If n=15, ¯xx¯(x-bar)=33, and s=15, construct a confidence interval at a 90% confidence level. Assume the data came from a normally distributed population.
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If n=15, ¯xx¯(x-bar)=33, and s=15, construct a confidence interval at a 90% confidence level. Assume the data came from a
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- If n = 11,(x -bar)= 31, and s = 19, construct a confidence interval at a 99% confidence level. Assume the data came from a normally distributed population. Give your answers to three decimal places.Find the critical t-value that corresponds to 70% confidence. Assume 7 degrees of freedom.Find the z critical value associated with 85% confidence.
- Out of 300 people sampled, 135 received flu vaccinations this year. Based on this, construct a 95% confidence interval for the true population proportion of people who received flu vaccinations this year. Give your answers as decimals, to three placesUse the given data to construct a confidence interval for the population proportion p of the requested level. =x53 , =n68 , confidence level 80% Round the answers to at least three decimal places.In each of Exercises, we have presented a confidence interval (CI) for the difference, µ1 - µ2, between two population means. Interpret each confidence interval. 99% CI is from -20 to 15.If n=24, (x-bar)=44, and s=18, construct a confidence interval at a 98% confidence level. Assume the data came from a normally distributed population. Give your answers to one decimal place.Find the critical values X7p and X2, for an 80% confidence when the sample size is 51.If n=31, (x-bar)=36, and S3D3, construct a confidence interval at a 95% confidence level. Assume the data came from a normally distributed population. Give your answers to one decimal place.The distribution of scores on a standardized test is approximately Normal with standard deviation of σσ = 22. If a sample of 150 students had a mean score of x¯x¯ = 74.3, use this to find a 99% confidence interval for the mean score of all students on this test. We are 99% confident that the the mean score of students on this test was in the interval What is the margin of error for your confidence interval? Margin of error =Translate the statement into a confidence interval. Approximate the level of confidence. In a survey of 1000 adults in a country, 78% think teaching is one of the most important jobs in the country today. The survey's margin of error is ±4%. **PLEASE give four decimal pointsIf n=-22, (x-bar)=50, and s-6, construct a confidence interval at a 90% confidence level. Assume the data came from a normally distributed population. Give your answers to one decimal place.SEE MORE QUESTIONSRecommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons Inc
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