SAT sample statistics and margin of error

What the questions look like

In these questions, someone measures a random part of a big group to estimate something about the whole group. Here's a typical setup. A parks department has 1,800 benches. It checks 150 of them at random and finds peeling paint on 30% of those. From that, it estimates that 30% of all 1,800 benches have peeling paint, with a margin of error of 7 percentage points.

The population is all 1,800 benches. The sample is the 150 that got checked.
The estimate is 30%, the best guess for all 1,800. An estimate can also be a mean (an average), like a mean delivery time of 28 minutes.
The margin of error, 7 points, says how far the estimate could reasonably be from the true value, in either direction.

So the plausible values for the true share of all 1,800 benches run from 23% to 37%. A plausible value is one the sample can't rule out.

The last line of the question usually reads like one of these:

“Which of the following is a plausible statement about the mean … for all …?”
“Which of the following is the most appropriate conclusion?”
“What is the margin of error?”
“Which of the following is the best estimate of the number of … who …?”
“Which of the following is a plausible value for the total number of people in the population who …?”
“Which of the following would be the most likely effect of using the larger random sample compared to the smaller random sample?”
“Which of the following is true about the margin of error associated with the estimate from the larger sample?”

Before you work anything out, find the population in the question. A right conclusion is about that whole group.

Desmos won't help here, since there's nothing to graph. The arithmetic is short, and the calculator is worth opening only to multiply a big population by a share.

Plausible values from the estimate and margin

Back to the benches: an estimate of 30% for all 1,800 benches, with a margin of error of 7 percentage points.

Low end: subtract the margin from the estimate.
$30-7=23$
High end: add the margin to the estimate.
$30+7=37$
The plausible values for the share of all 1,800 benches.
$23\%\text{ to }37\%$

The full margin goes on each side of the estimate.

Right: An estimate of 30% with a margin of 7 points gives plausible values from 23% to 37%.
Wrong: An estimate of 30% with a margin of 7 points gives plausible values from 26.5% to 33.5%.

That splits the margin and puts 3.5 on each side. The whole 7 goes on each side, so the range is 14 points wide.

A margin on a percent is counted in percentage points. A percentage point is one step on the percent scale, so going from 30% to 37% is a rise of 7 points. Questions often print the margin with a percent sign, as in “a margin of error of 7%.” When the estimate is a percent, read that as 7 points.

Trap: A percent of the percent

27.9% to 32.1% comes from taking 7% of 30, which is 2.1, and using 2.1 as the margin.

What to do: Use the margin exactly as printed. An estimate of 30% and a margin of 7% give 23% and 37%.

Pick the conclusion about the whole group

A pizza shop made 1,200 deliveries last month. The owner picked 36 of them at random and looked up how long each one took. Based on the sample, she put the mean delivery time for all 1,200 deliveries at 28 minutes, with a margin of error of 3 minutes.

Low end: subtract the margin.
$28-3=25$
High end: add the margin.
$28+3=31$

Each wrong choice below changes one part of the right conclusion.

The range describes the population.

Right: It's plausible that the mean time for all 1,200 deliveries was between 25 and 31 minutes.
Wrong: It's plausible that the mean time for the 36 sampled deliveries was between 25 and 31 minutes.

The owner looked up those 36 times, so she knows their mean exactly. It's 28. The range is about all 1,200 deliveries, and she checked only 36 of them.

Trap: Every single delivery

“Every delivery took between 25 and 31 minutes” keeps the right ends but applies them to each delivery. One delivery could take 12 minutes and another 50, and the mean could still be 28.

What to do: Ask which one number the range is for. Here it's the mean, so cross out any choice that puts each delivery inside the range.

Trap: One exact number

“The mean time for all 1,200 deliveries was exactly 28 minutes” repeats the estimate and throws the margin away.

What to do: If the question gives a margin of error, the answer has a range in it.

