In a random sample of 230 men , 18 or older , 142 were married. Construct and explain a 90% confidence interval estimate of a true population proportion of married men 18 or older.

Q: In a survey of 2279 ​adults, 729 say they believe in UFOs. Construct a 90% confidence…

A:

Q: In your work for a real estate company, you find that in a sample of 1762 home buyers, 722 found…

A: Given Data: x=722 n=1762   p^=7221762=0.4098

Q: A Pew Research Center report on gamers and gaming estimated that 49% of U.S. adults play video games…

A:

Q: The confidence interval was (41%, 45%) What was the margin of error? +/- % What was the summary…

A:

Q: In a survey of 2344 ​adults, 715 say they believe in

A: Given that there are 2344 adults and 715 say they believe in UFOs. Consider the sample proportion,…

Q: In a survey of 1,200 social media users, 56% said it is not okay to be friends with your boss.…

A:

Q: A university dean is interested in determining the proportion of students who receive some financial…

A: The point estimate=p^=118200=0.59n=sample size= 200z=critical value of 95% confidence…

Q: In a recent survey of 200 households, 172 had central air conditioning.  Find the 85% confidence…

A: Given that In a recent survey of 200 households, 172 had central air conditioning That is…

Q: Of 380 randomly selected medical students, 21 said that they planned to work in a rural community.…

A: Sample proportion=p^=21380z=critical value for 95% confidence interval=1.96n=sample size=380

Q: In a survey of 1005 adults 526 stated construct a 95% confidence interval

A:

Q: Construct the confidence interval for the population mean.

A: Answer:

Q: Find the margin of error for the given confidence level.

A: Margin of error :  In a confidence interval, the range of values above and below the sample…

Q: Of 127 randomly selected adults, 31 were found to have high blood pressure. Construct a 95%…

A: Given that, Of 127 randomly selected adults, 31 were found to have high blood pressure.  X = 31 N =…

Q: Construct a 95% confidence interval of the average.

A: Given : Y = 0.00039508 S = 0.0084646 n = 1,259

Q: A Gallup Youth Survey in 2001 found that of 501 randomly selected teenagers in the U.S., 271 of them…

A: Given Total number of teenagers selected = 501Number of teenagers said who get along with their…

Q: In a recent poll, 560 people were asked if they liked dogs, and 22% said they did. Find the margin…

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

Q: Based on a sample of approximately 600 nonsmokers and 200 smokers, estimate the true difference…

A:

Q: In a survey of 50,00 people, 80% indicated that they listened to their favorite radio station…

A: Hern= 5000 P^ = 0.80 We have to construct 95% ci for p

Q: A university dean is interested in determining the proportion of students who receive some sort of…

A:

Q: Of 380 randomly selected medical students, 21 said that they planned to work in a rural community…

A: Given that, Out Of 380 randomly selected medical students, 21 said that they planned to work in a…

Q: Of 380 randomly selected medical​ students, 21 said that they planned to work in a rural community.…

A: Given that 380 randomly selected medical​ students, 21 said that they planned to work in a rural…

Q: Find a 95% confidence interval for the proportion of renters.

A: It is given that  Favourable cases, X = 360 Sample size, n = 600 Confidence interval = 95%

Q: In a survey of 4061 ​adults, 738 say they have seen a ghost. Construct a​ 99% confidence interval…

A: In this question, it has been given that 738 people out of 4061 people have seen the ghost, we have…

Q: In a recent poll, 240 people were asked if they liked dogs, and 32% said they did. Find the margin…

A: Find z-critical value:…

Q: 5-Seventy percent of 50 randomly selected students from college of science and engineering were…

A:

Q: Of 91 adults selected randomly from one town, 60 have health insurance. Find a 90% confidence…

A: From the provided information, Sample size (n) = 91 Out of which 60 have health insurance. Sample…

Q: In a recent poll, 260 people were asked if they liked dogs, and 67% said they did. Find the margin…

A: Given: p = 67%=0.67 n= 260 The value of z at a 90% confidence level is 1.645, which can be found by…

Q: If a researcher uses 90% confidence level, what is the level of significance?

A: We have given that,  Confidence level  (C.I) = 90% = 0.90 Then,  We will find the level of…

Q: Construct a 95% confidence interval of the population proportion using the given information.

A: The sample proportion is, p^=xn=240300=0.8 From the table of critical values, the critical value for…

Q: In a survey of 2298 ​adults, 709 say they believe in UFOs. Construct a 95% confidence interval for…

A:

Q: In a survey of 1007 parents of children ages 8 to 14, 75% say they are willing to get a second job…

A: Given information- A proportion of parents say they are willing to get a second job or part-time…

Q: In a random sample of 145 students, we find that 53% have attended a baseball game at some point in…

A: Given,sample size (n)=145sample proportion(p^)=0.53α=1-0.95=0.05α2 =0.025z0.025 =1.96

Q: Of 254 randomly selected medical students, 25 said that they planned to work in a rural community.…

A: It is given that, the number of students said that they planned to work in natural community   (x)…

Q: What is z* for a 92% confidence level?

A: To find value of Z* for a 92% confidence level ??

Q: If 64% of a sample of 550 people leaving a shopping mall claims to have spent over $25 create and…

A: Given,   Sample size = 550 Sample proportion = 0.64   Critical value: Using z-table, the z-critical…

Q: Suppose we would like to estimate the proportion of all NBA fans who thought Aaron Gordon won the…

A:

Q: Construct a 95% confidence interval for the proportion of college students who say they own their…

A: Given : n=410 , X=155  , α=0.05 We want to find the 95% confidence interval for the proportion of…

Q: In a random sample of 250 students who enrolled in Science, 150 took notes during classes. Find an…

A: Given that n=250,x =150,alpha =0.05 Find an approximate 95% confidence interval for the population…

Q: A Gallup poll of 1487 adults showed that 43% of the respondents have Facebook pages.  Construct a…

A: Confidence Interval: In statistics, we can say a range of values of the specified population…

Q: identify the population proportion, the value of margin of error E, and construct the confidence…

A:

Q: In a random sample of 230 men, 18 or older, 142 are married. Construct and explain a 98% confidence…

A: Given Data Sample Size, n = 230 Number of successes, x = 142.0 Significance level,…

Q: Use the given degree of confidence and sample data to construct a confidence interval for the…

A: Given : Sample Size, n = 316 Number of successes, x = 112 confidence level,c=0.99…

Q: Construct a 99% confidence interval for the population proportion

A:

Q: A college administrator is interested in determining the proportion of students who receive some…

A:

Q: Car and Driver Magazine tested a random sample of 9 hybrid cars and found their MPG to be 42, 51,…

A:

Q: In a recent poll, 410 people were asked if they liked dogs, and 44% said they did. Find the margin…

A: n = 410 p^=0.44 Confidence interval = 95%

Q: When 258 college students are randomly selected and surveyed, it is found that 106 own a car. Find a…

A: The provided information is: It is given that 106 college students (s) own a car among 258 college…

Q: Aston’s Career Service wants to find out what proportion of Aston’s students work in a part-time…

A: From the given information, Population proportion = p = 60% = 0.60 Margin of error = E = 4% = 0.04…

Q: Of 380 randomly selected medical students, 21 said that they planned to work in a rural community.…

A: Here we need to construct a 95% confidence interval for the population proportion. Here the sample…

Q: In a survey of 2273 ​adults, 728 say they believe in UFOs.   Construct a 99% confidence interval for…

A: Given Information: Sample size (n) = 2273 Out of which X = 728 adults say they believe in UFOs. To…