Comparing Random Samples and Justifying Inferences | Grade 7 Math | LevelKite

Suppose you want to know how many students in a school bring lunch from home, but you cannot ask everyone. The whole group you want to know about is the population. The part you actually collect data from is a sample. In a random sample, every member of the population has the same chance of being chosen.

From a sample you calculate an estimate: a value for the population worked out from the sample data. You ask one random sample and get one estimate. A friend asks a different random sample from the same school and gets a different estimate. Who is right?

You already know that estimates vary from sample to sample. This lesson takes two more steps:

You will compare several estimates and say what they tell you together.
You will explain why random sampling makes that conclusion worth trusting.

Take a school with $N = 600$ students. Here $N$ is the number of members in the population. A random sample has $n = 50$ students. Here $n$ is the number of members in the sample, and $n$ is greater than $0$ . Each sampled student answers the question "Do you bring lunch from home?" Let $c$ be the number of sampled students who say yes, a whole number from $0$ to $n$ .

The fraction $\frac{c}{n}$ is the sample proportion. A sample is representative when it resembles the population in the ways that matter for the question. When a sample is random and representative, its proportion is close to the population's proportion. A fraction of a group of $N$ students is that fraction times $N$ , so you multiply the sample proportion by $N$ . Call the result $E$ , the estimated number of students in the school who bring lunch from home. It is an estimate, not an exact count. Here is that rule in symbols, followed by five random samples from this school.

E = \frac{c}{n} \times N

Five random samples of 50 students from the school of 600: who brings lunch from home?

Sample	Yes answers out of 50	Sample proportion	Estimate for 600 students
1	$14$	$\frac{14}{50} = \frac{28}{100} = 0.28$	$0.28 \times 600 = 168$
2	$17$	$\frac{17}{50} = \frac{34}{100} = 0.34$	$0.34 \times 600 = 204$
3	$12$	$\frac{12}{50} = \frac{24}{100} = 0.24$	$0.24 \times 600 = 144$
4	$16$	$\frac{16}{50} = \frac{32}{100} = 0.32$	$0.32 \times 600 = 192$
5	$16$	$\frac{16}{50} = \frac{32}{100} = 0.32$	$0.32 \times 600 = 192$

The five samples in the table are all random, yet their estimates are not all the same. That is expected, because chance decides who lands in each sample. You cannot tell which estimate is closest to the true count.

To compare estimates from random samples of the same size, ask two questions:

Where do they cluster? The mean (the sum of the estimates divided by how many there are) gives a center.
How far apart are they? The range (the greatest estimate minus the least) gives the spread.

The center suggests a value for the population. The spread shows how much an estimate can change from one sample to the next. A conclusion about a population drawn from sample data is called an inference. It is valid only when the samples are random and representative, and it stays approximate because estimates vary from sample to sample.

Worked example

Comparing five estimates

Five random samples of $50$ students from a school of $600$ students found $14$ , $17$ , $12$ , $16$ and $16$ students who bring lunch from home. The estimates for the whole school are $168$ , $204$ , $144$ , $192$ and $192$ students. What do the five estimates tell you together?

Find the least and the greatest estimate, 144 and 204, and subtract to get the range. The range measures how much the estimate changed from one random sample to another.
$204 - 144 = 60$
Find the mean: add the five estimates and divide by 5, the number of estimates. Adding in order gives 372, then 516, then 708, then 900. The mean marks the center of the estimates.
$\frac{168 + 204 + 144 + 192 + 192}{5} = \frac{900}{5} = 180$
Check the mean another way. The five samples hold 5 times 50, or 250, answers, and 14 + 17 + 12 + 16 + 16 = 75 of them are yes. Because every sample has the same size, the combined proportion gives the same estimate as the mean.
$\frac{75}{250} \times 600 = \frac{3}{10} \times 600 = 180$
Compare each estimate with the mean by subtracting 180. A negative result means the estimate is below the mean, and a positive result means it is above. The farthest estimate is 36 students from the mean.
$168-180=-12,\ 204-180=24,\ 144-180=-36,\ 192-180=12,\ 192-180=12$
State the inference in approximate words. The samples are random, so their estimates tend to cluster around the true count. They still differ from each other by up to 60 students, so the data support "about 180", not an exact number.

