Public Libraries and ChildrenA survey¹ asked a random sample of ŉ = 2752 US adults whether they had visited a public library in the last months. We see in thatdata that more females than males visit the library, but there is likely a confounding variable. The table below shows the survey resultsbroken down by whether or not the participant is a parent of a minor. Find a 99 % confidence interval for pc - PN, the difference inproportion visiting the libary between those with children in the house and those without.Round your answers to three decimal places.The 99% confidence interval is iChildren in houseNo children in houseTotalYesto iNo Total421 411832810 1110 1920Table 1: Have you visited a public library in the last year?1231 1521 2752

Given: Yes No Total Children in house 421 411 832 No children in house 810 1110 1920…

Answered: A survey¹ asked a random sample of n =…

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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Assume a nationwide random survey of 800 adults asked about attitudes toward “alternative medicine” such as acupuncture, massage therapy, and herbal therapy. Among the respondents, 448 said they would use alternative medicine if traditional medicine was not producing the results they wanted. Give a 99% confidence interval for the proportion of all adults who would use alternative medicine.
A poll reported that only 645 out of a total of 2029 adults in a particular region said they had a "great deal of confidence" or "quite a lot of confidence" in the public school system. This was down 5 percentage points from the previous year. Assume the conditions for using the CLT are met. Find an 80% confidence interval and interpret it. The 80% confidence interval for the proportion that express a great deal of confidence or quite a lot of confidence in the public schools is ____.
A survey conducted by the U.S. department of Labor found the 48 out of 500 heads of households were unemployed. Construct a 90% confidence interval estimate of the proportion of unemployed heads of households in the population.
According to the website www.collegedrinkingprevention.gov, "About 25 percent of college students report academic consequences of their drinking including missing class, falling behind, doing poorly on exams or papers, and receiving lower grades overall." A statistics student is curious about drinking habits of students at his college. He wants to estimate the mean number of alcoholic drinks consumed each week by students at his college. He plans to use a 95% confidence interval. He surveys a random sample of 66 students. The sample mean is 3-75 alcoholic drinks per week. The sample standard deviation is 3.68 drinks. Construct the 95% confidence interval to estimate the average number of alcoholic drinks consumed each week by students at this college. Your answer should be rounded to 2 decimal places.
An apartment hunter is shopping for either a one-bedroom or two-bedroom apartment in varying apartment buildings. She wants to estimate the additional amount she would have to pay, on average, for a two-bedroom apartment versus a one-bedroom apartment. The table shows the prices, in dollars, for one- and two-bedroom apartments from the nine buildings she visited. A 95% confidence interval for the true mean difference between two-bedroom and one-bedroom apartments for apartment buildings like the ones in the sample was (255.93, 508.51). A friend of the apartment hunter tells her to expect to pay about $500 more per month for a two-bedroom apartment than for a one-bedroom apartment. Does the confidence interval provide evidence for the friend’s claim? A. No, the confidence interval does not contain 0, so the friend’s claim is not plausible. B. No, the confidence interval contains many more numbers below 500 than above 500, so it is unlikely that the friend’s assertion is correct. C.…
Among 1815 randomly selected high school students surveyed by the Centers for Diseases Control, 300 said they were current smokers. d. Interpret the 95% confidence interval in this context.
Car and Driver Magazine tested a random sample of 9 hybrid cars and found their MPG to be 42, 51, 39, 44, 52, 48, 51, 44, and 47. Construct a 95% confidence interval for the population variance.
A college alumni Organization sent a survey to all recent graduates to ask their annual income. 20% of the alumni responded, and their mean annual income was $40,000 with a standard deviation of $10,000. Explain why these data should not be used to construct a confidence interval for the mean annual income of all recent graduates.
According to a poll, 81% of California adults (410 out of 506 surveyed) feel that education is one of the top issues facing California. We wish to construct a 90% confidence interval for the true proportion of California adults who feel that education is one of the top issues facing California.What is a point estimate for the true population proportion?
In a random sample of 2500 men being treated for obesity, the mean total cholesterol level is 175 with a standard deviation of 25. Construct a 95% confidence interval around the mean and interpret it. Use the context of the question please.
According to the website www.collegedrinkingprevention.gov, "About 25 percent of college students report academic consequences of their drinking including missing class, falling behind, doing poorly on exams or papers, and receiving lower grades overall." A statistics student is curious about drinking habits of students at his college. He wants to estimate the mean number of alcoholic drinks consumed each week by students at his college. He plans to use a 90% confidence interval. He surveys a random sample of 63 students. The sample mean is 3.7 alcoholic drinks per week. The sample standard deviation is 3.2 drinks. Construct the 90% confidence interval to estimate the average number of alcoholic drinks consumed each week by students at this college. 1. The t value is - 2. The standard error has a value of -- 3. The margin of error has a value of 4. Construct the 90% confidence interval to estimate the average number of alcoholic drinks consumed each week by students at this college.
Suppose we have only ten scores representing the ages (in years) of randomly selected U.S. commercial aircraft. The average age of the ten aircraft is 14.25 years and the sample standard deviation is 9.349 years. Construct a 95% confidence interval for the mean age of all U.S. commercial aircraft. I would appreciate to see what formulas you use and the work through because I messed up.

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