It clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects, Before treatment, 15 subjects had a mean wake time of 104.0 min. After treatment, the 15 subjects had a mean wake time of 81.9 min and a standard deviation of 23.6 min. Assume that the 15 sample values appear to be from a normally distributed population and construct a 95% confidence interval estimate of the mean wake time for a population with drug treatments. What does the result suggest about the mean wake time of 104.0 min before the treatment? Does the drug appear to be effective? Needs Complete solution with 100 % accuracy. clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before treatment, 15 subjects had a mean wake time of 104.0 min. After treatment, the 15subjects fad a mean wake time of 81.9 min and a standard deviation of 23.6 min. Assume that the 15 sample values appear to be from a normally distributed population and construct a 95%confidence interval estimate of the mean wake time for a population with drug treatments. What does the result suggest about the mean wake time of 104.0 min before the treatment? Does the drugappear to be effective?Construct the 95% confidence interval estamals of the mean wake time for a population with the treatment.(Round to one decimal place as needed)What does the result suggest about the mean wake time of 104.0 min before the treatment? Does the drug appear to be effective?The confidence intervaldrug treatmentan effectthe mean wake time of 104.0 min before the treatment, so the means before and after the treatmentThis result suggests that the

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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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It clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects, Before treatment, 15 subjects had a mean wake time of 104.0 min. After treatment, the 15 subjects had a mean wake time of 81.9 min and a standard deviation of 23.6 min. Assume that the 15 sample values appear to be from a normally distributed population and construct a 95% confidence interval estimate of the mean wake time for a population with drug treatments. What does the result suggest about the mean wake time of 104.0 min before the treatment? Does the drug appear to be effective?

Needs Complete solution with 100 % accuracy.

clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before treatment, 15 subjects had a mean wake time of 104.0 min. After treatment, the 15
subjects fad a mean wake time of 81.9 min and a standard deviation of 23.6 min. Assume that the 15 sample values appear to be from a normally distributed population and construct a 95%
confidence interval estimate of the mean wake time for a population with drug treatments. What does the result suggest about the mean wake time of 104.0 min before the treatment? Does the drug
appear to be effective?
Construct the 95% confidence interval estamals of the mean wake time for a population with the treatment.
(Round to one decimal place as needed)
What does the result suggest about the mean wake time of 104.0 min before the treatment? Does the drug appear to be effective?
The confidence interval
drug treatment
an effect
the mean wake time of 104.0 min before the treatment, so the means before and after the treatment
This result suggests that the

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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