B.4 Central Limit Theorem (Cookie Recipes)
Learning Objectives
By the end of this section, you will be able to:
- compute the mean and the standard deviation of a listed population, and use them as the \(\mu\) and \(\sigma\) the central limit theorem asks for;
- build the distribution of sample means from that same population twice, once from samples of size five and once from samples of size ten;
- state the theoretical distribution the central limit theorem predicts at each sample size, and check your own histograms against it;
- explain why the distribution of averages tightens as the sample size grows, and say by exactly how much.
B.4.1 Stats Lab: Central Limit Theorem (Cookie Recipes)
Class Time:
Names:
This is the second of the two central limit theorem labs, and it is deliberately not a repeat of the first. In §6.4 you went out and collected your own population by counting coins, which meant you never knew the true \(\mu\) and \(\sigma\) you were sampling from — you could only estimate them. Here the population is handed to you in full: sixty cookie recipes, each with the number of days its batch lasted, printed in a table you can read. Because you can see every value, you can compute \(\mu\) and \(\sigma\) exactly, and then check your sample means against a prediction rather than against a guess.
In real work you never see the whole population. A lab that hands you all sixty values is a rigged game on purpose — rigged so you can grade the theorem instead of only trusting it.
The other change is that you build the distribution of averages twice, at \(n = 5\) and again at \(n = 10\). One histogram of averages tells you that averaging did something. Two, drawn on the same axes from the same population, tell you what the sample size controls and what it leaves alone. That comparison is the whole reason the lab asks you to do the collection twice, and it is why the instruction to reuse the same axis scaling for all three pictures is not fussiness — a narrower distribution redrawn on a narrower axis looks exactly like the one before it, and the finding disappears.
Write down what you actually did, including anything irregular. If your random number generator repeated a recipe within one sample, if a group reported only three of its four means, if you rounded to the nearest tenth at a different step than your neighbor did, record it. A lab report that hides its own irregularities cannot be checked by anyone, and being checkable is most of what makes a result worth anything.
Wyatt tallied the sixty recipes before his group started sampling: 13 lasted one day and 14 lasted two days, while only 3 lasted three days and 5 lasted four days, and then 11 lasted five days and 13 lasted six days. One recipe lasted 11 days. Describe in words the shape of the population's histogram he is looking at, and say why a population shaped like this makes the lab worth doing.
Solution
Step 1 — group the counts by where they sit. The low end holds \(13 + 14 = 27\) recipes at one or two days. The high end holds \(11 + 13 = 24\) recipes at five or six days. The middle holds only \(3 + 5 = 8\) recipes at three or four days.
Step 2 — read the shape off those three groups. Two tall clumps at the ends with a deep dip between them. That is a bimodal distribution — it has two humps, not one. Off to the right, far from everything else, sits the single recipe that lasted 11 days.
Answer: the histogram Wyatt draws has a tall block at one and two days, a near-empty gap at three and four, a second tall block at five and six, and one lone bar out at 11. It is bimodal with an outlier, and it is about as far from a bell as a distribution can get. That is exactly why the lab uses it. If the population were already normal there would be nothing to demonstrate — sample means drawn from a normal population are exactly normal at every sample size, so all three of your histograms would look the same and the theorem would be invisible. The central limit theorem is interesting precisely because it promises a bell from populations like this one, and the only way to believe that is to watch it happen.
B.4.2 Given
For a population of \(N\) values \(x_1, x_2, \ldots, x_N\), the population mean is
$$ \mu = \frac{\sum x}{N} $$and the population standard deviation is
$$ \sigma = \sqrt{\frac{\sum (x - \mu)^2}{N}} = \sqrt{\frac{\sum x^2}{N} - \mu^2} $$Both divide by \(N\), the count of the whole population. This is what separates them from the sample statistics \(\bar{x}\) and \(s\), where \(s\) divides by \(n - 1\) because a sample is being used to estimate something it cannot see.
