Suppose a random sample of 142 graduates admissions officers were asked what role scores on standardized assessments play. In this sample, 87 members answered << a very important role >>. Find the 90% confidence interval of the population proportion of those responsible for this opinion. Suppose we calculate a confidence interval for the proportion of the population that pays for mail-in renewal that is 0.34 to 0.46. What is the confidence level of that interval? NOTE: The found Z value is rounded to the second decimal place

Q: In a survey of 2316 ​adults, 701 say they believe in UFOs.   Construct a 90 % confidence interval…

A: The objective of this question is to construct a 90% confidence interval for the population…

Q: According to a poll, 81% of California adults (409 out of 506 surveyed) feel that education is one…

A: Point estimate=0.81

Q: recent survey showed that among 725 randomly selected subjects who completed 4 years of college, 139…

A: Given that  n = 725 , Smoke = 139 , Not smoke = 586

Q: In a survey of 2865 ​adults, 1413 say they have started paying bills online in the last year.…

A:

Q: In a random sample of 100 customers, the items that customers purchased were tracked. It was found…

A: Given,sample size(n)=100sample proportion(p^)=0.171-p^=1-0.17=0.83α=1-0.90=0.1α2=0.05Z0.05=1.645

Q: There is an ever increasing number of sources of getting the daily news – traditional newspapers,…

A: Given info: A 98% confidence interval to estimate the true proportion of adults who would read…

Q: Insurance companies are interested in knowing the population percent of drivers who always buckle up…

A: It is given that, x is 311 and n is 400.

Q: n a survey of 2558 ​adults, 1413 say they have started paying bills online in the last year.…

A: Solution: Given information: n= 2558 Sample size of adults  x = 1413 adults started paying bills…

Q: In a survey of 3254 ​adults, 1433 say they have started paying bills online in the last year.…

A: Given X=1433 n=3254

Q: In a survey of 2703 ​adults, 1438 say they have started paying bills online in the last year.…

A: We have given that, Number of favorable cases(X) = 1438,   Sample size(n) = 2703 Sample proportion…

Q: Out of 400 people sampled, 32 had kids. Based on this, construct a 95% confidence interval for the…

A: Solution: Let X be the number of people had kids and n be the sample number of people. From the…

Q: a survey of 3447 adults, 1477 say they have started paying bills online in the last year.…

A:

Q: Suppose that the a college wants to have an idea of the average SAT score of the incoming freshman…

A: The confidence level is 0.90, the standard deviation is 310 and margin of error E is 50.

Q: An accounting firm has 1200 clients. From a random sample of 120 clients, 110 indicated very high…

A: Given : An accounting firm has 1200 clients So,  Population size (N) = 1200 Sample size (n) = 120…

Q: A surprising number of young adults (ages 19 to 25) still live in their parents' homes. A random…

A: Sample taken for ages 19 to 25 2253 men and 2629 women random sample In which 986 men and 923 women…

Q: Suppose a random sample of 142 graduates admissions officers were asked what role scores on…

A: Proportion is almost similar to the concept of probability. Proportion is a fraction of population…

Q: The General Social Survey asked 1294 people whether they performed any volunteer work during the…

A:

Q: Out of 200 people sampled, 42 had kids. Based on this, construct a 95% confidence interval for the…

A: For 95% confidence intervalα=1-CI=1-0.95=0.05So, zα2=z0.052=z0.025=1.96

Q: Out of sample of 500 people, 25% exercise daily. What is the confidence interval for the population…

A: Given Information:    n=500  p=0.25 q=1-p=1-0.25=0.75α=0.05

Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 500…

A:

Q: In a survey of 2552 ​adults, 1423 say they have started paying bills online in the last year.…

A: Given that, In a survey of 2552 adults, 1423 say they have started paying bills online in the last…

Q: According the the U.S. Census bureau, the percentage of Santa Ana residents living in poverty is…

A: Given: p^1=0.177p^2=0.133n1=1200n2=1500

Q: Out of 300 people sampled, 171 had kids. Based on this, construct a 95% confidence interval for the…

A:

Q: business conducts a random sample of 89 of its customers and find that 44 of them think the company…

A:

Q: ed 1294 people whether they performed any volunteer work during the past year. A total of 517 people…

A: The sample proportion is computed as follows,

Q: A surprising number of young adults (ages 19 to 25) still live in their parents' homes. A random…

A:

Q: Suppose that a health expert has claimed that 28% of college students smoke cigarettes. To…

A: Given 95% confidence interval for the percentage of college students who smoke is 19% to 24%. Given…

Q: According to a poll, 77% of California adults (389 out of 506 surveyed) feel that education is one…

A: Given that the number of favorable cases Favorable Cases XX = 389389 Sample Size NN = 506506…

Q: Construct a​ 99% confidence interval for the population proportion. Interpret the results.

A:

Q: ted below are student evaluation ratings of courses, where a rating of 5 is for "excellent." The…

A: The given data is  4 2.9 4 4.8 3 4.2 3.8 4.4 4.2 4 4.3 3.9 3.5 4.1…

Q: 9. A survey of 280 homeless persons showed that 63 were veterans. Construct a 90% confidence…

A: In a survey of 280 homeless persons showed that 63 were veterans.

Q: An internet service provider sampled 540 customers and found that 55 of them experienced an…

A:

Q: A random survey of 36 airline passengers indicates that twelve have never flown before. Find a 95%…

A: We have given that,  Sample size (n) = 36 and favorable cases         (X) = 12 Then,  We Will find…

Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 600…

A: According to the given information, we have Total number of adult residents sampled = 600 186 had…

Q: We wish to estimate what percent of 18 to 25-year old's that own a Samsung phone. Out of 400…

A: Solution: Let X be the number of adults own a Samsung phone and n be the sample number of adults.…

Q: If a 90%, a 95% and a 99% confidence interval are foung for a population mean, describe how the 3…

A: Population mean CI for known population standard deviation is, CI=x¯±Zσnwhere,σ=population standard…

Q: Out of 100 people sampled, 54 had kids. Based on this, construct a 90 % confidence interval for the…

A:

Q: poll reported that only 645 out of a total of 2029 adults in a particular region said they had a​…

A: According to the given information in this question We need to construct 80% confidence interval

Average

In mathematics, everyone must have heard the term average in one place or other. Almost all types of calculation require taking out an average. In everyday life, the term is used for denoting a certain amount that can be said to be a typical quantity for multiple people or objects. In other words, if a group of people possesses a certain object, then the average gives us an idea regarding the quantity of the object possessed by each person. However, this is solely based on mathematical calculation. It is important to note that most people will not strictly have the average value in a group.