In a survey funded by the University of Washington School of Medicine, 585 of 1,500 adult Seattle residents said they did not believe they could come down with a sexually transmitted infection (STI). Construct a 99% confidence interval estimating the proportion of adult Seattle residents that do not believe they can contract an STI. Round the answers to three decimal places. Verify nˆp(1−ˆp)≥10. Round your answer to one decimal place. nˆp(1−ˆp)= What is the 99% confidence interval to estimate the population proportion? Round your answer to three decimal places.

Q: Self-Reported Health Status Excellent 30 Very Good 80 Good 100 Fair 60 Рor 30 TOTAL 300

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A: Given Data: p^=0.53n=1510   z-value corresponding to 95% Confidence level=1.96

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Q: What is meant by saying that a 1-a confidence interval is a. exact?b. approximately correct?

A: a.Confidence Interval:Confidence interval is the interval estimate of the population parameter. It…

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A: Given that,X =  1320N = 2000

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A: Given : sample mean, x̄ = 13000  Margin of error , E = 1000

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Q: find a 99% confidence interval for the proportion of us adults favoring taxes on soda and junk food.…

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A: Sample size, n = 950Probability = 17% = 0.17 90% confidence level

Q: We would like to know: "What percentage of college students drink alcohol every day?" In a random…

A: Given that, N=500 X=75

Q: In a survey of 3234 adults aged 57 through 85​ years, it was found that 88.9% of them used at least…

A:

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A: Given information Sample size n = 928 Mean x̅ = 10.4 Standard deviation s = 28.5 Since significance…

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A: Given that, Critical values: By using the standard normal table of Z, the critical value at 95%…

Q: The Pew Forum on Religion and Public Life reported on Dec. 9, 2009, that in a survey of 2003…

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

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Q: In a random sample of 500 people eating lunch at a cafeteria on various Fridays, it was found that…

A: Given that Sample size n= 500  Favorable cases x=160 Sample proportion p^=x/n

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Q: The Pew Forum on Religion and Public Life reported on Dec. 9, 2009, that in a survey of 2003…

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Q: B. In a survey, 27 people were asked how much they spent on their child's last birthday gift. The…

A:

Q: In a survey, 19 people were asked how much they spent on their child's last birthday gift. The…

A: From the provided information, Sample size (n) = 19 Sample mean (x̄) = $46 Sample standard deviation…

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A: Obtain the critical value: The level of confidence is 1- α = 0.95. From the standard normal…

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A: The 95% confidence interval for p is (0.439, 0.457).

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A: Given: CI=95% p^=11241875=0.599 n=1875

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Q: In a survey funded by the University of Washington School of Medicine, 288 of 1,200 adult Seattle…

A:

Q: If appropriate, construct a 99.9% confidence interval for the ounce milkshakes sold at fast-food…

A: Data  504,450,399,613,608,700,472,510,642,591,375,580

Q: In a recent poll, 450 people were asked if they liked dogs, and 54% said they did. Find the Margin…

A: Solution

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A:

Q: It was found that in a sample of 90 teenage boys, 70% of them have received speeding tickets. What…

A: Given data ,n=90P=70%=0.7The z-value at 90% confidence is1.645=z∝12=1.645

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Q: A random sample of 340 college students were asked if they believed that places could be haunted,…

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Q: In a recent poll, 430 people were asked if they liked dogs, and 94% said they did. Find the Margin…

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Q: A survey was conducted to determine the proportion of adults who approve of attempts to clone a…

A: Given,   Lower bound = 0.09 Upper bound = 0.13   The objective is to find the point estimate of the…

2-Proportion Test

In a 2-proportion test, two separate proportions are measured at the same time, and their similarity is evaluated. It is also known as a 2-proportion z-test, and a comparison of two different proportions is done in this method.