A drug company has developed a new over the counter medicine intended to provide relief from headache pain before marketing the drug company would like to get an estimate of the population proportion of users of this drug whoa actually experience pain relief. They decide to select a random sample of 120 patients with headache pain and determine the proportion that experience pain relief. If 90 of the 120 patients experience pain relief determine a 99% confidence interval for the popu

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Q: You are interested in finding a 95% confidence interval for the mean number of visits for physical…

A: Given,mean(x¯)=13.6standard deviation (s)=9.1918sample size(n)=10degrees of…

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Q: f n=450 and X = 360, construct a 99% confidence interval.

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Q: In a sample of 460 adults, 262 had children. Construct a 95% confidence interval for the true…

A: Provided information: Total number of adults = 460 and 262 had children 95% confidence level

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Q: Out of 400 people sampled, 48 had kids. Based on this, construct a 95% confidence interval for the…

A: According to the given information, we have Total number of people sampled = 400 and 48 had kids.…

Q: A survey of 700 non-fatal accidents showed that 167 involved uninsured drivers. Construct a 99%…

A: Given data,n=700x=167Point estimate isp=xn=167700=0.2386Construct a 99% confidence interval?

Q: Out of 500 people sampled, 395 had kids. Based on this, construct a 90% confidence interval for the…

A: It is given that 395 out of 500 people sampled had kids. The sample proportion is 395/500 = 0.79.…

Q: A study found that for a sample of cars traveling on Interstate 70, the average speed was 70 mph…

A: We have to find margin of error.

Q: In a survey of 2000 adults in a recent​ year, 708 made a New​ Year's resolution to eat healthier.…

A: Confidence interval for population proportion is given by, p^ +- z* √(pq/n) At 90% z = 1.645 At 95%…

Q: A poll of 1060 adults in a country was conducted by telephone. Of those surveyed, 169 had…

A: The sample size is 1060 and number of success is 169.

Q: A medical director needs to know the proportion of patients who have a single bed in their room. A…

A: It is given that sample size (n) is 200 and p-hat value is 0.59.

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Q: Two years ago a public health researcher found that 27% of American teens regularly vaped a…

A: Answer:- Two years ago a public health researcher found that 27% of American teens regularly vaped…

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: A sample of 500 nursing applicants included 80 men.  Determine the 95% confidence interval for the…

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Q: In a study of 120 randomly selected families, 50 said they streamed their television content. Find…

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Q: In a survey of 2004 adults in a recent​ year, 717 made a New​ Year's resolution to eat healthier.…

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Q: A department chair wanted to know whose students performed better on a common pass/fail final exam:…

A: p1 :  proportion of students passed in smith's  class p2 :  proportion of students passed in Jones's…

Q: One personality test available on the World Wide Web has a subsection designed to assess the…

A: We have given that Mean(µ) = 140 Sample mean (x̅) = 134 Standard deviations (s) = 28

Q: A recent study of 85 workers found that 61 people rode the bus to work each day. Find the 95%…

A: Given that  A recent study of 85 workers found that 61 people rode the bus to work each day. Find…

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: 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

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Q: A research group asked 4488 Americans how they fall asleep, and 536 of them fall asleep on their…

A: Obtain the 90% confidence interval for the percentage of all Americans who fall asleep on their…

Q: In a random sample of 500 adults, 410 described themselves as vegetarians. Construct a 95%…

A: We have been provided with the following information about the adults who describe themselves as…

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: A report statesb that 44% of home owners had a vegetable garden. how large a sample is needed to…

A: Answer: Using the given data, A report stated that 44% of home owners had a vegetable garden, P =…

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A: In the given problem, we have to compare the outcomes of people that use the app to those that…

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Q: A recent survey suggests that 52% of Americans say dogs are their favorite animals. If 84 people are…

A: As per honor code, we will only answer first question: 1)  p=0.52 n=84, p^=0.42 Use the formula of…

Q: Solve this attachment.

A: Given: Margin of error (E)= 30 The 99% confidence interval is (0.20, 0.21) The value of p is…

Q: You are testing the claim that the proportion of men who own cats is smaller than the proportion of…

A: As per our company guidelines we are supposed to answer only first question. Kindly repost other…

Q: You want to get a 95% confidence interval for the difference between the percentage of females with…

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Q: To assess the proportion of the readers of a magazine being professional, a random sample of 1000…

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Q: A recent survey of 120 individuals revealed that 50 thought they were worse off financially than a…

A: Given,n=120x=50α=1-0.90=0.1α2=0.05Z0.05=1.645

Q: Last year, 535 out of 630 randomly selected U.S. firms tested their employees for drug use. Give a…

A: Given Data Sample Size, n = 630 Number of successes, x = 535

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

A: Solution

Q: What is the confidence interval? Select the correct choice below and, if necessary, fill in the…

A: Mean = 56.64 Standard deviation = 14.598 n=100

Q: of their personal information online if they could. Make a subjective estimate to decide

A: Given : Sample proportion = p̂ = 0.508 Sample size = n = 560 Claim : Most adults would erase all of…

Q: A researcher wanted to compare the percentages of men and women who play the lottery.  They randomly…

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