Use the t-distribution to find a confidence interval for a mean given the relevant sample results. Give the best point estimate for μ,the margin of error, and the confidence interval. Assume the results come from a random sample from a population that isapproximately normally distributed.point estimate = iA 99% confidence interval for using the sample resultsRound your answer for the point estimate to one decimal place, and your answers for the margin of error and the confidence intervalto two decimal places.margin of error = iThe 99% confidence interval is i=88.0, S = 31.4, and nto i=15

Sample size(n)=15 sample mean(x¯)=88.0 sample standard deviation(s)=31.4 degrees of freedom(df)=n-1…

Answered: Use the t-distribution to find a…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A student who took Stat 2 last semester learns that their overall course score was at the 85th percentile. Assuming that a histogram of all the Stat 2 course scores from last semester follows the normal curve, this means that this student had a score that was how many SDs above the average score? Choose the answer that is closest. Group of answer choices 1.15 1.05 1.10 1.0 PreviousNext
Tina catches a 14-pound bass. She does not know the population mean or standard deviation. So she takes a sample of five friends and they say the last bass they caught was 9, 12, 13, 10, and 10 pounds. Find the t and calculate a 95% (α = .05) confidence interval.
The distribution of heights in a population of women is approximately normal. Sixteen percent of the women have heights less than 62 inches. About 97.5% of the women have heights less than 71 inches. Use the empirical rule to estimate the mean and standard deviation of the heights in this population. Mean: K inches Standard Deviation: inches
Find the value:
A standardized exam's scores are normally distributed. In a recent year, the mean test score was 1463 and the standard deviation was 314. The test scores of four students selected at random are 1870, 1220, 2190, and 1340. Find the z-scores ← that correspond to each value and determine whether any of the values are unusual. F1 The z-score for 1870 is (Round to two decimal places as needed.) The z-score for 1220 is. (Round to two decimal places as needed.) The z-score for 2190 is (Round to two decimal places as needed.) The z-score for 1340 is. (Round to two decimal places as needed.) Which values, if any, are unusual? Select the correct choice below and, if necessary, fill in the answer box within your choice. OA. The unusual value(s) is/are. (Use a comma to separate answers as needed.) OB. None of the values are unusual. F2 80 F3 000 000 F4 F5 ^ MacBook Air F6 & A D F7 * DII F8 DD F9 ) J 운 F10 I a F11 + Next F12 delete
Use the t-table or any statistical tool and report the t-score that you multiple by the standard error to form the margin of error for a 95% confidence interval for a mean with 5 observations. a. 2.132 b. 2.776 c. 2.015 d. 2.571
We would like to create a confidence interval for the mean. Which of the following would produce the smallest margin of error? Select one: a.A 95% confidence level and a sample size of 150 subjects b. A 99% confidence level and a sample size of 600 subjects. c. A 99% confidence level and a sample size of 150 subjects. d.A 95% confidence level and a sample size of 600 subjects.
NEED ASAPIndicate the units and draw the graph. Round off to four decimals or in lowest term if fraction.
The 95% confidence interval of the variance of a population is computed as (12.367,21.761). Interpret this confidence interval.
Determine if the following statements are true or false. a. Jane took a sample of n = 100 observations from a population with mean and standard deviation. Mike also took a sample of n = 200 observations from the same population. Then, Mike’s sample mean will be closer to the true population mean than Jane’s sample. b. Jane took a sample of n = 100 observations from a population with mean and standard deviation. Mike also took a sample of n = 200 observations from a population with mean and standard deviation 2. Then, the sample mean from Mike’s sample is more likely to be close to the true mean than the sample mean from Jane’s sample. That is, the probability that X is between 90% and 110% of mean is higher for Mike’s sample than for Jane’s sample.
You know the mean and standard deviation of the population. You take a sample from this population and compute the 90% confidence interval for the mean. This interval contains values that are within how many standard deviations of the mean?

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