Setting Up a Test for a Population Mean or Population Mean Difference: Study Guide (Statistics 4.4)

From proportions to means

In Topic 3.5 you set up a test for a claimed population proportion. Now the variable is quantitative: a weight, a time, a score. The claim is about a population mean.

The structure of a test does not change. You assume the claimed value is true, then ask whether your sample would be surprising if it were. A full test has four parts: State (parameter and hypotheses), Plan (procedure and conditions), Do (test statistic and p-value) and Conclude. This topic covers State and Plan. Topic 4.5 does the rest.

What does change is the procedure. The population standard deviation σ is almost never known for a quantitative variable. So the test uses the sample standard deviation s, and the reference distribution is a t-distribution with n − 1 degrees of freedom, as in Topic 4.2.

Choosing the procedure

Ask three questions.

1. Is the variable quantitative, and is the question about a mean? If the variable is categorical, you need a proportion procedure.
2. Is the question asking whether the data give evidence against a claimed value? If it asks you to estimate, use an interval (Topics 4.2 and 4.3).
3. How were the data produced?
   • One sample, one measurement per individual → one-sample t-test for a population mean.
   • Two measurements on each individual, or natural pairs → matched pairs: take the differences and use a one-sample t-test for a population mean difference.
   • Two separate groups with no pairing → a two-sample t-test, which comes later in this unit (Topic 4.9).

Recognising matched pairs

Data are paired when each value in one set has a partner in the other set, and the partners belong together:

• the same person measured before and after something;
• the same person (or item) given both treatments, ideally in a random order;
• natural pairs, such as twins, or the left and right side of the same car.

Pairing removes much of the person-to-person variation, so the differences show the effect more clearly. Because each pair gives one difference, the analysis is a one-sample procedure on the differences. Treating paired data as two separate samples ignores the pairing and is a serious error.

Defining the parameter

The hypotheses are about a population value, so define it carefully.

Definition	Verdict
“μ = mean”	Too vague: mean of what, for whom?
“μ = the mean volume of the 12 bottles sampled”	Wrong: that is the statistic x̄, which we know.
“μ = the true mean volume (mL) of all bottles filled by machine 3 today”	Correct: parameter, variable, units and population.
“μd = the true mean difference in reaction time”	Incomplete: which way was the subtraction?
“μd = the true mean of (with music − without music) reaction time, in ms, for people like these volunteers”	Correct.

Writing the hypotheses

One-sample mean. H₀: μ = μ₀, where μ₀ is the claimed value. Hₐ is one of μ < μ₀, μ > μ₀ or μ ≠ μ₀.

Matched pairs. “No difference on average” is H₀: μd = 0. Hₐ is one of μd < 0, μd > 0 or μd ≠ 0.

The same rules as Topic 3.5 apply:

1. Hypotheses use μ or μd, never x̄ or x̄d.
2. μ₀ comes from the claim, not from the sample.
3. The direction of Hₐ comes from the question’s wording (“less than”, “increases”, “differs”), decided before looking at the data.
4. For an “at least” or “at most” claim, test at the boundary: “at least 500 mL”, with a suspicion of less, gives H₀: μ = 500 and Hₐ: μ < 500.

Direction for pairs depends on the order of subtraction. “The new method improves scores” with d = after − before gives Hₐ: μd > 0. With d = before − after, the same claim gives Hₐ: μd < 0. Choose an order, state it, and keep it.

The three conditions

Condition	What to check	Why it matters
Random	Data from a random sample, or from a randomized experiment (for pairs: random order of treatments, or a random sample of pairs)	Without randomness the t-distribution does not describe how x̄ varies
10%	When sampling without replacement, n ≤ 10% of N	Keeps the observations close to independent. Not needed for a randomized experiment with no sampling
Sample data	You are told the population is roughly normal; or n ≥ 30; or, when n is below 30, a graph of the sample shows no strong skewness and no outliers	Makes the sampling distribution of x̄ close enough to normal for the t-procedure

For matched pairs, apply the sample data condition to the differences: at least 30 differences, or fewer than 30 with no strong skew or outliers.

Notice what is not here: there is no “np₀ ≥ 10” check. That is for proportions only. When n < 30, you must look at the sample data, for example with a dotplot and the 1.5 × IQR rule. Saying “n < 30 so the condition fails” is wrong; small samples are fine if the data are not strongly skewed and have no outliers.

Worked example 1: a one-sided test for a mean

Question. A machine at a fictional bottling plant should fill bottles with 500 mL of water. An inspector suspects it is underfilling. She selects a random sample of 12 of the 8,000 bottles filled that day. The volumes (mL) are:

497.8, 501.2, 499.5, 498.1, 500.4, 496.9, 499.0, 498.6, 500.9, 497.5, 499.8, 498.3

Set up a significance test. Do not carry it out.

