Sampling Distribution of x and the Central Limit Theorem
Choose a sample size n, draw 30 random samples, and watch what happens to the 30 sample means. Work through the steps in order. Each step unlocks the next one.
Business context (Doane Figure 8.1). The MBA applicant pool at a medium-sized public university has
N = 2,637 GMAT scores with population mean μ = 520.78 and population standard deviation
σ = 86.80. The admissions office wants to know how reliable a sample mean is at different sample sizes
before it invests in a larger applicant survey.
1
Meet the population
This histogram shows every GMAT score in the applicant pool. This is the distribution of the individual scores X.
Population size
N = 2,637
Population mean
μ = 520.78
a parameter, fixed
Population std. dev.
σ = 86.80
spread of individual scores
Population distribution: all 2,637 GMAT scores
Individual GMAT scoresμ = 520.78
Notice the shape: the pool is not perfectly normal. It has a longer tail on the right (right-skewed). Keep this picture in mind. It will never change in this activity, whatever sample size you choose.
2
Choose your sample size n
Finish Step 1 to unlock.
How many applicants will the admissions office sample? The lab uses n = 5, 20, and 50. Start with n = 5.
n = 5
3
Draw one sample and compute x
Choose n in Step 2 to unlock.
Randomly select n applicants from the pool and compute their sample mean x (in Excel: =AVERAGE(range)).
Sample mean: x =
Click Draw another sample a few times. Does x stay the same? It changes from sample to sample because each sample contains different applicants. This is sampling variability, and it is why x is a random variable.
4
Take 30 samples of size n
Finish Step 3 to unlock.
Predict first. The individual scores have σ = 86.80. Will the 30 sample means be…
Sampling distribution: the 30 sample means (n = n)
30 sample means xNormal curve predicted by the CLTμ = 520.78
Bonus: 30 samples is a small number. See the shape with 1,000 samples
With only 30 means, the histogram is bumpy. Drawing 1,000 samples of the same size makes the shape easier to see (this chart is zoomed in). It is for the picture only; Steps 5 and 6 still use your 30 samples.
5
Summarize the 30 sample means
Finish Step 4 to unlock.
Treat the 30 values of x as a new data set. Copy them into Excel and compute:
their average, =AVERAGE(range), which estimates the center of the sampling distribution
their standard deviation, =STDEV.S(range), which is the observed standard error of x
Quantity
From your 30 samples
What theory predicts
Center: average of the 30 x values
μ = 520.78
Spread: standard error of x
6
Repeat with a larger n and compare
Save a run in Step 5 to unlock.
Go back to Step 2, pick a larger n, and repeat Steps 3 to 5. Save at least three runs: n = 5, 20, and 50. Each saved run appears below.
Run
n
Average of 30 x
Observed SE (STDEV.S of 30 x)
Theoretical SE σ / √n
No saved runs yet.
The 30 sample means for each saved run (same axis as the population)
Standard error shrinks as n grows
Theoretical SE = σ / √nObserved SE from your runs
Save runs for at least two different values of n to continue.
7
What did you discover?
Compare at least two sample sizes in Step 6 to unlock.
Answer each question on your own first, then check.
1. Did the population histogram in Step 1 change when you changed n?
No. The individual scores and their spread, σ = 86.80, stay the same. Larger samples do not reduce the spread of individual GMAT scores.
2. What happened to the spread of the 30 sample means as n increased?
It shrank. The standard error of the mean is σx = σ / √n: about 38.82 at n = 5, 19.41 at n = 20, and 12.28 at n = 50. Your observed values (STDEV.S of the 30 means) should be close to these.
3. Where was the histogram of sample means centered?
Near μ = 520.78 for every n. The sample mean is an unbiased estimator: E(x) = μ.
4. What happened to the shape of the sampling distribution as n increased?
It became more bell-shaped (closer to normal), even though the population is right-skewed. The bonus with 1,000 samples in Step 4 shows this most clearly.
5. Going from n = 5 to n = 20 multiplies the sample size by 4. What happens to the standard error?
It is cut in half, because √4 = 2: 86.80 / √5 ≈ 38.82 and 86.80 / √20 ≈ 19.41. Quadrupling n halves σx.
Central Limit Theorem (CLT). If a random sample of size n is drawn from a population with mean μ and standard deviation σ, the distribution of the sample mean X approaches a normal distribution with mean μ and standard error σx = σ / √n as the sample size increases.
Rule of thumb: n ≥ 30 is sufficient for most populations, even skewed ones. Symmetric populations require a smaller n.
"The raw data did not become normal. The sample means did."
The CLT changes the picture of X. It does not change the picture of X.