Which divisor you use is not a matter of taste here — it follows from what the sixty recipes are. The lab states that they are the population, so nothing is being estimated and there is no correction to make. Divide by 60. If you use the \(n - 1\) button on your calculator you will get 2.17 instead of 2.15, and while that is a small error in itself, it is a large error in understanding: it says you think these sixty recipes are a sample of some larger collection of recipes, which the lab has told you they are not. Later in the course you will meet the opposite situation constantly — a handful of measurements standing in for a population nobody can enumerate — and getting the distinction straight on a case where the answer is unambiguous is cheaper than getting it wrong on one where it is not.
Let \(X =\) length of time (in days) that a cookie recipe lasted at the Olmstead Homestead. Assume that each of the different recipes makes the same quantity of cookies.
| Recipe # | X | Recipe # | X | Recipe # | X | Recipe # | X |
|---|---|---|---|---|---|---|---|
| 1 | 1 | 16 | 2 | 31 | 3 | 46 | 2 |
| 2 | 5 | 17 | 2 | 32 | 4 | 47 | 2 |
| 3 | 2 | 18 | 4 | 33 | 5 | 48 | 11 |
| 4 | 5 | 19 | 6 | 34 | 6 | 49 | 5 |
| 5 | 6 | 20 | 1 | 35 | 6 | 50 | 5 |
| 6 | 1 | 21 | 6 | 36 | 1 | 51 | 4 |
| 7 | 2 | 22 | 5 | 37 | 1 | 52 | 6 |
| 8 | 6 | 23 | 2 | 38 | 2 | 53 | 5 |
| 9 | 5 | 24 | 5 | 39 | 1 | 54 | 1 |
| 10 | 2 | 25 | 1 | 40 | 6 | 55 | 1 |
| 11 | 5 | 26 | 6 | 41 | 1 | 56 | 2 |
| 12 | 1 | 27 | 4 | 42 | 6 | 57 | 4 |
| 13 | 1 | 28 | 1 | 43 | 2 | 58 | 3 |
| 14 | 3 | 29 | 6 | 44 | 6 | 59 | 6 |
| 15 | 2 | 30 | 2 | 45 | 2 | 60 | 5 |
Calculate the following:
a. \(\mu_X =\) _________
b. \(\sigma_X =\) _________
These two numbers are the only things about the population the central limit theorem needs, so work them out carefully and keep them somewhere you can find them — every prediction in the rest of the lab is built from this pair. Use the whole table, all sixty values, not the first row or a sample of them. The recipe numbers are labels, not data: the column you are summing is \(X\), and recipe #48 contributes an 11 to that sum just like recipe #1 contributes a 1.
Try it in rāSHio
Sixty values is a lot to key in twice, so get them into rāSHio once and reuse the column for everything below: choose File → Delimited List… and type the X column from Table B.4.1 separated by commas, spaces, or one per line. Enter the sixty X values only — the recipe numbers are labels, not data, and summing them by mistake is the most common way this step goes wrong.
Figure B.4.1 — Getting the sixty X values from Table B.4.1 into rāSHio: File → Delimited List… The walkthrough imports its own demo column; the steps are the ones you run on the cookie data.
Imani, who co-chairs the campus queer students' union and brought her study group along to this lab, has the totals for the sixty values in Table B.4.1 up on the board: the values total 214 and the squares of the values total 1,040. Work alongside her to find \(\mu_X\) and \(\sigma_X\). Then find the median and use it, together with the shape described in Try It Now B.4.1, to explain why a mean and a median that nearly agree do not mean the distribution is bell-shaped.
