3. Assuming that the samples comes from normal distributions, find the margin oferror E, given the following:a. n 12,X= 18,s = 3.0, 95% confidenceb. n 16,X= 48,s = 3.2, 95% confidenceC.n = 18,X= 78.3, s= 2.5, 95% confidenced. n= 23,X 80.4, s= 3.6, 99% confidencen = 26,X= 90.8, s= 2.8, 99% confidencee.

Given that ; By using the margin of error (m.e) formula to solve this problem Note ; according to…

Answered: 3. Assuming that the samples comes from…

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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The mean of the distribution of sample means would be equal to:
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Using this data set What is the standard error of the relevant distribution? Ignoring that the number of men in the 2012 Survey was about 17,675, and the number of women was about 18,634, assume the samples were 100 men and 100 women in each Survey My worked solution is: P1: men proportion change = 15.5/100 = 0.155 N1 = 100 P2: women proportion change = 57.9/100 = 0.579 N2 = 100 {[0.155 * (1 - 0.155) /] + [0.579 * (1 - 0.579) / 100]} = 0.0037 Standard Error = 0.0037 is this correct?
If we meet these conditions, the sampling distribution of the mean will have a normal shape and ... The mean of the sampling distribution will be u, (i.e., the same as the population mean) and the standard deviation of the sampling distribution will be (that's the population standard deviation divided by the square root of the sample size). In the AP Stats Guy video, he talks about the number of text messages his students send during class. Suppose the average number of text messages his students send during class is u = 30 text messages. If we take samples of say, n = 36 students at a time, we would expect the mean of the sampling distribution we create to be the same as the population mean, 30 text messages. If we can further say that standard deviation of the number of text messages is 12 text messages, by how much would we expect the sample means to vary? (hint, use the formula above) text messages
Answer this question for me mate. Much appreciated. :)
Find the quartiles of the Normal distribution having mean 60 and S.D 10.

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3. Assuming that the samples comes from normal distributions, find the margin of error E, given the following: a. n= 12, X 18, s = 3.0, 95% confidence b. n 16, X= 48, s = 3.2, 95% confidence %3D n = 18, X= 78.3, s= 2.5, 95% confidence C.