Module #6 Assignment
A.
First, let's address the calculations for the populations and sampla data
# Population data
population <- c(8, 14, 16, 10, 11)
# a. Compute the mean of the population mean
mean_population <- mean(population)
mean_population
Now, for selecting a random sample of size 2 out of the five members, you can use 'sample()' function:
# b. Select a random sample of size 2
sample_size <- 2
random_sample <- sample(population, size = sample_size)
random_sample
Next, compute the mean and standard deviation of your sample:
mean_sample <- mean(random_sample)
sd_sample <- sd(random_sample)
mean_sample
sd_sample
Finally, compare the mean and standard deviation of your sample to the entire population:
#d. Compare the Mean and Standard deviation of your sample to the entire population
mean_comparison <- mean_sample == mean_population
sd_comparison <- sd_sample == sd(population)
mean_comparison
sd_comparison
B.
1. To determine whether the sample proportion p has approximately a normal distribution, we need to check if both np and nq are greater than or equal to 5.
# Given valubes
n <- 100
p <- 0.95
# Calculate np and nq
np <- n * p
nq <- n * (1-p)
np
nq
If both np and nq are greater than or equal to 5, then the sample proportion p has approximately a normal distribution. You can check this by comparing the values of np and nq to 5.
2.
To find the smallest value of n which the sampling distribution of p os approximately normal, you need to solve for n when np and nq are both equal to or greater than 5. You can use trail and error or create a loop to find this value. Start with a small value of n and increment it until both np and nq are at least 5.
C.
The 'rbinom' function generates random binomial variates, which is more suitable for simulating coin tossing where there are two possible outcomes (heads and tails) with known probabilities (p and q). The 'sample' function is more general and doesn't take into account the specific probabilities associated with coin tossing.
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