Let's say you survey a random sample of 70 people. You ask the people in your survey if they prefer apples or oranges. You find that 53% prefer apples over oranges. You decide to make a 95% confidence interval for the proportion of people who prefer apples. What is the margin of error? (margin of error is z* times the standard error) Keep 3 places after the decimal place.

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

A: The hypothesized mean is 140.

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: State the hypotheses. H₂: All the means are equal. H₂: At least one mean is different from the…

A: The quality control department measures weekly values of different suppliers for the percentage…

Q: given as + Choose th

A: The margin of error is the amount of random sampling error in the results of a survey. The larger…

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A: The sample proportion is computed as follows, The critical value for 96% confidence level, i.e.,…

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A:

Q: We surveyed 53 people in an attempt to determine the proportion of Americans who lease a vehicle. 13…

A: The proportion of sample Americans who lease a vehicle is, Thus, the proportion of sample Americans…

Q: A television station surveys 200 customers and finds that 23% of customers watch the evening news.…

A: Given data n=200 p=23% confidence level=99%

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A: Probability is a chance of occurrence of any event.

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A: Test statistic formula for testing one sample proportion is, z=p^-pp1-pn  =p^-pSEp In the given…

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A: Given the data of weekly values of different suppliers for the percentage difference of waste…

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A:

Q: A survey company polls 1,233 voters about whether they plan to vote for the Liberal or the…

A: Sample size 52% of voters prefer the Liberal candidate i.e., Significance level

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: 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: Dr. Smith is investigating whether squeezing a stress ball throughout the day improves job…

A: μ=45, n=100, x̄=50,

Q: Suppose a professor wants to know if students in her statistics class have a good grasp of basic…

A: Given: Sample size n=6 Sample mean x¯=∑i=1nXin=62+92+75+68+83+956=79.167 Sample standard deviation…

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

A:

Q: BCD Travel wants to estimate the population proportion of adults who favor historical sites for…

A: p^ = 59% = 0.59Sample size = 24086% confidence level

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A: Obtain the margin of error. The margin of error is obtained below: The formula for the margin of…

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A:

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A: The random variable percentage of small incorporated businesses follows binomial distribution. The…

Q: aOf 500 random people surveyed, 72 were smokers. Find the 80% confidence interval for smokers.

A: It is given that, Out of 500 random people surveyed, 72 were smokers. So, the portion of the…

Q: lution is incorrect since the  confidence level is 90% margin of error is 3%

A: Given: Confidence level = 0.90 Margin of error (E) = 0.03 Assumed sample proportion  p^     = 0.50

Q: A researcher wanted to estimate the proportion of undergraduate college students (18– 29 years) with…

A:

Q: A research institute poll asked respondents if they felt vulnerable to identity theft. In the poll,…

A: It is given that x = 556n = 914Confidence level = 95%

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A:

Q: What happens to the spread of an interval if you increase the confidence level from 90% to 99%?

A:

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A: Obtain the large sample size at 95% confidence level.   The large sample size is obtained below as…

Q: If a 90%, a 95% and a 99% confidence interval are foung for a population mean, describe how the 3…

A: Population mean CI for known population standard deviation is, CI=x¯±Zσnwhere,σ=population standard…

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A:

Q: What are the z-scores for 90%, 95%, and 99% confidence?

A: the z-scores for 90%, 95%, and 99% confidence

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A: Solution: Given information: Margin of error = E = 0.05 Confidence level = 0.90= 0.90 Assumed…