(a) Construct a 99% confidence interval for the population proportion. (b) Interpret the results.
Q: A poll of 827 students at Alpha State College found that 61% of those polled preferred a quarter…
A: Given that Sample size n = 827 Sample proportion of students who prefer a quarter system = 0.61 The…
Q: A survey of randomly selected university students found that 95 of the 105 first-year students and…
A: Given that,for first-year students:the survey was conducted with 105 first-year students,The total…
Q: A telephone service representative believes that the proportion of customers completely satisfied…
A: See the handwritten solution
Q: A telephone service representative believes that the proportion of customers completely satisfied…
A: v
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A: Given, Sample mean = 60 Standard error = 3 The objective is to construct the 95% confidence…
Q: A telephone service representative believes that the proportion of customers completely satisfied…
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A: Given : Sample Size, n = 4019 Number of successes, x = 722 confidence level,c=0.9
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A: Number of Canadian adults in 2005= 75 Number of Canadian adults in 2009 = 120 significance level =…
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A: Given information: Number of babies that were born n=340 Number of girls that were born X=289…
Q: 90% confidence interval is narrower than a 95% confidence interval using the exact same data.…
A: Conditions on confidence interval: As the confidence level increases, width of the confidence…
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Q: A recent poll indicated that 42% of men supported raises taxes to fund additional park services.…
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Q: A random sample of high school students is used to estimate the mean time all high school students…
A: Solution: From the given information, a 99% confidence interval is 1.7 hours to 3.1 hours.
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A: Given Sample size =40
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A: given data sample 1 :children n1 = 53sample 2 : teenager n2 = 112ci for difference in…
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Q: A clinical trial tests a method designed to increase the probability of conceiving a girl. In the…
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Q: In a survey of 1002 people, 70% said that they voted in a recent presidential election. In…
A: n=1002
Q: What does the 95% represent in a 95% confidence interval
A: Solution CI=95%
Q: A 95% confidence interval is narrower than a 90% confidence interval for the same data set. O True O…
A: We have to find given probability.
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A: The proportion that students at Kent State live on campus is obtained below:From the given…
Q: A survey of randomly selected university students found that 95 of the 100 first-year students and…
A: (a) Obtain the 95% confidence interval estimate for the difference in the proportions of first year…
Q: Find the confidence level and a for an 88% confidence interval.
A: Here use confidence level and significance level as alternative of each other
Q: Let's say you survey a random sample of people. You find that 53% prefer apples over oranges. You…
A: You find that 53% prefer apples over oranges from survey. ∴ p^ = 0.53 90% ci for p = (0.448,0.612)…
Q: Each man in a random sample of n = 16 men between 30 and 39 years old is asked to do as many sit-ups…
A: We have given that, Sample mean (x̄) = 24.2, standard deviation (s) = 8 and sample size (n) = 16…
Q: What is the effect on the size of the confidence interval if a T score is used instead of a Z score?…
A: The formula for calculating confidence interval using Z score is as follow:
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A: Given : Sample Size, n = 220 Number of successes, x = 18 confidence level,c=0.95
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A: Given, The total sample size, n = 90 No. of blemishes, x = 12 Therefore p^=xn = 1290 = 0.133 Also,…
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Q: d) Construct a 90% confidence interval for Hcommunity college - Pno transfer to approximate the mean…
A: 100(1-α)% CI for μ1-μ2 is(x¯1-x¯2)-tα/2(n1-1)s12+(n2-1)s22n1+n2-2(1n1+1n2)≤…
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- A beverage company would like to determine if their drink is preferred to a rival company’s drink. They perform an experiment with 150 randomly selected individuals and have them taste test the two different drinks in a controlled environment so that each person's preference is independent of others. A 90% confidence interval for the proportion preferring their drink is (0.453, 0.587). Which of the following is the correct interpretation of this confidence interval? 90% of all consumers prefer this beverage company’s drink to their rival’s between 45.3% and 58.7% of the time. We’re 90% confident that between 45.3% and 58.7% of all consumers would prefer this beverage company’s drink to their rival’s drink. We’re 90% confident that there’s a 45.3% to 58.7% chance that a randomly selected person would prefer this beverage company’s drink to their rival’s drink. We’re 90% confident that the proportion of people in the sample who prefer this beverage company’s drink over their…A computer store near campus is stocking its shelves and wants to know the relative proportions of students who use PCs vs Macs. A random sample of 60 students is surveyed, and each student is asked whether they use a PC or Mac more often. 