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Module 10 · Sampling 2 of 5 ~40 min ⚡ +90 XP available

The Sample Mean

Lesson 21 left you with a sample and an unknown population. Now you compute something from that sample. The sample mean $\bar{x}$ is the obvious thing to compute and the arithmetic takes one line, so the whole lesson turns on the second idea: draw a fresh sample the same size and you get a different $\bar{x}$ . Not because anyone made a mistake. That variation is not noise to be apologised for, it is the object the next three lessons study.

Today's hook, Two students each draw a random sample of $40$ from the same population and compute the sample mean. One gets $6.1$ , the other gets $6.9$ . Before reading on, write down who made the error, and how you would decide.
0/5QUESTS
01
Recall, your gut answer first

Write your answer before you read on.

Two students each draw a random sample of $40$ from the same population and compute the sample mean. One reports $6.1$ , the other reports $6.9$ . Who made the arithmetic error, and how would you decide?

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02
The two moves for this lesson
  1. Compute $\bar{x}$ . Add the sample values, divide by how many there are. One line, and it is the same arithmetic you have done since Year 7.
  2. Ask what a different sample would have given. This is the move nobody makes unprompted, and it is the entire point of the lesson.
Move 1 is arithmetic. Move 2 is the mathematics. Everything from here to the central limit theorem is built on taking move 2 seriously: $\bar{x}$ is not a fact about the population, it is a fact about the sample you happened to draw.
03
What you'll master
  • Define the sample mean for a sample drawn from a population, and compute it.
  • Use the correct notation, $\bar{x}$ for the sample mean against $\mu$ for the population mean.
  • Recognise that sample means obtained from repeated sampling may be different , even when every sample is the same size.
  • Explain why two different sample means are not evidence that anybody made a mistake.
04
Key terms
Sample mean ($\bar{x}$)The arithmetic mean of the values in one sample. Like this: a sample of $3, 4, 6, 7$ has $\bar{x} = 20 \div 4 = 5$ .
$\sum$ notationA shorthand for "add these up". Like this: $\sum x_i$ means $x_1 + x_2 + \cdots + x_n$ , so for $3, 4, 6, 7$ it is $20$ .
Repeated samplingDrawing more than one sample from the same population. Like this: four different groups of four students, all drawn from the same class.
Sampling variabilityThe fact that a statistic changes from sample to sample. Like this: four samples of the same size giving means of $5$ , $6$ , $8$ and $11.75$ .
05
Defining the sample mean
core concept

For a sample of $n$ values $x_1, x_2, \ldots, x_n$ drawn from a population, the sample mean is

$$\bar{x} = \frac{x_1 + x_2 + \cdots + x_n}{n} = \frac{1}{n}\sum_{i=1}^{n} x_i$$

Read it as: total the sample values, divide by how many you have. The bar over the $x$ is doing important work, and it is not decoration.

SymbolNameBelongs toKnown?
$\bar{x}$sample meanone samplealways, you computed it
$\mu$population meanthe whole populationgenerally not

A worked computation. Five students record their weekly paid work hours: $12, 15, 9, 14, 20$ . Then $n = 5$ and

$$\bar{x} = \frac{12 + 15 + 9 + 14 + 20}{5} = \frac{70}{5} = 14 \text{ hours}$$
Units and rounding. A sample mean carries the units of the data, so $14$ is $14$ hours, not $14$ . And do not round mid-calculation: divide the exact total, then round once at the end to the accuracy the question asks for.

The sample mean of a sample of n values is x̄ = (x₁ + x₂ + ... + xₙ)/n, that is (1/n)Σxᵢ. It is computed from one sample, so it is always known. The population mean μ is a different quantity and is generally unknown.

Pause, copy the formula in both forms, and write beside it which of $\bar{x}$ and $\mu$ you can always compute and which you generally cannot.

Quick check: A sample of six values totals $92.4$ . What is $\bar{x}$ , to one decimal place?

06
Repeated samples give different means
core concept

Here is a population small enough to see all of it, which is a luxury real sampling never gives you. Eight students, and their weekly paid work hours:

Hours1346791024

The eight values total $64$ , so the population mean is $\mu = 64 \div 8 = 8$ hours. Because this population is tiny we know $\mu$ exactly, which lets us do something you can never do in practice: check the samples against the truth .

