Module 6. Assignment:

Question A:

A1. Compute the mean of this population:

A2. Select a random sample of size 2 out of the five members:

 I have selected (14,10)

A3. Compute the mean and standard deviation of your sample:

Mean = (14+10)/2 = 12

SD = sqrt((((14-12)^2)+((10-12)^2))/(2-1)) = sqrt(8) = 2.83

A4. Compare the Mean and Standard deviation of your sample to the entire population of this set (8,14, 16, 10, 11):

 

Comparison:

The population mean is 11.8, whereas the sample mean is 12. Both values are close but differ due to sampling variability. The population standard deviation is about 2.86 whereas the sample standard deviation is 2.83. I believe the discrepancy is due to the sample having fewer observations and its SD using "n - 1", while population SD is only using "n". 

Question B:

B1. Does the sample proportion p have approximately a normal distribution? Explain:

n = 100

p = 0.95

q = 1 - p = 0.05

np = 100(0.95) = 95

nq = 100(0.05) = 5

Standard deviation of p is normal if:

np > 10 or nq > 10

Since nq = 5, then the standard deviation is not normal and the condition fails. 

B2. What is the smallest value of n for which the sampling distribution of p is approximately normal?:

a) np > 10

  0.95n > 10

 n > 10.53

b) nq > 10

0.05n > 10

n > 200

Therefore, the condition would have to be n > 200, where that value is the smallest it can be to meet the condition. 

Question C:

C1. Please explain the reason why rbinom is better than sample in the coin tossing simulation:

In this circumstance, rbinom is better for the coin tossing simulation because rbinom is built specifically for binomials trials, which is perfect for coin flips. The sample method can imitate the coin flips, but it is a general sampling method where it does not directly model the process like the binomial method. For example, we can have values like c(0,1), which can work, but for probability simulations for coin tossing, rbinom is the more appropriate and efficient method.

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