55% select ”PC.” (a) What is the population and what is the sample in this study? (b) Calculate a 95% confidence interval for the proportion of UCI students that are PC users. (c) Provide an interpretation of this confidence interval in the context of this problem. (d) The confidence interval is pretty wide and leaves a lot of uncertainty over the proportion of UCIundergarduates who use PCs. With the goal to estimate a narrower 95% confidence interval, what is a simple change to this study that you could suggest for the next time that a similar survey is conducted?The dean of students at a large college is interested in learning about their opinions regarding the percentage of first-year students who should be given parking privileges in the main lot. He sends out an email survey to all students about this issue. A large number of first-year students reply but very few sophomores, juniors, and seniors reply. Based on the responses he receives, he constructs a 90% confidence interval for the true proportion of students who believe first-year students should be given parking privileges in the main lot to be (0.71, 0.79). Which of the following may have an impact on the confidence interval, but is not accounted for by the margin of error? response bias O nonresponse bias O sampling variation O undercoverage bias
- Suppose the English department of a local university has a total of 105 faculty members. Of this number, 24 (about 23%) are part-time instructors. Is it appropriate to find a confidence interval for the percentage of part-time instructors in the department? Find a 95% confidence interval or explain why you should not find a confidence interval for the percentage of part-time instructors in the English department.During a flu vaccine shortage in the United States, it was believed that 45 percent of vaccine-eligible people received flu vaccine. The results of a survey given to a random sample of 2,350 vaccine-eligible people indicated that 978 of the 2,350 people had received flu vaccine. (a) Construct a 99 percent confidence interval for the proportion of vaccine-eligible people who had received flu vaccine. Use your confidence interval to comment on the belief that 45 percent of the vaccine-eligible people had received flu vaccine. (b) Suppose a similar survey will be given to vaccine-eligible people in Canada by Canadian health officials. A 99 percent confidence interval for the proportion of people who will have received flu vaccine is to be constructed. What is the smallest sample size that can be used to guarantee that the margin of error will be less than or equal to 0.02 ?A study is conducted by a physical therapist to determine whether there is a difference in the proportion of men and women in the U.S. that participate in regular sustained physical activity. It is found that 22% of menout of 550 and 19.5% of women out of 450 participated in regular sustained physical activity. Construct a 90% confidence interval for the difference in proportions of men and women that participate in regular sustained and state your conclusion.
- Call me: A sociologist wants to construct a 99% confidence interval for the proportion of children aged 8 - 10 living in New York who own a smartphone.(a) A survey by the National Consumers League estimated the nationwide proportion to be 0.40 . Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.02 ?A recent poll indicated that 42% of men supported raises taxes to fund additional park services. The poll also found that 64% of women supported such a tax. The sample consisted of 29 men and 25 women. Construct a 90% confidence interval around the difference in proportions.How to calculate a 90% confidence interval?
- Two brands of tires were compared for the proportion of defective tires. Random samples of tires were collected from each brand, and the number of defective tires from each brand was recorded. The data is in the following table. Number of defective tires (x) 41 32 n Continental Michelin 320 180 a.Construct a 97% confidence interval for the difference between the two proportions. b.For part a, find the desired sample sizes if you want the Continental sample to be 1.5 times Michelin sample with a sampling error equal to 0.03. c. Conduct a test of hypothesis to determine if the defective rates are the same. Use the P-value to interpret the results (a is not given).A telephone service representative believes that the proportion of customers completely satisfied with their local telephone service is different between the South and the Northeast. The representative's belief is based on the results of a survey. The survey included a random sample of 700700 southern residents and 580580 northeastern residents. 49%49% of the southern residents and 42%42% of the northeastern residents reported that they were completely satisfied with their local telephone service. Find the 80%80% confidence interval for the difference in two proportions. Step 2 of 3: Find the value of the standard error. Round your answer to three decimal places.You want to find a 92% confidence interval for the proportion of PLM students who get a student scholarship allowance from the government of the City of Manila. You found an information from the city administrator that states that around 75% of PLM students are to get the allowance. At a maximum error of 5%, how many samples do you need?