Answer

Together the five random samples suggest that about $180$ of the $600$ students bring lunch from home. The estimates ran from $144$ to $204$ , so the true count could be a few dozen above or below $180$ .

Worked example

Is the lead bigger than the variation?

A school of $600$ students will elect a president. Each student will vote for exactly one of two candidates, Ana or Ben. The student council asks five random samples of $50$ students each how they plan to vote. The numbers who choose Ana are $27$ , $24$ , $29$ , $26$ and $24$ . Do the samples show who is ahead?

Estimate Ana's votes from each sample. The estimate is the count for Ana, divided by 50, times 600. Dividing by 50 and then multiplying by 600 is the same as multiplying by 600 divided by 50, which is 12. So each sampled student stands for 12 students, and you multiply each count by 12.
$12 \times 27 = 324,\ 12 \times 24 = 288,\ 12 \times 29 = 348,\ 12 \times 26 = 312,\ 12 \times 24 = 288$
Compare each estimate with the halfway mark. Half of 600 is 300, so Ana is ahead only if more than 300 students choose her. Subtract 300 from each estimate: a positive result puts Ana ahead in that sample, and a negative result puts her behind.
$324-300=24,\ 288-300=-12,\ 348-300=48,\ 312-300=12,\ 288-300=-12$
Find the mean of the five estimates to locate their center. Adding in order gives 612, then 960, then 1272, then 1560.
$\frac{324 + 288 + 348 + 312 + 288}{5} = \frac{1560}{5} = 312$
Find the range to see how much the estimates vary from one random sample to another.
$348 - 288 = 60$
Compare the lead with the variation. The mean puts Ana only 12 votes above the halfway mark, but the estimates differ from each other by up to 60 votes and fall on both sides of 300. A lead that small could come from chance in who was sampled, so it does not settle the question.
$312 - 300 = 12$

Answer

The random samples lean slightly toward Ana, with a mean estimate of about $312$ of $600$ votes. But the estimates run from $288$ to $348$ and fall on both sides of $300$ , so the election is too close to call from these samples.

Both examples relied on random samples. Here is why that matters.

Multiplying a sample proportion by $N$ gives a sound estimate only when the sample proportion is close to the population's proportion. In other words, the sample has to be representative.

Random selection gives every member of the population the same chance of being chosen. The method favors no group and leaves no group out. So each group tends to appear in the sample in about the same share as in the population, which means a random sample tends to be representative. It need not match the population exactly, because chance decides who is picked. But chance has no favorite direction, so random samples tend to land a little high sometimes and a little low at other times.

A biased method gives some members a better chance of being chosen than others. When those members differ from the rest on the trait you are counting, the samples tend to miss in the same direction, and repeating the method does not remove the miss.

Key concept

Random samples make inferences valid, within limits

An inference about a population is valid only when it comes from a random, representative sample. Random selection gives every member the same chance of being chosen, so a random sample tends to resemble the population and its estimate tends to land near the true value.

Estimates still vary from sample to sample. So compare several random samples of the same size by their center (where the estimates cluster) and their spread (how far apart they are), and state the conclusion as an approximate value, not an exact one.

You can test this reasoning on a question whose true answer is already known. A school has $N = 600$ students, and its records show that exactly $150$ of them borrowed a library book this month. Call these students borrowers.

A class uses two methods, five times each, to choose $n = 50$ students:

Random: draw $50$ names from the list of all $600$ students, so every student has the same chance.
Library at lunch: ask $50$ students who are in the library at lunch.

Let $c$ be the number of borrowers in a sample. Each estimate is $E = \frac{c}{50} \times 600 = 12c$ students, because dividing by $50$ and then multiplying by $600$ is the same as multiplying by $12$ . In a real survey you would not know the true count. Here you do, so you can measure how far off each method is. The table shows what each method found.