Solution
Step 1 — divide the total by the population size.
$$ \mu_X = \frac{214}{60} \approx 3.5667 $$Step 2 — use the shortcut form of the population standard deviation.
$$ \sigma_X^2 = \frac{\sum x^2}{N} - \mu_X^2 = \frac{1040}{60} - (3.5667)^2 = 17.3333 - 12.7211 = 4.6122 $$ $$ \sigma_X = \sqrt{4.6122} \approx 2.1476 $$Step 3 — locate the median. With 60 sorted values the median averages positions 30 and 31. Counting up from the bottom, 13 values are 1s and 14 are 2s, which reaches position 27; the next three values are 3s, reaching position 30. So position 30 holds a 3 and position 31 holds a 4.
$$ \text{median} = \frac{3 + 4}{2} = 3.5 $$Answer: Imani gets \(\mu_X \approx 3.57\) days and \(\sigma_X \approx 2.15\) days, with a median of 3.5 days. The mean and the median differ by less than a tenth of a day, which for a unimodal distribution would be decent evidence of symmetry — and symmetry is one of the things a bell has. But this distribution is bimodal, and the agreement here comes from the two clumps roughly balancing each other across the middle, not from a single peak sitting in the center. The middle of this distribution is nearly empty; the median lands at 3.5 because that is where the two halves meet, not because that is where the data piles up. A mean and a median that agree rule out strong skew. They say nothing at all about how many humps there are, and that is why the shape question in Try It Now B.4.1 had to be answered by looking at the counts rather than by comparing two summary numbers.
Try it in rāSHio
Stats → Summary Statistics in rāSHio returns the mean, the standard deviation and the median of your column in one step. Use it to CHECK the \(\mu_X\) and \(\sigma_X\) you worked out by hand, not to replace them. One catch worth knowing: if the standard deviation it reports is 2.17 rather than 2.15, it is dividing by \(n - 1\) — multiply by \(\sqrt{59/60}\) to get the population value this lab asks for.
Figure B.4.2 — Checking your hand-computed \(\mu_X\) and \(\sigma_X\) in rāSHio: Stats → Summary Statistics. The walkthrough runs on rāSHio's built-in sample data, so the numbers it returns are not yours; the steps are.
B.4.3 Collect the Data
Fix a population and a sample size \(n\). Take samples of size \(n\), compute the sample mean \(\bar{x}\) of each one, and collect those means into a data set of their own. The distribution of that collection is the sampling distribution of the sample mean, written \(\bar{X}\).
It is a distribution of averages, not of individuals. One value in it is built from \(n\) values of the original population, so it is a different object with its own shape, its own center, and its own spread.
Use a random number generator to randomly select four samples of size \(n = 5\) from the given population. Record your samples in Table B.4.2. Then, for each sample, calculate the mean to the nearest tenth. Record them in the spaces provided, and record the sample means for the rest of the class.
- Complete the table:
Try it in rāSHio
rāSHio will draw the recipe numbers for you: File → Random Numbers… with a minimum of 1, a maximum of 60 and 5 draws gives one sample of five — run it four times for this round, then four more with 10 draws for the \(n = 10\) round. Look each number up in Table B.4.1 to get the X value it stands for.
Figure B.4.3 — Drawing each sample's recipe numbers in rāSHio: File → Random Numbers… The walkthrough uses its own range; the steps are the ones you run with a minimum of 1 and a maximum of 60.
| Sample 1 | Sample 2 | Sample 3 | Sample 4 | Sample means from other groups |
|---|---|---|---|---|
| \(\bar{x} =\) \_\_\_\_\_ | \(\bar{x} =\) \_\_\_\_\_ | \(\bar{x} =\) \_\_\_\_\_ | \(\bar{x} =\) \_\_\_\_\_ |
- Calculate the following:
a. \(\bar{x} =\) _________
b. \(s_{\bar{x}} =\) _________
- Again, use a random number generator to randomly select four samples from the population. This time, make the samples of size \(n = 10\). Record the samples in Table B.4.3. As before, for each sample, calculate the mean to the nearest tenth, record them in the spaces provided, and record the sample means for the rest of the class.
