13. In another pre-election poll, a pollster finds that 438 out of 600 people polled favor Smith. Construct the 95% confidence interval for the proportion (to nearest 0.001) of people in the whole population who
Q: In a sample of 300 adults, 228 had children. Construct a 90% confidence interval for the true…
A: Given C=90% Confidence level X=228 N=300
Q: Out of a sample of 600 adults aged 18 to 30, 162 still lived with their parents. Based on this,…
A:
Q: In a sample of 310 adults, 205 had children. Construct a 99% confidence interval for the true…
A: Given data, n=310 x=205 p=x/n=205/310=0.661 z-value at 99% confidence is Zc=2.576
Q: 47. Out of 400 people sampled, 372 had kids. Based on this, construct a 99% confidence interval for…
A:
Q: In a sample of 120 adults, 114 had children. Construct a 99% confidence interval for the true…
A:
Q: A graduate Statistic Student wants to estimate the true proportion of county college ( "c.c.")…
A: we have given that sample size N=100 ,favorable cases X=64 sample proportion P^=X/N
Q: In a sample of 350 adults, 277 had children. Construct a 99% confidence interval for the true…
A: The sample size, i.e., the sampled number of adults, n=350.The sampled number of adults with…
Q: In a sample of 270 adults, 151 had children. Construct a 90% confidence interval for the true…
A: Given that Sample size n =270 Favorable cases x =151 Sample proportion p^=x/n =151/270 =0.5593
Q: An accounting firm has 1,220 clients. From a random sample of 122 clients, 110 indicated very high…
A: From the given information, the sample size is 122 clients. Among 122 clients, 110 indicated very…
Q: In a sample of 500 adults, 340 had children. Construct a 99% confidence interval for the true…
A: Proportion is almost similar to the concept of probability. Proportion is a fraction of a population…
Q: An accounting firm has 1,190 clients. From a random sample of 119 clients, 108 indicated very high…
A: Given:
Q: Ten medical universities in Turkey were surveyed. The sample shows that from 250 resident students,…
A: Data for resident students:The sample size is The number of successes (being women) is Data for…
Q: A report statesb that 33% of home owners had a vegetable garden. how large a sample is needed to…
A: p=33%=0.33, E=5%=0.05
Q: In a sample of 290 adults, 191 had children. Construct a 95% confidence interval for the true…
A: 0.604 < p < 0.713 Explanation:Approach to solving the question:To construct a 95% confidence…
Q: A random sample from a population of voters consists of 152 democrats and 175 non- democrats. (a)…
A: The sample size n is 152+175=327. The number of democrats (x) is 152.
Q: In a sample of 270 adults, 173 had children. Construct a 99% confidence interval for the true…
A: The sample size n= 270. The sample proportion is 173/270. The absolute value of Zα/2 is 2.5758 using…
Q: In a sample of 240 adults, 180 had children. Construct a 95% confidence interval for the true…
A:
Q: 324 out of a simple random sample of 400 students hated their online math course. Find a 99%…
A: Given: Number of students that hated the online math course (y) = 324 Number of students (n) = 400…
Q: In a sample of 360 adults, 245 had children. Construct a 99% confidence interval for the true…
A:
Q: In a sample of 290 adults, 261 had children. Construct a 95% confidence interval for the true…
A: Given : Sample Size, n = 290 Number of successes, x = 261
Q: In a sample of 350 adults, 277 had children. Construct a 99% confidence interval for the true…
A: Sample size, Number of adults having children, Confidence level is .
