(b)Point estimate:0.25critical value:1.96Calculate0.0000.1270.250Error range:0.123Confidence interval of95%:0.25 +0.1230.373Based on the sample, plot the confidence interval of the95%for the population proportion of all winning scratching posts.• Enter the upper bound and lower bound values on the graph to display your confidence interval.• Under the point ( ◆), indicate the statement of the ad,0.42.95% Confidence Interval:Critical values70.005=2,5760.010=2,32620.025 19600.050 1,6450.100=1,2821.000 (b): Your answer is wrong.There is a popular lottery in which a ticket is called a scratcher. An ad for this lottery claims that the42%of the population all scratching posts are winners. Youwant to investigate this claim by selecting a random sample of48scratching postsFollow these steps to construct a confidence interval for the95%for the population proportion of all winning tickets. Then indicate whether the confidenceinterval you constructed contradicts the ad claim. (If necessary, you can refer to a list of formulas.)(to) Click on "Take sample" to see the results of the random sample.Take a samplewinning scratchingSample size:48postloser scratcherNumber1236Proportion0.250.75Enter the values for the sample size, the point estimate of the population proportion, and the critical value you need to construct the confidenceinterval of the95%. (Choose the correct critical value from the critical value table provided.) When finished, click on "Calculate".Standard error:0.063XŚ

The random sample of size, The given data is,Number ProportionWinning scratching post120.25Loser…

Answered: There is a popular lottery in which a…

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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A random sample of n1 278 people who live in a city were selected and 98 identified as a "dog person." A random sample of n2 = 109 people who live in a rural area were selected and 65 identified as a "dog person." Find the 90% confidence interval for the difference in the proportion of people that live in a city who identify as a "dog person" and the proportion of people that live in a rural area who identify as a "dog person." Round answers to to 4 decimal places. < pi – p2 < Submit Question
A university health services physician is concerned about how much sleep freshman are getting. She asked a simple random sample of 100 students if they got at least 8 hours of sleep the previous night. then she constructs a 90% confidence interval for the true proportion of all freshman college students who got at least 8 hours of sleep the previous night. Explain what would happen to the width of the interval if the sample size would decrease to 25 students
A recent survey showed that 42% of US employers were likely to require higher employee contributions for health care coverage. Suppose the survey was based on a sample of 500 companies. Find the 99% confidence interval for the proportion of all companies likely to require higher employee contributions for health care coverage. (Round to four decimal places)
In a preelection poll, a pollster finds that 442 out of 650 people polled favor Smith. Construct the 95% confidence interval for the proportion (nearest to 0.0001) of people in the whole population who favor Smith.
CNN is conducting a poll, and wants to produce a 90% confidence interval for the proportion of all New York residents who favor legalizing recreational marijuana. Of the 602 randomly selected New Yorkers surveyed, 394 were in favor of the initiative. a. With 90% confidence, the proportion of New York residents who favor legalizing recreational marijuana is between BLANK and BLANK. b. Suppose that we survey 602 randomly selected Americans many times, and calculate a confidence interval for each group. About Blank . percent of these confidence intervals will contain the true population proportion of New Yorkers who favor legalizing recreational marijuana, and about BLANK percent will not contain the true population proportion.
If the number of smokers in a sample of 40 students selected from university A is 14 and the number of smokers in a sample of 380 students selected from university B is 84. Then the 99% confidence interval for the difference between proportions of smokers in the two universities (PB- PA) is closest to O a. (-0.1450,-0.1130) O b. (-0.3310,0.0729) O c. None of these o d. (-0.3110,0.0534) e. (-0.1430,-0.1150)
In a survey of 508 adults, 317 indicated that emails can be easily misinterpreted. If we wish to construct a 95% confidence interval to estimate the true proportion of adults who believe that emails can be easily misinterpreted we would substitute a. P hat is equal to b. Z is equal to
Wildlife biologists inspect 200 deer taken by hunters and find 45 of them carrying Lyme disease ticks. What would be the margin of error for this study with a 90% confidence interval?
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.)
A team of researchers aims to understand whether a sleep treatment drug has an effect on the number of hours of REM sleep. Five subjects were randomly selected and their time in REM sleep (in minutes) was measured before treatment and after treatment. The before treatment measurements are: (30, 45, 90, 60, 100). The after-treatment measurements for the same five patients are: (35, 30, 80, 70, 110). Your boss looks at the confidence interval you computed and says that it is too wide to be useful for decision making. You propose a list of changes to your methodology that may result in a narrower interval. Which option should NOT be on your list: Group of answer choices increase the confidence level recruit more subjects into the study decrease the confidence level
If the number of smokers in a sample of 40 students selected from university A is 14 and the number of smokers in a sample of 380 students selected from university B is 84. Then the 99% confidence interval for the difference between proportions of smokers in the two universities (PB-PA) is closest to O a. (-0.3110,0.0534) O b. (-0.3310,0.0729) O c. (-0.1430,-0.1150) O d. (-0.1450,-0.1130) Oe. None of these
8.2 question 5). In 2013, the Gallup poll asked 1,039 American adults if they believe there was a conspiracy in the assassination of President Kennedy and found that 634 believe there was a conspiracy ("Gallup news service," 2013). Estimate the proportion of American’s who believe in this conspiracy using a 98% confidence interval.

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