| Sample 1 | Sample 2 | Sample 3 | Sample 4 | Sample means from other groups |
|---|---|---|---|---|
| \(\bar{x} =\) \_\_\_\_\_ | \(\bar{x} =\) \_\_\_\_\_ | \(\bar{x} =\) \_\_\_\_\_ | \(\bar{x} =\) \_\_\_\_\_ |
- Calculate the following:
a. \(\bar{x} =\) _________
b. \(s_{\bar{x}} =\) _________
- For the original population, construct a histogram. Make intervals with a bar width of one day. Sketch the graph using a ruler and pencil, and scale the axes. Put the number of days on the horizontal axis and the count on the vertical axis.
- Draw a smooth curve through the tops of the bars of the histogram. Use one to two complete sentences to describe the general shape of the curve.
Try it in rāSHio
Draw the pencil histogram first — choosing the intervals is the point of step 5 — then check your bar heights against rāSHio: Graph → Histogram with the bin start at 0 and a bin width of 1 puts one bar on each number of days. Write down the bin start and width you used, because both histograms of averages have to reuse the same axis.
It is the picture the other two get measured against. If you skip it or draw it on a different scale, the next two pictures will still look fine and will still prove nothing.
Notice the order the steps come in. You draw the population histogram in step 5, after collecting the samples, but it is the picture that everything else is compared with — so it is worth drawing carefully and worth scaling generously. Choose a horizontal axis that runs from 0 out to at least 11, so the outlying recipe has somewhere to sit, and pick a vertical scale tall enough for a bar of 14. Both of the histograms you draw later will reuse those same axes. That reuse is the single most important instruction in the lab, and it is also the easiest one to break, because a distribution of averages occupies a much narrower stretch of the horizontal axis than the population does. Redrawing it on a tighter axis to "fill the page" is exactly the move that hides the result, and it is a tempting one, because the tighter drawing genuinely looks better. Resist it: the empty space on the left and right of your second and third histograms is the finding.
Thao generated four samples of size five for her group and looked up these recipe numbers: sample 1 gave recipes 3, 19, 27, 41, 55; sample 2 gave 9, 14, 33, 52, 58; sample 3 gave 6, 22, 35, 44, 57; sample 4 gave 12, 30, 38, 46, 60. Read the four sets of \(X\) values off Table B.4.1, find the four sample means, then find the mean and standard deviation of those four means and compare each with the population figures from Try It Now B.4.2.
Solution
Step 1 — look up the values and average each sample.
$$ \text{Sample 1: } 2, 6, 4, 1, 1 \ \Rightarrow \ \bar{x} = \frac{14}{5} = 2.8 $$ $$ \text{Sample 2: } 5, 3, 5, 6, 3 \ \Rightarrow \ \bar{x} = \frac{22}{5} = 4.4 $$ $$ \text{Sample 3: } 1, 5, 6, 6, 4 \ \Rightarrow \ \bar{x} = \frac{22}{5} = 4.4 $$ $$ \text{Sample 4: } 1, 2, 2, 2, 5 \ \Rightarrow \ \bar{x} = \frac{12}{5} = 2.4 $$Step 2 — average the four means.
$$ \bar{x} = \frac{2.8 + 4.4 + 4.4 + 2.4}{4} = \frac{14.0}{4} = 3.5 $$Step 3 — find the standard deviation of the four means. These four are a sample of the sampling distribution, so divide by \(n - 1 = 3\).