Q: In a sample of 530 adults, 445 had children. Construct a 99% confidence interval for the true…
A: We have given that, Favorable cases (X) = 445 and sample size (n)=530 Then, We will find the 99%…
Q: In another pre-election poll, a pollster finds that 442 out of 650 people polled favor Smith.…
A: Given: Total no of people polled , n=650 No. of people favor Smith, x=442 Confidence level=95%
Q: A researcher wanted to check whether the percentage of Canadian adults with obesity has decreased in…
A:
Q: In a sample of 300 adults, 267 had children. Construct a 90% confidence interval for the true…
A: To construct a 90% confidence interval for the population proportion, we'll use the formula for the…
Q: Use the data from this survey to construct a 91% confidence interval estimate of the proportion of…
A:
Q: When 328 college students are randomly selected and surveyed, it is found that 122 own a car. Find…
A: We want to find 95% CI for population proportion (p)
Q: 5. Asample of 350 engineering applicant included 140 women. Find the 95% confidence interval of the…
A: Given: Sample size, n=350 No. of women, x=140
Q: In a sample of 400 adults, 220 had children. Construct a 90% confidence interval for population…
A:
Q: 7. A recent survey of 500 women indicated that 29% of women over age 55 in the study were widows.…
A: Givensample size(n)=500p^=29%=0.29confidence level=90%
Q: 6. The hospital would also like to estimate the Blue Shield as their insurance provider. A ra used…
A:
Q: The American Heart Association is about to conduct in anti smoking campaign and wants to know the…
A: Solution: Out of a sample of 1421 Americans over 35, 412 smoke. Sample size: n = 1421 The number of…
Q: Suppose we wish to construct a 95% confidence interval for the population proportion of fans in…
A: Given that, Total fans = 200 130 answered yes 95% confidence level
Q: You are trying to figure out if there is interest in a new restaurant opening in town. You speak to…
A: Solution: 26. Let X be the number of people that would visit the new restaurant and n be the sample…
Q: 2.) A survey of 200 college students revealed that 160 of them eat dessert daily. Construct a 95%…
A:
Q: A group of 20 students were given an achievement test under two conditions: (i) when tense (ii) when…
A: Given a group of 20 students were given an achievement test under two conditions: (i) when tense(ii)…
Q: In a sample of 470 adults, 306 had children. Construct a 99% confidence interval for the true…
A:
Q: In a sample of 460 adults, 281 had children. Construct a 90% confidence interval for the true…
A:
Q: . In a random sample of 250 television viewers in a large city, 190 had seen a certain controversial…
A: 7. Given that the sample size is 250. Among 250 television viewers, 190 had seen a certain…
Q: 6. In a survey of 2563 adults from France, 1666 said that they believed that the activities of…
A: Solution:Here, p be the true population proportion of adults in France who thinks humans are…
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- In a sample of 570 adults, 485 had children. Construct a 95% confidence interval for the true population proportion of adults with children.Give your answers as decimals, to three places < p <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.
- 5. In a pre-election poll, a pollster finds that 552 out of 1025 people polled favor Smith. Construct the 95% confidence interval for the proportion of people in the whole population who favor Smith.You recently sent out a survey to determine if the percentage of adults who use social media has changed from 56%, which was the percentage of adults who used social media five years ago. Of the 2335 people who responded to survey, 1725 stated that they currently use social media.a) Use the data from this survey to construct a 93% confidence interval estimate of the proportion of adults who use social media. Record the result below in the form of (#,#)(#,#). Round your final answer to four decimal places.In November 2018, 50,000 people living in the United States were asked whether they had a computer at home. Suppose 48,121 answered yes, they had a computer at home. There were 326,766,748 people living in the United States at the time of the survey. The 99% confidence interval for the proportion of people living in the United States who have a computer at home is 0.9602 to 0.9646. Determine an interval with 99% confidence for the number (how many) of people living in the United States who have a computer at home. O 310,787,854 to 318,172,783 O 313,761,431 to 315,199,205 O 313,957,491 to 315,035,822 O 314,022,845 to 314,937,792
- In a sample of 580 adults, 348 had children. Construct a 95% confidence interval for the true population proportion of adults with children. Give your answers as decimals, to three placesYou recently sent out a survey to determine if the percentage of adults who use social media has changed from 62%, which was the percentage of adults who used social media five years ago. Of the 2283 people who responded to survey, 1635 stated that they currently use social media.a) Use the data from this survey to construct a 94% confidence interval estimate of the proportion of adults who use social media. Record the result below in the form of (#,#)(#,#). Round your final answer to four decimal places.10. Suppose we sample 300 UNI students and find the 27 regularly wear a watch. Find a 98% confidence interval for p, the percentage of all UNI students who regularly wear a watch.Out of 200 people sampled, 76 had kids. Based on this, construct a 90% confidence interval for the truepopulation proportion of people with kids.Give your answers as decimals, to three placesIn 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 RectanIn a sample of 360 adults, 266 had children. Construct a 95% confidence interval for the true population proportion of adults with children. Give your answers as decimals, to three placesSEE MORE QUESTIONSRecommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons Inc
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