Now take four different samples, every one of them of size $n = 4$ :

SampleValuesTotal$\bar{x}$
S1$3, 4, 6, 7$$20$$20 \div 4 = 5$
S2$1, 4, 9, 10$$24$$24 \div 4 = 6$
S3$1, 3, 4, 24$$32$$32 \div 4 = 8$
S4$6, 7, 10, 24$$47$$47 \div 4 = 11.75$

Four samples, one population, identical sample size, and four answers: $5$ , $6$ , $8$ and $11.75$ . Nobody made a mistake. Every division above is correct, and every sample is a legitimate sample of size $4$ .

  • S3 landed exactly on $\mu = 8$ . That is possible, and it is luck. Nothing about S3's method was better.
  • S4 came out at $11.75$ , nearly $50\%$ high , because it happened to catch the extreme value $24$ alongside three moderate values.
  • S1 missed the $24$ entirely and came out low at $5$ .
This answers card 01. Two students getting $6.1$ and $6.9$ from samples of $40$ may both have done the arithmetic correctly. Different samples give different means, and the disagreement alone is not evidence of an error. Asking "who was wrong?" is asking the wrong question about the wrong object.
Which one is the right answer? None of them, and all of them. Each $\bar{x}$ is a legitimate estimate of the same fixed $\mu$ . You cannot tell from a single sample how close you landed, because in any real problem $\mu$ is exactly the thing you do not know. What you can study is how $\bar{x}$ behaves across all possible samples, and that is the next lesson.

Sample means obtained from repeated sampling may be different, even when every sample is the same size. That variation is not an error: each sample contains different members. One sample mean may land exactly on μ by chance, and from a single sample you cannot tell how close you are.

Pause, copy the four-sample table with its four different means, and beside it the sentence "different samples give different means, and that is not a mistake".

True or false: If two random samples of the same size from the same population give different sample means, at least one of them must contain a calculation error.

PROBLEM 1 · COMPUTE AND NAME

A random sample of $8$ delivery times, in minutes, is $22, 31, 27, 19, 35, 28, 24, 30$ . Find the sample mean, and state what it estimates. (2 marks)

1
$\sum x_i = 22 + 31 + 27 + 19 + 35 + 28 + 24 + 30 = 216$
Total the sample values first, and keep the total exact rather than rounding anything yet.
PROBLEM 2 · TWO SAMPLES DISAGREE

From the eight-student population above ($1, 3, 4, 6, 7, 9, 10, 24$ hours, $\mu = 8$), sample A is $3, 4, 6, 7$ and sample B is $6, 7, 10, 24$ . Compute both sample means and explain the difference to someone who thinks one of them is wrong. (3 marks)

1
$\bar{x}_A = \dfrac{3 + 4 + 6 + 7}{4} = \dfrac{20}{4} = 5$ hours
Sample A drew the four smallest-but-one values and missed the 24 entirely.
PROBLEM 3 · WORK BACKWARDS

A random sample of $12$ households has a mean weekly recycling weight of $6.5$ kg. (a) Find the total weight for the sample. (b) A thirteenth household, recycling $19$ kg, is added. Find the new sample mean, and comment. (3 marks)

1
$\sum x_i = n \bar{x} = 12 \times 6.5 = 78$ kg
Rearranging the definition: if x̄ = Σx/n then Σx = n x̄. Being able to run the formula backwards is worth a mark on its own.

Complete: A sample of $n$ values has sample mean $\bar{x}$ . The total of the sample values is therefore , and repeated samples of the same size may give sample means.

Trap 01
Reading disagreement as error
Two samples, two means, and the instinct is that somebody divided wrongly. Different samples contain different members, so different means are the expected outcome, not the surprising one. A question asking you to "explain the difference" wants sampling variability named, not arithmetic rechecked.
Trap 02
Averaging the wrong thing when a value is added
To add a value to a sample, add it to the total and increase $n$ by one. Averaging the old mean with the new value, $(6.5 + 19) \div 2$ , treats one household as equal in weight to twelve and is wrong by a wide margin.
Trap 03
Dropping the bar, or the units
Writing $x = 27$ for a sample mean loses the distinction the notation exists to carry, and $\mu = 27$ claims something the sample cannot support. Write $\bar{x} = 27$ minutes: the bar says which mean, and the unit says of what.

True or false: In card 06, sample S3 gave $\bar{x} = 8$ , which equals $\mu$ . This shows S3 was drawn by a better method than the other three samples.

1

Find the sample mean of $14, 9, 22, 17, 13$ .

2

A sample of $20$ values has $\bar{x} = 4.85$ . Find the total of the sample values.

3

From the population $1, 3, 4, 6, 7, 9, 10, 24$ , write down any sample of size $4$ other than the four in card 06, and compute its mean. Is your answer above or below $\mu = 8$ ?

