I need help with this assignment For each sample, you can quickly calculate samp
ID: 3182630 • Letter: I
Question
I need help with this assignment
For each sample, you can quickly calculate sample mean and standard deviation by loading in the data into R in the form of a vector and then using the functions mean ( ) and sd( ) to calculate x- and S for the sample. For example for the first sample I can use the commands in R as follows to load in the data into a vector x and then calculate the mean x- and standard deviation S x = c(97, 117, 140, 73, 93, 148, 108, 135, 126, 121) x bar = mean(x) s = sd(x) print (x bar) print (s) (a) Calculate the sample mean x- and the sample standard deviation s for the first sample. (b) Assume the population standard deviation is sigma = 22. For the first sample, construct a 95% confidence interval for the population mean mu based upon the central limit theorem and the normal distribution. (c) For the first sample, construct a 99%- confidence interval for the population mean mu. Which interval is wider, the 95% interval or the 99% interval? (d) Assuming the population standard deviation is sigma = 22. For the second sample, construct a 95% confidence interval for the population mean mu based upon the central limit theorem and the normal distribution. How does the 95% confidence interval for the second sample compare with the 95% confidence interval for the first sample in terms of width and position? (e) For the 3^rd, 4^th and 5^th samples construct the 95% confidence intervals assuming sigma = 22. Why do you get different answers for different samples? (f) In the previous problems we assumed that we know what sigma was. Now let us drop that assumption. Calculate the 95% confidence interval for the first and second sample using the T distribution confidence interval formula. An important indicator of lung function is forced expiratory volume (FEV), which is the volume of air that a person can expire in one second. Dr. Hernandez plans to measure FEV in a random sample of n young women from a certain population, and to use the sample mean y as an estimate of the population mean. Let E be the event that Hernandez's sample mean will be within plusminus 100 ml of the population mean. Assume that the population distribution is normal with mean mu = 3000 ml and standard deviation sigma = 400 ml. Find Pr{E} if (a) n = 15 (b) n = 60 (c) How does Pr{E} depend on sample size? At a large university, the mean age of the students is 22.3 years, and the standard deviation is 4 years. A random sample of 64 students is drawn. What is the probability that the average age of these students is greater than 23 years?Explanation / Answer
a)
Sample mean xbar = 116.8
Sample Standard deviation (s) = 21.7909
Population Standard deviation (Sig) = 22
The formula for Confidence interval for mu when sig is known is,
xbar - E < mu < xbar + E
E = Zc * Sig / sqrt(n)
= 1.960 * 22 / sqrt(10)
= 13.6357
xbr - E = 103.1643
xbr + E = 130.4357
Therefore 95 % confidence interval for population mean mu is,
(103.1643 , 130.4357)
b)
Sample mean xbar = 116.8
Sample Standard deviation (s) = 21.7909
Population Standard deviation (Sig) = 22
The formula for Confidence interval for mu when sig is known is,
xbar - E < mu < xbar + E
E = Zc * Sig / sqrt(n)
= 2.576 * 22 / sqrt(10)
= 17.9213
xbr - E = 98.8787
xbr + E = 134.7213
Therefore 99 % confidence interval for population mean mu is,
(98.8787 , 134.7213)
The 99 % confidence interval is wider.
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