A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than examine the records for all students, the dean randomly selects 200 students and finds that 118 of them are receiving financial aid. Use a 99% confidence interval to estimate the true proportion of students on financial aid. Express the answer in the form p± E and round to the nearest thousandth. OA. 0.59 ± 0.623 OB. 0.59 ± 0.090 OC. 0.59 ± 0.007 OD. 0.59 ± 0.003

Q: e: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the…

A: Sample size n =47 Sample mean=109 Population standard deviation =32

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Q: Find the indicated margin of error. Out of 260 trials, 150 turned out positive. Find the margin of…

A: It is given that the sample size (n) is 260 and the number of positive trials (x) is 150.

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A:

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Q: d. FIND t a n = 10 for the 90% confidence interval for the mean. %3D

A: d. Given: n=10 α=0.1 The degrees of freedom is obtained as below: df=n-1=10-1=9 Critical value: The…

Q: Assume that you want to construct a 95% confidence interval estimate of a population mean. Find an…

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Q: If n = 108, a = 29 and o = 2.6, construct a 91% confidence interval. Give your answers to three…

A: Since we know population standard deviation, use z-distribution to find z-critical value.…

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Q: Find a 95% confidence interval for a mean, µ, where n = 324, x̄ = 2.7, s = .91, and se = .06.

A: Given: n = 324, x̄ = 2.7, s = .91, se = .06.

Q: You are given the sample mean and the population standard deviation. Use this information to…

A: Solution-: Given: x¯=83.97,n=76,σ=13.89  Use this information to construct the​ 90% and​ 95%…

Q: 95% confidence interval for the true percent of workers in this city who have full health insurance…

A: The question is about confidence interval.Given :Randomly selected no. of workers ( n ) = 1700No. of…

Q: Find the margin of error for the 95% confidence interval used to estimate the population proportion.…

A: Confidence level = 95% Sample size (n) = 2440 Sample proportion (p̂) = 0.70

Q: A statistics practitioner took a random sample of 56 observations from a population whose standard…

A: We have given that,  Sample size (n) = 56, population standard deviation (σ) = 35 and sample mean…

Q: Determine the t score used for the confidence interval. Report to 3 decimal places. n 3 6 and 95%…

A: Solution: 23. From the given information, n=6 and the confidence level is 0.95.

Q: Use the given confidence interval to find the margin of error and the sample proportion.…

A: Margin of error (E) for proportion is defined as below                                E = Width2…

Q: Determine the margin of error and the confidence interval for each based the data given below. Show…

A: We have given that.  1)     n = 58; x̄ = 75; σ = 10, 95% confidence interval.  2)      n = 23; x̄ =…

Q: Assume that population mean is to be estimated from the sample described. Use the sample results to…

A: Given95% confidence interval N=361 x = 59.1 seconds. z= 4.9 SecondsFind the margin of error?

Q: Construct the confidence interval for the population mean y. C= 0.90, x=16.6, o= 10.0, and n= 80

A: Here, C=0.90, x=16.6, σ=10.0, and n=80.

Q: Use the sample data and confidence level given below to complete parts (a) through​ (d).   A…

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

Q: The estimated prevalence is 11%, and the desired confidence level is 91% with a margin of error of…

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Q: Find ??/2 for the 93% confidence interval. Select one: A. 1.83 B. 1.81 C. 1.80 D. 1.82

A: Solution: Given information: Confidence level = 93% = 0.93 α=1-confidence level α=1-0.93= 0.07…

Q: You sample 40 dogs and find a sample mean weight of 60 lbs with a sample standard deviation of 5.…

A: From the provided information, Sample size (n) = 40 Sample mean (x̅) = 60 lbs Sample standard…

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A: Suppose, p is the population proportion of all adults, who say that they believe in UFOs.

Q: If n=27, x=48, and s-6, construct a confidence interval at a 80% confidence level. Assume the data…

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Q: You measure 37 textbooks' weights, and find they have a mean weight of 59 ounces. Assume the…

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Q: A 95% confidence interval for the population proportion is (☐☐ ). (Round to three decimal places as…

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Q: A statistics practitioner took a random sample of 43 Note: For each confidence interval, enter your…

A: n = 43 standard deviation = 27 mean = 104 95% confidence interval for mean is given by  x±zα2*σn 104…

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Q: FIND t a n = 15 for the 98% confidence interval for the mean.

A: Provided information that, n=15 Alpha (α)=1-0.98=0.02 α/2=0.01

Q: If n = 41, ã=30, and s = 6, construct a confidence interval at a 80% confidence level.

A: It is given that Sample size n = 41 Sample mean = 30 Sample SD s = 6 The critical value of t at df =…

Q: You measure 37 dogs' weights, and find they have a mean weight of 35 ounces. Assume the population…

A:

Q: If n=14, (x-bar)=46, and s=18, construct a confidence interval at a 80% confidence level. Assume the…

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A: To construct 90% and 95% confidence intervals and answer the sub-parts.

Q: A survey asked, "How many tattoos do you currently have on your body?" Of the 1229 males surveyed,…

A: From the provided information,

Q: Rosetta wants to estimate the percentage of people who rent their home. She surveys 250 individuals…

A:

Q: Use a t-distribution to find a confidence interval for the difference in means μ = μ₁-₂ using the…

A: From the provided information, Sample size (n) = 8

Q: A poll of 1021 U.S. adults found that 46% regularly used Facebook as a news source. Find the…

A: Given, n = 1021 p^=0.46 q^ = 1 - p^ = 1 - 0.46 = 0.54 α=0.1

Q: In a survey of 2318 adults, 747 say they believe in UFOs. Construct a 99% confidence interval for…

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A: Given:- simple random sample of 1200 young people  so, n= 1200 86% had earned a high school diploma…

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A:

Q: A statistics practitioner took a random sample of 59 observations from a population whose standard…

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Q: Express the confidence interval (13.8 %, 20 %) in the form of p + E.

A: Here Given  Confidence Interval (13.8%, 20%)

Q: A statistics practitioner took a random sample of 43 observations from a population whose standard…

A: We have given that,  sample mean (x̄) = 109, standard deviation (σ) = 30 and sample size (n) = 43

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A: The formula of the confidence interval is,