State.

• μ = the true mean volume (mL) of all bottles filled by the machine that day.
• H₀: μ = 500. Hₐ: μ < 500. The direction comes from “suspects it is underfilling”.

Plan. One-sample t-test for a population mean (σ is unknown).

1. Random: the 12 bottles were a random sample of that day’s bottles. ✓
2. 10%: 12 ≤ 10% of 8,000 = 800. ✓
3. Sample data: n = 12 < 30, so check the data. Ordered: 496.9, 497.5, 497.8, 498.1, 498.3, 498.6, 499.0, 499.5, 499.8, 500.4, 500.9, 501.2. Q1 = 497.95, Q3 = 500.1, IQR = 2.15. Fences: 497.95 − 3.225 = 494.725 and 500.1 + 3.225 = 503.325. No value is outside the fences, and the values are spread fairly evenly, with no strong skew. ✓

All conditions are met, so the t-test is appropriate.

Check. The sample mean (499.0 mL) was not used in the set-up. It is needed in the Do step.

Worked example 2: a matched-pairs test where a condition fails

Question. A psychology student asks whether background music affects reaction time. Each of 14 volunteers did a reaction test with music and without music, in an order decided by a coin toss. The differences, d = with music − without music, in milliseconds, are:

12, −8, 5, 15, 3, −2, 9, 7, 11, 4, 6, −5, 10, 68

(a) Explain why this is a matched-pairs design. (b) Define the parameter and write the hypotheses. (c) Check the conditions.

(a) Each volunteer was measured under both conditions, so the two times for one person form a pair. The analysis uses one difference per person.

(b) μd = the true mean difference in reaction time (with music − without music), in ms, for people like these volunteers. H₀: μd = 0. Hₐ: μd ≠ 0. “Affects” does not give a direction, so the test is two-sided.

(c)

1. Random: the order of the conditions was randomly assigned for each volunteer. ✓
2. 10%: not needed: the volunteers were not sampled from a population. (So any conclusion applies to people like these volunteers.)
3. Sample data: only 14 differences, so check them. Q1 = 3, Q3 = 11, IQR = 8. Upper fence = 11 + 1.5 × 8 = 23 ms. The difference of 68 ms is an outlier (Figure 2). ✗

With fewer than 30 differences and a clear outlier, the t-test is not appropriate as it stands. The outlier would have a large effect: the mean difference is 9.64 ms with it and 5.15 ms without it.

What could the student do? First, investigate the 68 ms value. If it was a recording error, correct it. If that trial was spoiled (for example, the volunteer was distracted), the student could repeat or remove it, and must report this. If it is genuine, collect more pairs: with at least 30 differences the sample data condition is met.

Worked example 3: which procedure?

For each study, name the procedure and write H₀.

1. A gym records the resting heart rate of 20 randomly chosen members before and after a 12-week programme, to see whether it falls. Same people measured twice: one-sample t-test for a population mean difference. With d = before − after, H₀: μd = 0, Hₐ: μd > 0.
2. A café claims its average wait is 4 minutes. A reviewer times 35 randomly chosen orders, suspecting longer waits. One sample, one value each: one-sample t-test for a population mean. H₀: μ = 4, Hₐ: μ > 4 (minutes).
3. A teacher compares test scores of 25 students taught by method A with 25 different students taught by method B. Two separate groups with no pairing: not this topic; a two-sample t-test (Topic 4.9).

Common misconceptions

• Hypotheses about x̄. “H₀: x̄ = 500” tests a number you already know. Use μ.
• Analysing paired data as two independent samples. If the same individuals give both values, take differences.
• Not stating the order of subtraction, or switching it between the definition and Hₐ.
• Using np₀ ≥ 10 for a mean. That check is for proportions.
• “n < 30, so the conditions fail.” Small samples are fine if the data show no strong skew or outliers.
• Checking the two sets of measurements instead of the differences for matched pairs.
• Choosing Hₐ from the data. “Affects” means two-sided even if most differences are positive.
• Applying the 10% condition to a randomized experiment with volunteers. It is about sampling without replacement.

Where this leads

With the test set up, Topic 4.5 calculates the t test statistic and p-value and writes the conclusion: Carrying Out a Test for a Population Mean or Population Mean Difference. First, try the practice questions, then use the revision notes and the checklist to consolidate. For the interval approach to the same kind of claim, see the Topic 4.3 study guide.