Trap: Past one end

“It's plausible that the mean time was less than 25 minutes” uses one of the real ends, but it points away from the range. Values below 25 are outside what the sample supports.

What to do: Write both ends before you read the choices, then look for the choice that has both.

A choice may also object to the sample itself.

A random sample supports an estimate for the whole population, including the members nobody measured.

Right: The 36 random deliveries give plausible values for the mean of all 1,200.
Wrong: No conclusion is possible, because only 36 of the 1,200 deliveries were checked.

Picking at random is what lets a sample stand in for the whole group. A small sample shows up as a wider margin of error, and the question already gave you the margin.

Here the question tells you the sample was random. Questions about how a study was set up, and which group its results apply to, are in evaluating statistical claims.

Worked example

Otto manages a farm with an irrigation schedule for $64$ fields. He selected $14$ fields at random and recorded the amount of water scheduled for each field. From this sample, Otto estimated that the mean scheduled amount of water per field for all $64$ fields was $84$ cubic meters, with an associated margin of error of $6$ cubic meters.

Using this estimate and margin of error, which of the following is the most appropriate conclusion about the mean for all $64$ fields?

A
It is plausible that the mean scheduled amount per field was between $78$ and $90$ cubic meters.
B
The estimate shows that the mean scheduled amount per field was exactly $84$ cubic meters.
C
The estimate shows that every field was scheduled to receive between $78$ and $90$ cubic meters.
D
The estimate shows that the mean scheduled amount per field was between $81$ and $87$ cubic meters.

Show the answer

A
It is plausible that the mean scheduled amount per field was between $78$ and $90$ cubic meters.
Correct answer
Subtract and add the to the estimate: $84-6=78$ and $84+6=90$ . The resulting interval gives plausible values for the mean scheduled amount across all $64$ fields.
B
The estimate shows that the mean scheduled amount per field was exactly $84$ cubic meters.
The estimate of $84$ cubic meters does not establish the exact population mean. The supports an interval of plausible values.
C
The estimate shows that every field was scheduled to receive between $78$ and $90$ cubic meters.
The estimate and concern the mean across all $64$ fields. They do not establish the scheduled amount for every individual field.
D
The estimate shows that the mean scheduled amount per field was between $81$ and $87$ cubic meters.
This halves the stated from $6$ to $3$ cubic meters, producing $84-3=81$ and $84+3=87$ . The estimate supports an interval from $78$ to $90$ cubic meters; it does not establish the narrower interval.

Solution

The estimated population mean is $84$ cubic meters, and the is $6$ cubic meters. Subtract the margin to find the lower endpoint: $84-6=78$ . Add it to find the upper endpoint: $84+6=90$ .

Therefore, the estimate supports choice A: it is plausible that the mean scheduled amount per field for all $64$ fields was between $78$ and $90$ cubic meters.

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Finding the margin from the two ends

Sometimes you get the range and have to work back. A garden center measured tomato seedlings picked at random from its greenhouse. It reports that plausible values for the mean height of all the seedlings in the greenhouse run from 31.4 to 34.2 centimeters. What's the margin of error?

Width of the range: high end minus low end.
$34.2-31.4=2.8$
Halve it. The margin reaches from the middle to one end, so it's half the width.
$2.8\div2=1.4$
The estimate is the middle: add the ends and halve.
$\frac{31.4+34.2}{2}=32.8$
Check the low end: estimate minus margin.
$32.8-1.4=31.4$
Check the high end: estimate plus margin.
$32.8+1.4=34.2$

The margin is 1.4 centimeters.

Trap: The whole width

2.8 is the whole width. Used as the margin, it would put the ends at 30.0 and 35.6. A choice of 0.7 halves the width twice, and 32.8 is the estimate.

What to do: Halve the width, then check that the middle minus and plus your margin gives both ends.

Estimating a count

A hotel has 480 rooms. Staff check 60 rooms picked at random and find that 15 of them need a new lamp. What's the best estimate of the number of rooms in the hotel that need one?

