Question

Q4. (Sampling distributions and the central limit theorem) [10 points Sup- pose you programmed a computer to do the following: Step 1: Randomly choose an integer number from 1-5 (with equal proba bility of choosing each value). Do this 147 times to get a sample of n=147 randoin numbers Step 2: Using the sample in step 1, calculate μ = x and σ-82 Step 3: Repeat steps 1-2 another 9,999 times to get a total of 10,000 differ- ent sample means μ and variances σ2 calculated in the same way a. Suppose you created a histogram using your 10,000 values of μ. What does this histogram te us [2 points b. What do you expect would be the mean of the 10,000 values of μ? What do you expect would be the variance? [2 points] c. What can you say about the approximate shape of the histogram? [2 points] d. Suppose you increased the sample size in step 1 above to n-289 observa- tions. How would this change (or not change) your answers to (a)-(c)? In (c), how does the histogram change (or not change)? (4 points

Answer 1

R codes

> Step 1

>
> random_sample = sample(1:5,1)
> sample = rep(random_sample,147)
> length(sample)
[1] 147
>
>
> Step 2

>
> mean = mean(sample)
> variance = var(sample)
>
> Step3

>
> mean1 = variance1 = c()
> for (i in 1: 9999)
+ {
+ random_sample = sample(1:5,1)
+ sample = rep(random_sample,147)
+ mean1[i] = mean(sample)
+ variance1[i] = var(sample)
+ }
>

a)

From this we can say that all numbers are drawn in equal proportion.

b)

Expected value of mean of all 10000 values of mean is (1+2+3+4+5)/5 = 3

From sample Value of mean for 10000 values of mean is 2.9999

Expected value of variance of all 10000 value of mean is var of 1,2,3,4,5 which is 2.5

From sample Value of variance for 10000 values of mean is 2.013

c)

Shape of histogram has uniform distribution. All points between 1 and 5 are uniformly distributed.