In a sample of 5 courses with no support course the average final of 200 students was 70%with a standard deviation of 10%. Calculate a 95% confidence interval for the mean ofstudents who did not take a support course. (Type numbers in Excel without percent sign)(68.61, 71.39)(68.61, 71.38)(68.60, 71.39)(69.96, 70.04)5 is not larger than 40 so we cannot compute a confidence interval.

Critical Value: In test of hypotheses, Critical Values are the cut-off values that provide with the…

Answered: In a sample of 5 courses with no…

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 janitorial service wants to estimate the amount of time it takes to change a light bulb. A random sample of 100 light bulb change jobs yields a mean of 22 minutes and a standard deviation of 5 minutes, The 95% confidence interval for the average time it takes to replace a light bulb is (20.1775 , 23.8225) (21.902 , 22.098) O (21.02 , 22.98) (12.2 , 31.8)
The researcher recorded the amount of time each bird spent in the plain chamber during a 60-minute session. Suppose the study produced a mean of M = 44.62 minutes on the plain chamber with SS = 206 for a sample of n = 10 birds. Construct the 99% confidence interval to estimate the mean amount of time spent on the plain side for the population of birds. Please type only the lower interval with three decimal places. For example if CI= ( 10.25, 15), then type 10.25 as your answer.
A researcher is interested in studying the use of hypnosis to relieve pain. She took a random sample of 16 subjects and measured their sensory ratings. The mean of the sensory ratings was found to be 7.5 and standard deviation to be 2.3. It is known that the sensory ratings are normally distributed. Construct a 99% confidence interval for the mean sensory ratings of the population. a) (5.81, 9.19) b) (6.27, 8.73) c) (6.36, 8.64) d) (6.02, 8.98)
Suppose the mean height in inches of all 9th grade students at one high school is estimated. The population standard deviation is 3 inches. The heights of 6 randomly selected students are 71, 73, 69, 72, 71 and 64. x= Margin of error at 99%= confidence level = 99% confidence interval = [smaller value, larger value]=
100 students take a test in 2018. The average score was 75 and the standard deviation of the final exams scores was 5. The following year (2019), the professor changed her teaching style, and then taught the class to 90 students, gave the same exam, and found an average score of 80 and a standard deviation of 8.. Compute the 95% confidence interval for the final exam score in ECON 540 for 2018. Interpret the interval.
A random sample of 83 bicycle enthusiasts found the mean number of miles ridden over the summer is 684 with a standard deviation of 154.5. The 90% confidence interval for the true mean number of miles ridden over the summer for all bicycle enthusiasts is (655.79, 712.21). Which statement below is the correct interpretation of the confidence level? (A) There is a 90% chance the true mean number of miles ridden is between 655.79 and 712.21. (B) 90% of all bicycle enthusiasts ride between 655.79 and 712.21 miles over the summer. (C) We are 90% confident that a randomly selected cyclist will ride between 655.79 and 712.21 miles over the summer. (D) If 50 random samples of size n=83 were taken from the population of bicycle enthusiasts and a 90% confidence interval was calculated for each sample, then the true mean number of miles ridden would be contained within 45 of the calculated intervals, on average. (E) In repeated sampling, the mean number of miles ridden over the summer will be…
Suppose the mean height in inches of all 9th grade students at one high school is estimated. The population standard deviation is 3 inches. The heights of 7 randomly selected students are 72, 61, 74, 68, 64, 66 and 64. x¯ = Margin of error at 95% confidence level = 95% confidence interval = [ , ]
The mean score on the math portion of the ACT for the year 2000 is μ = 20.7. A sample of 20 students in a local school district had a mean of 22.95. Using the sample, a 95% confidence interval estimate for the mean is (20.76, 25.14). State the point estimate for the mean ACT score for students. Based on this sample information , do you believe that the local students have a mean different from the general population? Why? 22.95 is in the center of the interval. The means are definitely not different. 22.95 is just outside the 95% confidence interval. Can't say the means are really different. 22.95 is not equal to 20.7. The means are definitely different. 20.7 is just outside the 95% confidence interval. Can't say the means are really different.
Researchers are studying whether the proportion of homeschooled children in district 1 (considered 1) is less than the proportion of homeschooled children in neighboring district 2 (considered as 2). To test the theory info was gathered: district 1 had 600 students of which 90 were homeschooled. District 2 had 550 students of which 99 were homeschooled. What is the 95% confidence interval for the difference proportion between district 1 and 2? A. (-0.086, 0.026) B. (0.143, 0.217) C. (0.113, 0.187)
Dr. Higney, an academic researcher in an English department, wants to learn about the average length of a novel. In order to do this, Dr. Higney collects data on the word counts of 200 randomly selected novels. The sample mean of this data set is = 85252 with a sample standard deviation of 8413 words. Using this data, construct a 90% confidence interval for the true mean length of novels. (84,273.50, 86,230.51) (82719.67,89784.33) (94086.04, 96417.96) (80937.53, 90566.47) S
Use the given confidence interval to find the margin of error and the sample mean. (13.2,20.6) The sample mean is The margin of error is (Type an integer or a decimal ) (Type an integer or a decimal)

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