Test on a known answer: exactly 150 of the 600 students are borrowers

Sample	Random: borrowers out of 50	Random: estimate	Library: borrowers out of 50	Library: estimate
1	$11$	$12 \times 11 = 132$	$32$	$12 \times 32 = 384$
2	$14$	$12 \times 14 = 168$	$34$	$12 \times 34 = 408$
3	$12$	$12 \times 12 = 144$	$33$	$12 \times 33 = 396$
4	$15$	$12 \times 15 = 180$	$34$	$12 \times 34 = 408$
5	$13$	$12 \times 13 = 156$	$32$	$12 \times 32 = 384$

Worked example

Testing two methods against a known answer

Exactly $150$ of the $600$ students borrowed a library book this month. The five random samples gave the estimates $132$ , $168$ , $144$ , $180$ and $156$ . The five library-at-lunch samples gave $384$ , $408$ , $396$ , $408$ and $384$ . Which method supports a valid inference about the school, and why?

Define the error of an estimate as the estimate minus the true count of 150, in students. A positive error means the estimate is too high, and a negative error means it is too low. Find the errors of the random samples.
$132-150=-18,\ 168-150=18,\ 144-150=-6,\ 180-150=30,\ 156-150=6$
Find the mean error of the random samples: add the five errors in order, then divide by 5. The errors have both signs, so they partly cancel. A mean error of 6 means the random estimates average 150 + 6 = 156 students, close to the true count.
$-18+18=0,\ 0-6=-6,\ -6+30=24,\ 24+6=30,\ \frac{30}{5}=6$
Find the errors of the library samples in the same way. Every error is positive, so nothing cancels.
$384-150=234,\ 408-150=258,\ 396-150=246,\ 408-150=258,\ 384-150=234$
Find the mean error of the library samples. Adding the errors in order gives 492, then 738, then 996, then 1230. A mean error of 246 means the library estimates average 150 + 246 = 396 students, far above the true count.
$\frac{234 + 258 + 246 + 258 + 234}{5} = \frac{1230}{5} = 246$
Compare sample proportions with the school's proportion of borrowers, which is 150 out of 600, or 0.25. The random samples had 11 to 15 borrowers out of 50, close to 0.25. The library samples had 32 to 34 borrowers out of 50, far above 0.25.
$\frac{150}{600}=\frac{1}{4}=0.25,\quad \frac{11}{50}=0.22,\quad \frac{15}{50}=0.30,\quad \frac{32}{50}=0.64,\quad \frac{34}{50}=0.68$
Give the reason. Drawing names from the full list gives borrowers and non-borrowers the same chance, so a sample tends to hold them in about the school's shares. In the library at lunch, students who borrow books are more likely to be present, so the samples over-count them and the estimates are pushed up.

Answer

Only the random method supports a valid inference. Its estimates were all within $30$ students of the true count of $150$ , some too high and some too low. Every library estimate was more than $230$ too high. Even the random estimates are approximate, not exact.

A common mistake: trusting the samples that agree. The library estimates have a range of $408 - 384 = 24$ students. The random estimates have a range of $180 - 132 = 48$ students. It is tempting to think: "The random samples disagree more, so random sampling is less reliable. I will say about $396$ students." That answer is $246$ too high. Agreement shows only that a method gives similar results each time. It cannot show that the method is fair: the library samples agree with each other, yet they all over-count the same group. Variation between random samples is not a fault. It comes from chance, and comparing the samples lets you see how large it is. An inference is valid only from random, representative samples, and even then it stays approximate.

Takeaway. Estimates from random samples of the same size vary from sample to sample, and you cannot tell which single one is closest to the true value. So compare them: find where the estimates cluster and how far apart they are, then state your conclusion as "about" a value, not as an exact number. You can trust that conclusion only because the samples are random. Random selection gives every member of the population the same chance, so the samples tend to be representative, and their misses come from chance instead of from a method that favors one group.