residents who are parents in this county.
- Suppose a new treatment for a certain disease is given to a sample of 180 patients. The treatment was successful for 153of the patients. Assume that these patients are representative of the population of individuals who have this disease. Calculate a 98% confidence interval for the proportion successfully treated. (Round the answers to three decimal places.)The manager of a fleet of automobiles is testing two brands of radial tires and assigns one tire of each brand at random to the two rear wheels of eight cars and runs the cars until the tires wear out. The data (in kilometers) follow. Find a 99% confidence interval on the difference in the mean life. Car Brand 1 Brand 2 1 36,925 34,318 2 45,300 42,280 3 36,256 35.548 4 32,100 31,950 5 37,210 38,015 6 48,360 47,800 7 38.200 37,810 8 33,500 33,215 Round your answer to 2 decimal places. Do not use commas.We wish to estimate what percent of adult residents in a certain county are parents. Out of 200 adult residents sampled, 34 had kids. Based on this, construct a 90% confidence interval for the proportion, p, of adult residents who are parents in this county. Give your answers as decimals, to three places.The average commute to work (one way) is 25 minutes according to the 2005 American Community Survey. If we assume that commuting times are normally distributed and that the standard deviation is 6.1 minutes, calculate the probability that a randomly selected commuter spends for the following cases. Round the final answers to at least four decimal places and intermediate z-value calculations to two decimal places. Part 1 of 2 (a) More than 31 minutes commuting one way =P>X31 0.1627 Part 2 of 2 (b) Less than 10 minutes commuting one way =P<X10If n = 420 and p (p-hat) = 0.2, construct a 95% confidence interval. Give your answers to three decimals < p <In a sample of 300 adults, 198 had children. Construct a 90% confidence interval for the true population proportion of adults with children. Give your answers as decimals, to three places RectanSEE MORE QUESTIONSRecommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons Inc
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