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So I'm trying to study for a test and I'm stuck on two textbook questions. I'm having trouble grasping the concept of sampling distributions and when to apply certain rules/when to categorize a population as normal.

I have some points in the testbook and I'm trying to apply this knowledge to these questions but I'm just confused. I attempted 1 but would like clarification, and I don't know how to do 2.

Textbook states:

Central Limit Theorem – regardless of the shape of the population from which a sample was drawn, the sampling distribution of the mean of the sample will have a mean µ (which is equal to the population from which we’re sampling) and a standard deviation σ equal to σ/√n.

When n is large enough (n ≥ 30), the sampling distribution of the sample mean x̄ is approximately normally distributed (roughly mound shaped). And thus, has mean µ and standard deviation σ/√n.

QUESTIONS:

  1. Random samples of size n were selected from populations with the means and variances given here. Find the mean and standard deviation of the sampling distribution of the sample mean in each case:

a. n = 36, μ = 10, σ^2 = 9

So I'm assuming because n is greater than 30, I can apply the central limit theorem and do μ = 10, and σ = 3/√36 = 1/2 which is the correct answer in the textbook.

b. n = 100, μ = 5, σ^2 = 4

Since n ≥ 30, μ = 5 and σ = 1/5 which is also right.

c. n = 8, μ = 120, σ^2 = 1

But over here, since n is not greater or equal to 30, I'm guessing I can't use central limit theorem and don't really know what to do. The textbook says μ = 8 and σ = 1/√8. I don't know why the same standard deviation formula is used but the mean isn't the original value of μ like the others? Why 8? What rule is it following if it's not the central limit theorem?

And then I'm just confused in general how to get the answers for 2, as it's not given in the textbook. I don't really understand how to determine if a sampled population is normal/not normal and how that changes the sampling distribution.

  1. Refer to exercise 1. a. If the sampled populations are normal, what is the sampling distribution of x̄ for parts a, b, and c? b. According to the Central Limit Theorem, if the sampled populations are not normal, what can be said about the sampling distribution of x̄ for parts a, b, and c?
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    $\begingroup$ Regrettably, the CLT says nothing like what you are claiming it states. It is true that the sampling distribution of the mean of a sample drawn from a population with mean $\mu$ and standard deviation $\sigma$ has mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$, but the result is not because of the CLT at all. Note that no assertion has been made about the population except that it has finite mean $\mu$ and finite standard deviation $\sigma$: in particular, the word "normal" has not been used anywhere. $\endgroup$
    – Dilip Sarwate
    Aug 17, 2019 at 4:43
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    $\begingroup$ Perhaps also worth noting that the threshold of 30 for when "to apply CLT" is considerably arbitrary and ignores any properties of the specific problem you would be tackling (e.g. is a sample size of 30 the same whether it comes from a binomial with variance 0.5 than, say, 0.005?) $\endgroup$
    – Kuku
    Aug 17, 2019 at 5:02
  • $\begingroup$ @DilipSarwate I suspect the OP meant those 2 paragraphs are how the book defines CLT -- including approx normality for large n -- not just the 1st paragraph you mention. I've set them in a blockquote to clarify. $\endgroup$
    – civilstat
    Aug 17, 2019 at 14:33
  • $\begingroup$ I expect that Dilip is aware that it would be a quote from a book -- that's why it's so regrettable (a single person making such an error is something we can solve, but when a book so badly mis-teaches tens of thousands of students, it's a disaster we cannot hope to rectify). Worse, many, many basic books promulgate exactly the same nonsense, making the numbers likely well into the millions worldwide. Vastly more people get this wrong information than actually see what the CLT says. $\endgroup$
    – Glen_b
    Aug 18, 2019 at 2:04
  • $\begingroup$ I'm not saying Dilip is unaware it is a quote, nor that the book is flawless. Rather, I'm saying the book quote does in fact discuss approx normality, yet his comment claims it doesn't, which may be confusing instead of helpful to the OP and future readers. A better version of that original comment might be: (ctd) $\endgroup$
    – civilstat
    Aug 19, 2019 at 0:05

1 Answer 1

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Regarding Q1c, that is an obvious error in the solution in the book.

Looking at Q2, we can say that for large n, the sampling distribution will always be approximately normally distributed.

But what a large n is can vary depending on the variable. That is because when you have a variable with natural bounds, then a population mean close to a bound will yield an asymmetric, non-normal sampling distribution.

Look at the following example, where I drew 10000 samples for four different sample sizes and plotted the means for the draws. The underlying variable is binary with p = .7. We can see that in this case the sampling distribution is approximately normal for n = 30:

sampling distributions for p = .7

But if we change the population mean to a value close to the natural upper bound (p = .95 in the plot), then the picture changes. The sampling distribution is nowhere near to normal for n = 30 or even for n = 100, but for larger sample sizes it eventually will.

enter image description here

So the threshold of n = 30 is quite arbitrary as Kuku pointed out. To give a full answer for Q2: For normally distributed variables the sampling distribution will be normal even for 1c.

If the variables aren't normally distributed, then you are probably supposed to apply the threshold in the textbook. So then 1a and 1b will be normally distributed and 1c not. But in truth it really depends on the characteristics of the underlying variable.

Code: 

library(tidyverse)
share <- .7
population_size <- 10000
sample_size <- 30

binary_var <- c(rep(T, population_size * share), 
            rep(F, population_size * (1-share)))

many_sample_means <- function(x, sample_size, n_samples = 10000) {
  map(1:n_samples, ~sample(x, sample_size)) %>% 
    map(mean) %>% 
    unlist %>% 
    table
}


sample_sizes <- c(10, 30, 50, 100)
xlabs <- str_c("sample size: ", sample_sizes)

l <- c("n = 10" = 10, n_30 = 30, n_50 = 50, n_100 = 100) %>% 
  map(~many_sample_means(binary_var, .))

par(mfrow = c(2,2))

for(i in seq_along(l)) {
  plot(l[[i]], main = xlabs[i], ylab = "count of realizations", xlab = "mean", xlim = c(0,1))
}

par(mfrow = c(1,1))
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