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 152 students and finds that 101 of them are receiving financial aid. Use a 95% confidence interval to estimate the true proportion of students who receive financial aid.  Write the LOWER BOUND for the confidence interval.  Round to

Q: Wildlife biologists inspect 200 deer taken by hunters and find 45 of them carrying Lyme disease…

A:

Q: A recent study of high school students shows the percentage of females and males who have at LEAST…

A: The (1-α)% confidence interval for the difference in population proportions p1-p2 is given by;…

Q: Use the given confidence interval to find the margin of error and the sample mean. .734, .760

A: From the given information, the confidence interval is (0.734, 0.760). It can be expressed in,

Q: In a recent poll, 300 people were asked if they liked dogs, and 54% said they did. Find the Margin…

A: Obtain the margin of error for the poll at the 95% confidence level. The margin of error for the…

Q: Explain how the Choosing of a Level of Confidence is done.

A: The level of confidence represents how often the true percentage of the population lies within the…

Q: 3. A survey of 2,950 Filipinos showed that 2,655 trust the Supreme Court decisions on legal issues.…

A: Let X be the number of trust the supreme court decision on legal issues  n be the sample size  Given…

Q: those polled answered the economy. Construct a 93% confidence interval for the proportion of voters…

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Q: A researcher is interested to find out the proportion of registered nurses in Manitoba who are more…

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Q: Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use…

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Q: A new fertilizer was applied to the soil of 199 bean plants. 48% showed increased growth. Find the…

A: Margin of error for proportion is given by, MOE= z*√pq/n Where P Is proportion and Q= 1-P n is…

Q: A random sample of 900 car owners in a particular city found 252 car owners who received a speeding…

A: Given,n=900x=252p^=xnp^=252900=0.281-p^=1-0.28=0.72α=1-0.95=0.05α2=0.025Z0.025 =1.96   (from…

Q: Compute a 98% confidence interval for the true percent of students who are against the new…

A: Given: n=521 Confidence level=98%

Q: Find the condence level and a for a. a 90% confidence interval.b. a 99% confidence interval.

A: a.The confidence level and α for a 90% confidence interval is obtained below:The confidence level is…

Q: In a recent poll, 130 people were asked if they liked dogs, and 24% said they did. Find the margin…

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Q: Emilio wants to estimate the percentage of people who have a financial adviser. He surveys 350…

A: given data  n = 350 x = 76 we have to interprete 99% confidence intervl.

Q: A poll of 2929 U.S. adults found that 85% regularly used Facebook as a news source. Find the…

A:

Q: In a survey of 323 adults over 50, 60% said they were taking vitamin supplements. Find the margin of…

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Q: Form 90%, 95%, and 99% confidence intervals for the proportion of voters who favor the candidate.…

A: Given that Sample size n = 1000 Number of respondents favor the candidate, X = 214 The critical…

Q: A student polls his school to see if students in the school district are for or against the new…

A: From the provided information, Sample size (n) = 395 Out of which 210 are against the new…

Q: A poll of 2977 U.S. adults found that 28% regularly used Facebook as a news source. Find the margin…

A: Solution: Given information: n = 2977 Sample size  p^= 0.28  Sample proportion.

Q: In a recent poll, 260 people were asked if they liked dogs, and 80% said they did. Find the Margin…

A: Given:

Q: How

A: solution :  [ option B is correct 1068 ] given data Margin of Error E=0.03 Significance Level alpha…

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Q: Translate the statement into a confidence interval. Approximate the level of confidence. In a survey…

A: The sample size is 1072, 67% say they are willing to get a second or part-time job to pay for their…

Q: t poll, 240 people were asked if they liked dogs, and 35% said they did. Find the Margin of Error…

A: Given: n = 240 p^=35%=0.35 The value of z at 95% confidence level is 1.96, which can be obtained by…

Q: In a recent poll, 289 people were asked if they liked dogs, and 17% said they did. Find the margin…

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Q: Use the confidence interval to find the margin of error and the sample mean.   (0.676,0.800)

A: The margin of error is the most common term which we use in a confidence interval. It is the degree…

Q: A survey showed that 20.2% of the children that participated in the survey like flavor O If the…

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Q: Suppose that insurance companies did a survey. They randomly surveyed 450 drivers and found that 310…

A: Obtain 95% confidence interval for the population proportion that claim to always buckle up. The 95%…

Q: In a recent poll, 360 people were asked if they liked dogs, and 55% said they did. Find the Margin…

A: It is given that Sample size n = 360 Sample prpoprotion = 0.55 The critical value of Z at 99%…

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 recent poll, 390 people were asked if they liked dogs, and 33% said they did. Find the Margin…

A: Given p^=Sample proportion=0.33 n=sample size=390 Confidence interval = 99%

Q: Question Emilio wants to estimate the percentage of people who have a financial adviser. He surveys…

A: we have to find all values for given questions.....

Q: Translate the statement into a confidence interval. Approximate the level of confidence. In a survey…

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

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

A: Solution

Q: A survey was conducted that asked 1016 people how many books they had read in the past year. Results…

A: Given x-bar=13.7 S=16.2 n=1016

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

A: Given; x=561n=935The level of significance is;α=1-0.95=0.05 (a) The best point estimate of the…

Q: Use the confidence interval to find the margin of error and the sample mean. (1.77, 2.01) The margin…

A: Margin of error= ?

Q: Bert computes a 95% confidence interval for p and obtain the interval [0.600,0.700]. Bert boss says,…

A: Given - 95% confidence interval for p by Bert = [0.600,0.700]

Q: A poll of 2612 U.S. adults found that 15% regularly used Facebook as a news source. Find the margin…

A: The margin of error is defined as MOEγ=zγ×σn,γ is the confidence level,σ is the sample standard…

Q: A poll of 2368 U.S. adults found that 40% regularly used Facebook as a news source. Find the margin…

A: Solution: From the given information, n=2368, p-hat = 40%. The confidence level is 0.90.

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.