4

Explain in one sentence why two correct sample means from the same population can differ.

5

A sample of $9$ has mean $12$ . One value, $30$ , is removed. Find the mean of the remaining $8$ .

Which does NOT belong? True statements about the sample mean $\bar{x}$ :

11
Revisit your thinking

Earlier you were asked which of two students, reporting $6.1$ and $6.9$ from samples of $40$ , had made the error.

Neither result is necessarily wrong. Two random samples of the same size drawn from the same population contain different members, so they can produce different means. The difference of $0.8$ alone does not establish an error. The question "who was wrong?" assumes $\bar{x}$ is a property of the population. It is a property of the sample. Both numbers may be correct, both estimate the same unknown $\mu$ , and from these two alone you cannot say which is closer to it.

What you can now ask is the useful question: how much do sample means vary, and is $0.8$ a lot? That question has an answer, and the next lesson builds the object that supplies it.

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Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

02
Short answer
ApplyBand 33 marks

Q1. A random sample of $10$ batteries has lifetimes, in hours, of $42, 51, 47, 39, 55, 44, 48, 50, 46, 38$ . (a) Find the sample mean. (b) State what it estimates, and explain why you cannot assume it equals that quantity. (3 marks)

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AnalyseBand 43 marks

Q2. Two random samples, each of size $5$ , are drawn from the same population. One has $\bar{x} = 6.2$ and the other $\bar{x} = 7.8$ . Explain why this is not evidence of an error, and state what extra information would be needed to decide which sample mean is closer to $\mu$ . (3 marks)

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AnalyseBand 54 marks

Q3. A population of eight values is $1, 3, 4, 6, 7, 9, 10, 24$ , with $\mu = 8$ . (a) Give a sample of size $4$ whose mean is below $\mu$ , and one whose mean is above, showing both means. (b) Explain why a sample of size $4$ containing the value $24$ is likely to have a mean above $\mu$ . (c) Give a sample of size $4$ containing $24$ whose mean is not above $\mu$ , and explain what that shows. (4 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. Total $= 14 + 9 + 22 + 17 + 13 = 75$ , $n = 5$ , so $\bar{x} = 75 \div 5 = 15$ .

2. $\sum x_i = n\bar{x} = 20 \times 4.85 = 97$ .

3. Any sample of size $4$ from the eight values, other than the four in card 06. For example $1, 6, 7, 9$ gives $\bar{x} = 23 \div 4 = 5.75$ , which is below $\mu = 8$ . Samples containing $24$ tend to sit above it.

4. Because the two samples contain different members. Each value in a sample of size $n$ contributes $1/n$ of the mean, so swapping even one member changes $\bar{x}$ .

5. Old total $= 9 \times 12 = 108$ . Removing $30$ leaves $78$ across $8$ values, so the new mean is $78 \div 8 = 9.75$ .

Q1 (3 marks): (a) Total $= 460$ , $n = 10$ , so $\bar{x} = 460 \div 10 = 46$ hours [1 for the total and n, 1 for the mean with units]. (b) It estimates $\mu$ , the mean lifetime of the whole population of batteries [1]. You cannot assume $\bar{x} = \mu$ because it was computed from ten batteries rather than the whole population, and a different random sample could give a different value. It could equal $\mu$ by chance, but the sample alone cannot establish that.

Q2 (3 marks): The two samples contain different members, and sample means obtained from repeated sampling may differ even when the samples are the same size, so a gap between them is expected rather than evidence of a mistake [1]. Both are legitimate estimates of the same fixed $\mu$ [1]. To decide which is closer you would need to know $\mu$ itself, and $\mu$ is precisely what is unknown, so nothing computable from the two samples alone can settle it [1].

Q3 (4 marks): (a) Below: $1, 3, 4, 6$ gives $\bar{x} = 14 \div 4 = 3.5$ . Above: $7, 9, 10, 24$ gives $\bar{x} = 50 \div 4 = 12.5$ [1]. (b) In a sample of size $4$ each value carries a weight of $\tfrac{1}{4}$ , and $24$ sits $16$ above $\mu$ , so on its own it lifts the sample mean by $16 \div 4 = 4$ relative to a sample of otherwise typical values. Unless the other three are unusually small, the mean lands above $\mu$ [1 for the weighting argument, 1 for the conclusion]. (c) $1, 3, 4, 24$ gives $\bar{x} = 32 \div 4 = 8$ , exactly $\mu$ . The three smallest values in the population together offset the $24$ completely. This shows the reasoning in (b) is about what is likely across samples, not what is guaranteed for any particular one [1].

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