$$ s_{\bar{x}} = \sqrt{\frac{(-0.7)^2 + (0.9)^2 + (0.9)^2 + (-1.1)^2}{3}} = \sqrt{\frac{3.32}{3}} = \sqrt{1.1067} \approx 1.05 $$Answer: Thao's four means are 2.8, 4.4, 4.4, and 2.4, averaging 3.5 with a standard deviation of about 1.05. Both numbers land where the theorem says they should. The average of the means, 3.5, is within seven hundredths of \(\mu_X \approx 3.57\) — the center did not move. The spread, 1.05, is less than half of \(\sigma_X \approx 2.15\) — the averages are packed far more tightly than the individuals they came from. Look at the ranges to see the same thing without any arithmetic: the twenty individual values the group looked up ran from 1 to 6, a span of five days, while their four means ran from 2.4 to 4.4, a span of two. Nothing was thrown away and nothing was corrected; the extreme values are all still in there, each one now sharing a slot with four companions that pull it back toward the middle. One caution belongs in her write-up and in yours: four means is far too few to judge a spread from, so do not read 1.05 as a measurement of anything. The theoretical value is computed in Try It Now B.4.4, and being within about ten percent of it on four values is luck as much as evidence.
B.4.4 Repeat the Procedure for n = 5
Draw random samples of size \(n\) from a population with mean \(\mu\) and standard deviation \(\sigma\). As \(n\) grows, the sampling distribution of the sample mean approaches a normal distribution:
$$ \bar{X} \sim N\!\left(\mu,\ \frac{\sigma}{\sqrt{n}}\right) $$The conclusion holds whatever the shape of the original population — skewed, flat, bimodal, or full of gaps. Only \(\mu\) and \(\sigma\) enter the statement; no other feature of the population appears anywhere in it.
- For the samples of \(n = 5\) days averaged together, construct a histogram of the averages — your means together with the means of the other groups. Make intervals with bar widths of \(\frac{1}{2}\) a day. Sketch the graph using a ruler and pencil, and scale the axes to match the histogram you drew for the population.
- Draw a smooth curve through the tops of the bars of the histogram. Use one to two complete sentences to describe the general shape of the curve.
Read the theorem as three separate claims, because your histogram tests them separately and can pass one while failing another. The center does not move: the average of the averages is the same \(\mu\) you computed from the sixty recipes, so the middle of this picture should sit at about 3.57, right where the middle of the population histogram sat. The spread shrinks by a factor of \(\sqrt{n}\), so this picture should be about \(\sqrt{5} \approx 2.24\) times narrower than the population's. And the shape moves toward a bell, which is the strangest of the three claims, because the population you drew a moment ago had two humps and a gap in the middle and this one is supposed to have a single peak. Watch what happens to the gap in particular. An average of five lands at 3 or 4 in plenty of ways — a couple of low values and a couple of high ones will do it — so the middle of this distribution fills in even though almost nothing in the population lived there. The hole closing is not a rounding artifact. It is the mechanism.
Almost no recipe lasted three or four days. But a sample holding two short recipes and three long ones averages right into that gap, and most samples of five are mixed. The middle fills because mixtures are common and extremes are not.
The bar width the lab specifies, half a day, is worth a moment's thought too, since it is half the width you used for the population. That is not an inconsistency: a distribution packed into a narrower stretch of the axis needs finer bins to show any structure at all, and with a spread near one day, one-day bins would leave you with about four bars total. The horizontal scale stays the same — same axis, same tick marks, same physical distance per day — while the bins get finer. Keeping the scale and changing the bin width is exactly the right combination, and confusing the two is the most common way this lab goes wrong. If you find yourself redrawing the axis rather than subdividing it, stop. One consequence worth expecting: the population histogram and the two histograms of averages will not end up with the same number of bars, and they are not supposed to. Same axis, finer bins, three pictures you can lay on top of one another and compare bar for bar at any given number of days. If instead you find that the three drawings only match when you slide the axis around, what you have built is three unrelated pictures rather than one comparison, and no amount of care in the arithmetic will rescue the conclusion from that.
Travis and his husband Owen are in the same section and check each other's predictions before they draw anything. Using \(\mu_X \approx 3.5667\) and \(\sigma_X \approx 2.1476\) from Try It Now B.4.2, state the theoretical distribution of \(\bar{X}\) for \(n = 5\) as they did. Then find the fraction of sample means the theorem predicts will land between 3 and 4 days, and compare it with the fraction of individual recipes that lasted 3 or 4 days.
