A researcher wants to estimate the proportion of depressed individuals taking a new anti-depressant drug who find relief. A random sample of 250 individuals who had been taking the drug is questioned; 202 of them found relief from depression. Based upon this, compute a 90% confidence interval for the proportion of all depressed individuals taking the drug who find relief. Then find the lower limit and upper limit of the 90% confidence interval. Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places. (If necessary, consult a list of formulas.) Lower limit: ? Upper limit: ?

Q: during production of detergent. In people with no lung dysfunction, a typical air flow rate is 0.80.…

A: As the population standard devoation is known, we will use z distribution. The value of z at the…

Q: test. It was found that 80 of them scored higher than 100. Based on this, compute a 95% confidence…

A: According to the given information in this question We need to construct 95% confidence interval

Q: Out of a sample of 600 adults aged 18 to 30, 162 still lived with their parents. Based on this,…

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Q: In 2016 the CDC wanted to estimate the proportion of adults who are smokers in the U.S. The…

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Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 400…

A: The objective of this question is to construct a 95% confidence interval for the proportion of adult…

Q: Many high school students take the AP tests in different subject areas. In one year, of the 144,140…

A: given data  n1 = 144140   x1 = 84545p^1 = x1n1 = 84545144140 = 0.5865 n2 = 216298    x2 = 126501p^2…

Q: A poll of n = 500 randomly selected elementary school students resulted in 110 of them stating…

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Q: A random sample of 230 people living in Seattle found that 83 of them had a pet. A random sample of…

A: Solution: Given information: n1= 230 Sample size of people iving in Seattle n2= 172  Sample size of…

Q: Students at Green River plans a study to stats Test scores. Previous research has shown the…

A: We have given that  Margin of error =E =0.5 Population Standard deviation = 1 99%confidence…

Q: n a survey of 2000 American adults age 60 and over, 1320 received a flu shot. Construct a 90%…

A: Given , n =2000 X=1320 P=x/mn =1320/2000  =0.66 E-value at 90% coincidence is 7c=1.645

Q: If n=12, (x-bar)=33, and s=16, construct a confidence interval at a 80% confidence level. Assume the…

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Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 400…

A: Solution: It is given here:

Q: The data for an independent –measures research study produced a sample mean difference of 6 points…

A: Following information is given:

Q: A study investigated ways to prevent staph infections in surgery patients. The researchers examined…

A: From the provided information, Sample size (n) = 100 From which 88 patients tested positive for…

Q: According to a report on consumer fraud and identity theft, Indiana had 1671 complaints of identity…

A:

Q: Explain why you should get a much better prediction for μ by using x from a sample size n rather…

A: The confidence interval on mean is given as x-Zα2sdn≤μ≤x-Zα2sdn where  μ=population meanx=sample…

Q: Scientists tested each in a sample of 6 alleles of antigen-produced protein for level of peptide.…

A: The sample size n=6 The sample mean x¯=1.54 The sample standard deviation s=0.13 The confidence…

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

A: Given X=25 n=500 95% confidence interval

Q: one with a 95% confidence level and the other with a 99.7% confidence level. Which interval is…

A: Confidence interval is probability, which refers that population parameter will fall within set of…

Q: A study investigated ways to prevent staph infections in surgery patients. The researchers examined…

A: Given that, x=86,n=100 The sample proportion is, p^=xn  =86100  =0.86

Q: In a random sample of 500 Americans, 25% indicated that they have a cat for a pet. Estimate with 97%…

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Q: In a random sample of 500 Americans, 50% indicated that they have a cat for a pet. Estimate with 95%…

A: Given information

Q: Free of CVD       Prevalent CVD    Total Men          1548                 244                 1792…

A: Given The tota number of men with CVD =244The tota number of women with CVD =135The tota number of…

Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 100…

A: According to the given information, we have Out of 100 adult residents sampled, 88 had kids. 99%…

Q: If n = 15, the sample mean = 150, SD = 10, what is the confidence interval (using 99% confidence…

A: Given data,n=15x=150σ=10z-value at 99% confidence is Zc=2.576

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

A: Solution

Q: What proportion of Law School applicants are accepted into Law School? In a survey of a random…

A: According to the given information in this question  We can say that the We are 95% confident the…

Q: Out of 300 people sampled, 24 had kids. Based on this, construct a 95% confidence interval for the…

A:

Q: A smart phone manufacturer is interested in constructing a 99% confidence interval for the…

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Q: The police department wants to reduce the margin of error for their 95% confidence interval. If they…

A: Here Given  The police department wants to reduce the margin of error for their 95% confidence…

Q: A sample of 1100 observations taken from a population produced a sample proportion of 0.36 . Make a…

A: Given information Sample proportion = 0.36 Sample size (n) = 1100

Q: A study investigated ways to prevent staph infections in surgery patients. The researchers examined…

A: given data  n = 100 x = 86 80% ci for p.  p^ = xn = 86100 = 0.86

Q: A student found that in a random sample of 300 people in a town, 105 play sports. Based on this…

A: Given that, X = 105 N = 300 90% confidence level

Q: If n=25, z-46, and s=5, construct a confidence interval at a 95% confidence level. Assume the data…

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Q: In a sample of 480 adults, 384 had children. Construct a 99% confidence interval for the true…

A: The random variable adults with children follow normal distribution. We have to construct 99%…

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: Elvin wants to know how the school views him as their leader. He asked his friend, Marcus, to…

A:

Q: You are comparing men and women on hours spent watching TV. You pick a sample of 12 men and 14 women…

A:

Q: According to a poll, 86% of California adults (434 out of 506 surveyed) feel that education is one…

A:

Q: Out of 200 people sampled, 50 had kids. Based on this, construct a 90% confidence interval for the…

A: Given that, Total number of people sampled = 200 Number of people had kids = 50 Confidence level =…

Q: Now you are interested in the percent of dogs that were neutered in your sample. For 40 dogs, 30 of…

A:

Q: If n=17, (x-bar)=48, and s=2, construct a confidence interval at a 80% confidence level. Assume the…

A: We have to construct confidence interval for mean of single normal population.

Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.