The sample's share: the count in the sample over the sample size.
$\frac{15}{60}=0.25$
Multiply the whole hotel by that share.
$480\times0.25=120$

Best estimate of a count

\text{population size}\times\frac{\text{count in the sample}}{\text{sample size}}

Multiply the population size by the sample's share, a number from 0 to 1.

Right: The best estimate is 480 × 0.25 = 120 rooms.
Wrong: The best estimate is 480 × 15 = 7,200 rooms.

15 is a count from the sample. The share, 0.25, is what carries over to the whole hotel. And 7,200 is more rooms than the hotel has.

Trap: The sample's own count

15 is a choice because it's the count among the 60 rooms that got checked. The question asks about all 480.

What to do: Check which group the question asks about, then multiply that group's size by the share.

If the question gives the share as a percent, write it as a decimal first: 25% of 480 is $0.25\times480=120$ .

A plausible count

When the last line asks for a plausible value instead of the best estimate, you need the whole range. Back to the benches: an estimate of 30% of all 1,800 benches, with a margin of 7 percentage points. Which of these is a plausible number of benches with peeling paint?

A. 45
B. 126
C. 600
D. 700

The percent range: 30 minus 7 and 30 plus 7.
$23\% \text{ to } 37\%$
Scale the low end to all 1,800 benches.
$1{,}800\times0.23=414$
Scale the high end.
$1{,}800\times0.37=666$
Only C falls inside the range.
$414\le600\le666$

C is correct. The best estimate, $1{,}800\times0.30=540$ , isn't a choice, and you don't need it to be. Any count from 414 to 666 is plausible.

Trap: Counts that miss the range

45 is 30% of the 150 benches in the sample. 126 is 7% of 1,800, the margin used as if it were the share. 700 is above the high end, 666.

What to do: Multiply the population by a percent from the range, then check that your answer sits between the two scaled ends.

Worked example

Noor manages a floral studio that prepared $900$ custom bouquets during a monthlong promotion. She selected $750$ of the bouquets at random and reviewed their prices.

From the sample, Noor estimated that $62\%$ of all $900$ bouquets cost at least $50, with an associated margin of error of $4$ percentage points.

Which of the following is a plausible value for the total number of the $900$ bouquets that cost at least $50?

A
$465$
B
$36$
C
$558$
D
$630$

Show the answer

A
$465$
This applies the estimated $62\%$ to the sample of $750$ : $0.62(750)=465$ . The question asks for the total among all $900$ bouquets.
B
$36$
This multiplies the $4\%$ by the population: $0.04(900)=36$ . That is an error count, not the estimated number of bouquets costing at least $50.
C
$558$
Correct answer
Apply the estimated to all $900$ bouquets: $0.62(900)=558$ . The estimate is within the interval from $0.58(900)=522$ to $0.66(900)=594$ , so $558$ is plausible.
D
$630$
This treats the full interval width of $8$ as the , producing $62\%+8\%=70\%$ and $0.70(900)=630$ . The given margin of error is $4$ percentage points.

Solution

The estimated percentage is $62\%$ , with a of $4$ . The plausible percentage range is $62\%-4\%=58\%$ through $62\%+4\%=66\%$ . Apply this range to the population of $900$ bouquets:

$0.58(900)=522$

$0.66(900)=594$ A plausible total must therefore be between $522$ and $594$ . The point estimate is $0.62(900)=558$ , which lies in this interval.

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How many more

A library picks 200 of its 4,500 patrons at random and asks which of two new reading-room layouts they prefer. 110 choose layout A, 70 choose layout B, and 20 have no preference. About how many more of all 4,500 patrons would be expected to prefer layout A than layout B?

The gap in the sample.
$110-70=40$
Turn the gap into a share of the whole sample, all 200 patrons asked.
$\frac{40}{200}=0.2$
Scale that share to all 4,500 patrons.
$0.2\times4{,}500=900$

Find the gap as a share of the whole sample, then scale it to the population once.