Solution
Step 1 — write the distribution. The center is \(\mu_X\) and the spread is \(\sigma_X / \sqrt{5}\).
$$ \frac{\sigma_X}{\sqrt{n}} = \frac{2.1476}{\sqrt{5}} \approx \frac{2.1476}{2.2361} \approx 0.9604 $$ $$ \bar{X} \sim N(3.5667,\ 0.9604) $$Step 2 — convert 3 and 4 to \(z\)-scores against that spread.
$$ z_1 = \frac{3 - 3.5667}{0.9604} \approx -0.59 \qquad z_2 = \frac{4 - 3.5667}{0.9604} \approx 0.45 $$Step 3 — take the difference of the areas.
$$ P(3 < \bar{x} < 4) \approx 0.6736 - 0.2776 = 0.3960 $$Step 4 — count the individuals in the same interval. From Try It Now B.4.1, 3 recipes lasted three days and 5 lasted four days.
$$ \frac{3 + 5}{60} = \frac{8}{60} \approx 0.133 $$Answer: Travis and Owen get \(\bar{X} \sim N(3.5667, 0.9604)\), and about 40% of samples of five should average between 3 and 4 days, against only 13% of individual recipes that lasted 3 or 4 days. The same interval, on the same quantity, holds three times as much of the distribution of averages as it does of the population — and it does so because that interval was the population's emptiest region. Nothing moved into the middle; the middle is simply where mixtures land, and a sample of five is almost always a mixture. This one comparison is the lab's whole result in a single pair of numbers, and it is worth writing both of them in your report. It also tells you what to expect from your histogram: a clear peak somewhere near 3.5, and roughly two of every five of the class's means falling in the two bars that straddle it.
Try it in rāSHio
Distributions → Normal in rāSHio does that two-step table lookup in one: set the mean to 3.5667 and the standard deviation to 0.9604, ask for the area between 3 and 4, and read off about 0.396. Then change the standard deviation to 0.6791 and ask again — the same interval holds about 0.536 of the sample means once \(n = 10\).
Figure B.4.4 — Reading the area between two values off a normal curve in rāSHio: Distributions → Normal. The walkthrough uses its own mean and standard deviation; the steps are the ones you run with 3.5667 and 0.9604.
B.4.5 Repeat the Procedure for n = 10
The standard deviation of the sampling distribution of the sample mean:
$$ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} $$It measures how far a typical sample mean lands from the population mean \(\mu\). It is not the spread of the individuals — that is \(\sigma\) — and it shrinks toward zero as \(n\) grows, which is why a larger sample gives a more reliable estimate of \(\mu\).
- For the samples of \(n = 10\) days averaged together, construct a histogram of the averages — your means together with the means of the other groups. Make intervals with bar widths of \(\frac{1}{2}\) a day. Sketch the graph using a ruler and pencil, and scale the axes to match your first two histograms.
- Draw a smooth curve through the tops of the bars of the histogram. Use one to two complete sentences to describe the general shape of the curve.
The formula tells you in advance what the third picture should look like next to the second, and the answer is: the same, but narrower by a factor of \(\sqrt{2}\). Doubling the sample size from five to ten does not halve the spread — it divides it by \(\sqrt{10}/\sqrt{5} = \sqrt{2} \approx 1.41\), so a standard error near 0.96 becomes one near 0.68. That is a real improvement and a visible one on a shared axis, but it is much less than the doubling of work you paid for it, and knowing that exchange rate is what stops you from over-promising precision later in the course. It also sets your expectation for the shape. Going from \(n = 5\) to \(n = 10\) should make the bell a little more convincing, not perfectly convincing — this population is bimodal with an outlier at 11 days, and ten is still a small sample to lean the theorem on. Expect the third curve to be a better bell than the second, and the second to be a better bell than the population, without any of the three being textbook-perfect.
