An insurance company checks police records on 557 accidents selected at random and notes that teenagers were at the wheel in89 of them. Complete parts a) through d).a) Construct the 95% confidence interval for the percentage of all auto accidents that involve teenage drivers.95% CI = (% %)(Round to one decimal place as needed.)

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Answered: an insurance company checks police…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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In a poll of 7 randomly selected students showed that 31.3 % of the respondents believe in global warming. Based on the sample results, the margin of error E that corresponds to a 95% confidence level is equal to: O a. 0.45 O b. 0.34 O c. 0.29 O d. 0.22
A newsgroup is interested in constructing a 99% confidence interval for the proportion of all Americans who are in favor of a new Green initiative. Of the 542 randomly selected Americans surveyed, 447 were in favor of the initiative. a. With 99% confidence the proportion of all Americans who favor the new Green initiative is between and . b. If many groups of 542 randomly selected Americans were surveyed, then a different confidence interval would be produced from each group. About . percent of these confidence intervals will contain the true population proportion of Americans who favor the Green initiative and about percent will not contain the true population proportion.
The following table shows 1000 Engineer school applicants classified according to scores made on a college entrance examination and the quality of the high school from which they graduated, as rated by a group of educators: Quality of High Schools Superior Scores Average Low 300 100 High 260 340 A 95% confidence interval for the population proportion of applicants that a low score on the examination and graduated from an average high school is: (0.081 , 0.118) None of these (0.271,0.328) O (0.310, 0.369)
Answer Part B
You are interested in constructing a 99% confidence interval for the proportion of all caterpillars that eventually become butterflies. Of the 431 randomly selected caterpillars observed, 50 lived to become butterflies. Round answers to 4 decimal places where possible. a. With 99% confidence the proportion of all caterpillars that lived to become a butterfly is between___ and___ . b. If many groups of 431 randomly selected caterpillars were observed, then a different confidence interval would be produced from each group. About____ percent of these confidence intervals will contain the true population proportion of caterpillars that become butterflies and about ____ percent will not contain the true population proportion.
Example 5: Find a 90% confidence interval for the mean difference in the mileage obtained for tires with the new material and tires with the old material. Following are the paired differences (new minus old) for each of the 17 riders, chosen at random, who took part in the experiment: 50, 45, 50, 50, 100, 100, 100, -10, 75, -25, 0, 25, 75, 25, 50, 40, 35
In a random sample of 500 people eating lunch at a cafeteria on various Fridays, it was found that 160 preferred seafood. A 99% confidence interval for the actual proportion of people eating seafood on Fridays (rounded off to two decimals) is: A. [0.28; 0.36] B. [0.29; 0.35] C. [0.27; 0.37] D. [0.26; 0.36]
You were interested to know how long do students of mathematics in the university spend studying their math lessons in a day. You asked 10 random students and obtained the following data (in hours): 2, 1.5, 3, 2, 3.5, 1. 0.5, 3, 2, 4. Construct a 99% confidence interval for the average time that students spend studying math in a day. O 1.46 hours to 3.04 hours O none of the choices. O 1.31 hours to 3.19 hours O 0.95 hours to 3.55 hours
A study examined parental influence on teenagers' decisions to smoke. A group of students who had never smoked were asked about their parents' attitudes to smoking. The students were questioned two years later to see if they had started smoking. The study found that, of the 287 students whose parents disapproved, 58 became smokers. Of the 48 students with lenient parents, 11 smoked. Complete parts a through c below. a) Create a 95% confidence interval for the difference between the proportion of children with lenient parents who later smoked, p1, and the proportion of children with disapproving parents who later smoked, p2. The confidence interval for p1−p2 is −0.101negative 0.101,0.1550.155. (Round to three decimal places as needed.) b) Interpret your interval in this context. Select the correct choice below and fill in the answer boxes within your choice. (Round to one decimal place as needed.) A. There is 95% confidence that the proportion of teens…
The following numbers represent five randomly selected individuals’ ages: 20, 32, 36, 42, 45. If the age distribution for the whole population follows a normal distribution, which one represents the 90% confidence interval for the mean age based on this particular sample? A ) (20.26, 49.74) B ) (25.66, 44.34) C ) (14.83, 55.17) D ) (28.28, 41.72) E ) (23.72, 46.28)
In a poll, 69% of the people polled answered yes to the question "Are you in favor of the death penalty for a person convicted of murder?" The margin of error in the poll was 2%, and the estimate was made with 96% confidence. At least how many people were surveyed?
In a survey of 3279 adults aged 57 through 85 years, it was found that 81.9% of them used at least one prescription medication. Complete parts (a) through (c) below. How many of the 3279 subjects used at least one prescription medication? (Round to the nearest integer as needed.) Construct a 90% confidence interval estimate of the percentage of adults aged 57 through 85 years who use at least one prescription medication

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