Right: About 0.2 × 4,500 = 900 more patrons would be expected to prefer layout A.
Wrong: About (40 ÷ 180) × 4,500 = 1,000 more patrons would be expected to prefer layout A.

180 leaves out the 20 patrons with no preference. The 4,500 includes people like them, so the share has to be out of all 200 patrons asked.

Trap: A share read as a count

20 is the gap in percentage points, since 40 out of 200 is 20%. That's a share of the sample, and the question asks how many patrons.

What to do: Check what the last line counts. It asks for patrons, so multiply the share by 4,500.

A poll asks the same thing in election words, “If … people vote, by how many votes would … be expected to win?”, and you work it the same way.

What a bigger sample changes

A transit agency surveys 320 riders picked at random. It estimates that 58% of all riders want later trains, and its margin of error is 5 percentage points. Next month it runs the same survey on a new random sample of 640 riders and works out the margin the same way. What's the most likely effect of the larger sample?

When the margin is worked out the same way, a larger random sample most likely gives a smaller margin of error.

Right: The new margin of error is most likely less than 5 percentage points.
Wrong: The new margin of error is exactly 2.5 percentage points.

Doubling the sample usually makes the margin smaller, but nothing in the question tells you by how much, so you can't pick an exact new value.

The margin also depends on things other than sample size, such as how spread out the answers are. That's why the question says the margin was worked out the same way, or asks for the “most likely” effect.

Trap: Sample size pointed the wrong way

“The margin will grow, because more riders bring more unusual answers” gets the direction backward. “The estimate will rise above 58%” predicts something nobody can know before the survey runs.

What to do: Pick the choice that says the margin gets smaller. The new estimate could land above or below 58%, so cross out any choice that says which.

Worked example

A gym keeps records of the number of towels washed in each laundry load. Luka selected $150$ loads at random from these records and estimated that the mean number of towels per load across all the recorded loads was $64$ , with an associated margin of error of $6$ .

For a second study, Luka selected $300$ loads at random from the same records. The two samples had the same standard deviation of towel counts. Both studies used the same estimation method and confidence level.

Which of the following would be the most likely effect of using the larger sample?

A
The margin of error would increase because more extreme towel counts would be included.
B
The estimated mean number of towels per load would increase.
C
The margin of error would be less than $6$ .
D
The margin of error would be exactly $3$ .

Show the answer

A
The margin of error would increase because more extreme towel counts would be included.
This assumes that including more extreme towel counts in a larger random sample increases the . A larger random sample generally gives a more precise estimate of the population mean, even if it includes more extreme counts.
B
The estimated mean number of towels per load would increase.
This assumes that increasing the sample size increases the estimated mean. The new estimate could be above or below $64$ ; a larger sample does not determine the direction of that change.
C
The margin of error would be less than $6$ .
Correct answer
With the same sample , estimation method, and confidence level, increasing the random sample from $150$ to $300$ loads would most likely make the less than $6$ .
D
The margin of error would be exactly $3$ .
This halves the from $6$ to $3$ because the sample size doubles. Doubling the sample size does not imply that the margin of error is cut in half.

Solution

The sample size increases from $150$ to $300$ randomly selected loads from the same records. The two samples have the same of towel counts.

Because the sample standard deviation, estimation method, and confidence level stay the same, the would most likely decrease from $6$ . Doubling the sample size does not establish an exact new margin of error, so the supported conclusion is that it would be less than $6$ .

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Practice

Before you start each one, find the population and decide what the last line wants: a range, a margin, a count, a gap, or what a bigger sample does.

Try it

5 questions on this skill, with every choice explained.

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Related lessons

Inference from sample statistics and margin of error

What the questions look like

Plausible values from the estimate and margin

Pick the conclusion about the whole group

Finding the margin from the two ends

Estimating a count

A plausible count

How many more

What a bigger sample changes

Practice