Precision costs \(\sqrt{n}\), not \(n\). To halve the spread you had at \(n = 5\), you would need samples of 20 — four times the data for twice the precision.
That square root is the reason sample sizes in published studies look the way they do. A study that wants to cut its margin of error in half has to quadruple its sample, and a study that wants one-tenth the margin needs a hundred times the data, which is why nobody promises a hundredfold improvement no matter how much money is available. It is also why a small increase in \(n\) at the low end buys so much more than the same increase further out: going from 5 to 10 removes about 29% of the spread, while going from 100 to 105 removes about 2%. If you have time and enough classes pooling data, drawing a fourth histogram at \(n = 20\) makes both halves of this point visible at once — the picture tightens again, and it tightens by less per person surveyed than the last one did. That exchange rate is also the reason this lab bothers with a second round at all instead of just drawing the first picture twice. Doubling a sample of five is the cheapest doubling available to you, and it is the last one that buys a visible amount of precision — every doubling after it costs more people for exactly the same proportional gain. Keep that in mind when you write up question 4, because the answer there is not simply that a bigger sample is better; it is that a bigger sample is better by a specific and shrinking amount you can predict in advance.
Alexis and their boyfriend Tomás are pooling their group's \(n = 10\) results. Find the standard error for \(n = 10\) and state the theoretical distribution of \(\bar{X}\) as they need it. Show that the ratio of the two standard errors, at \(n = 5\) and \(n = 10\), is \(\sqrt{2}\). Then find the sample size that would halve the \(n = 5\) standard error.
Solution
Step 1 — compute the standard error at \(n = 10\).
$$ \sigma_{\bar{x}} = \frac{2.1476}{\sqrt{10}} \approx \frac{2.1476}{3.1623} \approx 0.6791 $$ $$ \bar{X} \sim N(3.5667,\ 0.6791) $$Step 2 — take the ratio of the two standard errors.
$$ \frac{0.9604}{0.6791} \approx 1.414 \approx \sqrt{2} $$The \(\sigma_X\) cancels, leaving \(\sqrt{10}/\sqrt{5} = \sqrt{2}\) — the ratio does not depend on the population at all.
Step 3 — solve for the \(n\) that halves the \(n = 5\) standard error. We want \(\sigma_X/\sqrt{n} = 0.9604/2 = 0.4802\).
$$ \sqrt{n} = \frac{2.1476}{0.4802} \approx 4.472 \qquad n \approx 20 $$Answer: Alexis and Tomás find that at \(n = 10\) the standard error is about 0.68 days and \(\bar{X} \sim N(3.5667, 0.6791)\); the two standard errors differ by a factor of \(\sqrt{2}\); and halving the \(n = 5\) spread takes samples of 20. Read the last two answers together, because they are the same fact stated twice. Going from 5 to 10 gets you a factor of \(\sqrt{2}\), and doing it again — 10 to 20 — gets you another \(\sqrt{2}\), and two factors of \(\sqrt{2}\) multiply to 2. Every doubling of the sample size buys the same fixed 29% reduction, so the reductions compound rather than adding up, and the cost in people surveyed doubles each time while the payoff stays the same size. That is what "precision costs \(\sqrt{n}\)" means in practice, and it is worth stating in your report as a number rather than as a shape: their third histogram is narrower than their second by a factor they predicted before they drew it.
B.4.6 Discussion Questions
Answer these with your group, in complete sentences, pointing at your own three histograms rather than at what the theorem says ought to have happened.
- Compare the three histograms you have made — the one for the population and the two for the sample means. In three to five sentences, describe the similarities and differences.
- State the theoretical distributions, according to the central limit theorem, for the sample means.
a. \(n = 5\): \(\bar{X} \sim\) _________(_________, _________)
b. \(n = 10\): \(\bar{X} \sim\) _________(_________, _________)
- Are the sample means for \(n = 5\) and \(n = 10\) "close" to the theoretical mean, \(\mu_X\)? Explain why or why not.
- Which of the two distributions of sample means has the smaller standard deviation? Why?
- As \(n\) changed, why did the shape of the distribution of the data change? Use one to two complete sentences to explain what happened.
Question 3 is the one worth arguing about in your group, because "close" has no fixed meaning until you supply one. The honest way to answer it is to compare each gap against the standard error that applies to that sample size, rather than against nothing. A mean that sits 0.4 days away from \(\mu_X\) is well inside a standard error of 0.96 and is unremarkable at \(n = 5\), while the same 0.4 days is well over half a standard error at \(n = 10\) and is worth a comment. Notice that the yardstick tightens as \(n\) grows, so a fixed number of days is not equally close at both sizes. Question 5 has a trap in the other direction: the shape of the data did not change at all, because the population is the same sixty recipes it always was. What changed is which distribution you were drawing.
Lupe is writing up the discussion for her group. Answer questions 2 and 4 with her, for the given population, using \(\mu_X \approx 3.5667\) and \(\sigma_X \approx 2.1476\). Then say what you would expect to see if the class had also built averages of groups of 30 instead of stopping at 10.
Solution
Step 1 — answer question 2 for both sample sizes. The family is normal, the center is \(\mu_X\) in both cases, and only the spread changes.
$$ n = 5: \quad \bar{X} \sim N\!\left(3.5667,\ \frac{2.1476}{\sqrt{5}}\right) = N(3.5667,\ 0.9604) $$ $$ n = 10: \quad \bar{X} \sim N\!\left(3.5667,\ \frac{2.1476}{\sqrt{10}}\right) = N(3.5667,\ 0.6791) $$Step 2 — answer question 4. The \(n = 10\) distribution has the smaller standard deviation, 0.68 against 0.96, because the standard error is \(\sigma_X/\sqrt{n}\) and only the denominator changed. A larger sample gives every extreme value more companions to be averaged against, so extreme averages get rarer and the distribution packs in tighter around \(\mu_X\).
Step 3 — extend to \(n = 30\).
$$ \frac{2.1476}{\sqrt{30}} \approx \frac{2.1476}{5.4772} \approx 0.3921 $$Answer: Lupe finds both distributions centered at 3.5667 with spreads of 0.9604 and 0.6791; the \(n = 10\) one is narrower, by a factor of \(\sqrt{2}\), because \(n\) sits under a square root in the denominator of the standard error. At \(n = 30\) the spread would fall to about 0.39, less than a fifth of the population's 2.15, and the bell would be the most convincing of the four — 30 is the rule-of-thumb sample size at which the theorem is usually treated as reliable even for a badly behaved population, and this population, bimodal with an outlier at 11 days, is a fair test of that rule. Two things would stay put no matter how large \(n\) got. The center would still be 3.5667, because averaging never moves the middle. And the population would still be bimodal, because nothing you do to your samples changes the sixty recipes. The bell you are watching appear belongs to the averages, not to the cookies.
Key Terms
central limit theorem — the result that the sampling distribution of the sample mean approaches \(N(\mu, \sigma/\sqrt{n})\) as the sample size \(n\) grows, whatever the shape of the population.
population mean — the mean of every value in the population, \(\mu = \sum x / N\), computed by dividing by the population size \(N\).
population standard deviation — the standard deviation of every value in the population, \(\sigma\), computed by dividing the squared deviations by \(N\) rather than by \(n - 1\).
sampling distribution of the sample mean — the distribution formed by the sample means of samples of a fixed size \(n\) drawn from a population, written \(\bar{X}\).
standard error of the mean — the standard deviation of the sampling distribution of the sample mean, \(\sigma_{\bar{x}} = \sigma/\sqrt{n}\).