Constructing a Confidence Interval for the Difference Between Two Means: Study Guide (Statistics 4.7)

From one mean to two

In Topic 4.2 you estimated one population mean with x̄ ± t* (s/√n). Many questions compare two groups instead. Do parcels from one depot take longer to arrive than parcels from another? Does a new fertiliser grow heavier plants than the standard one?

To answer, you take a sample from each group and compare the sample means. The difference x̄₁ − x̄₂ is your best single estimate of the true difference μ₁ − μ₂. But another pair of samples would give a different difference. In Topic 4.6 you saw how much x̄₁ − x̄₂ varies from sample to sample. Now you turn that knowledge into a confidence interval: a range of plausible values for μ₁ − μ₂.

The structure is the same as every interval in the course:

point estimate ± margin of error, where margin of error = critical value × standard error.

Step 1: name the procedure and define the parameter

Which procedure? Use a two-sample t-interval for μ₁ − μ₂ when:

• the response variable is quantitative (a measurement, not a yes/no answer);
• there are two groups; and
• the two samples are independent: knowing a value in one group tells you nothing about any particular value in the other.

Independent samples come from two separate random samples, or from a randomized experiment in which each subject receives one treatment only.

Watch for paired data. If each value in one group is linked to one value in the other (the same person before and after, twins, two measurements on the same car), the samples are not independent. Then you find the differences within each pair and use a one-sample t-interval for the mean difference (Topic 4.2). The test is simple: can you match each value in group 1 to exactly one value in group 2 for a real reason? If yes, the data are paired.

Define the parameter. A full definition names three things:

1. the difference in means, with the order of subtraction;
2. the response variable, with units; and
3. both populations (or both treatments) in context.

Template: μ₁ − μ₂ = the difference (group 1 minus group 2) in the mean [response, units] of all [population 1] and all [population 2].

Choose the order of subtraction at the start and keep it to the end. Either order is correct; only the signs of the interval change.

Step 2: check the three conditions

Condition	What to check	Notes
Randomization	The data come from two independent random samples, or from a randomized experiment	Name the random process in context
10%	n₁ ≤ 10% of N₁ and n₂ ≤ 10% of N₂	Only when sampling without replacement. Not needed for a randomized experiment
Sample data	Both n₁ ≥ 30 and n₂ ≥ 30, or both populations are stated to be approximately normal	If either sample has fewer than 30 values, look at both samples: each must be free from strong skewness and outliers

Each condition has its own job. Randomization makes the samples representative and the two groups independent. The 10% condition keeps the values within each sample close to independent, so the standard error formula works. The sample data condition makes the sampling distribution of x̄₁ − x̄₂ approximately normal, which is what the t critical values assume.

Two common slips: mixing up the randomization and 10% conditions, and checking only the smaller sample. Each condition is checked for both groups.

Step 3: point estimate, standard error and critical value

Point estimate: x̄₁ − x̄₂.

Standard error: the population standard deviations are almost never known, so you replace σ₁ and σ₂ from Topic 4.6 with the sample standard deviations:

SE = √(s₁²/n₁ + s₂²/n₂)

Variances add, not standard deviations. Square each s, divide by its own n, add, then take one square root.

Critical value t*: for a C% interval, t* cuts off the central C% of a t-distribution. The degrees of freedom (df) for two samples come from a long formula that technology applies for you. The value is usually not a whole number. It always lies in this range:

smaller of (n₁ − 1, n₂ − 1) ≤ df ≤ n₁ + n₂ − 2

If you cannot use technology for df, use the smaller of n₁ − 1 and n₂ − 1. This conservative df gives a slightly larger t* and so a slightly wider interval. It never claims more precision than the data support.

Margin of error: t* × SE. Interval:

(x̄₁ − x̄₂) ± t √(s₁²/n₁ + s₂²/n₂)*

On a graphing calculator, use the two-sample t-interval function and choose not to pool. Pooling assumes the two population standard deviations are equal, which the course does not assume.

Worked example 1: courier delivery times (summary statistics)

Context. A fictional courier company has two depots. In one month, Depot North sent 6,000 parcels and Depot South sent 4,500. The company selected a random sample of 40 parcels from North and, separately, a random sample of 35 from South, and recorded each delivery time in hours.

Depot	n	x̄ (hours)	s (hours)
North	40	26.4	6.8
South	35	22.9	5.1

Question. Construct a 95% confidence interval for the difference in mean delivery times.

1. Procedure and parameter. Two-sample t-interval. Let μ_N − μ_S = the difference (North minus South) in the mean delivery time, in hours, of all parcels sent from Depot North and all parcels sent from Depot South that month.

2. Conditions.

• Randomization: two independent random samples, one from each depot. ✓
• 10%: 40 ≤ 10% of 6,000 = 600 and 35 ≤ 10% of 4,500 = 450. ✓
• Sample data: n_N = 40 ≥ 30 and n_S = 35 ≥ 30, so the sampling distribution of x̄_N − x̄_S is approximately normal. ✓

3. Calculate.

1. Point estimate: 26.4 − 22.9 = 3.5 hours.
2. Variance terms: 6.8²/40 = 1.1560 and 5.1²/35 = 0.7431. Sum = 1.8991.
3. SE = √1.8991 = 1.3781 hours.
4. Technology gives df = 71.41 (between 34 and 73, as expected) and t* = 1.9937.
5. Margin of error = 1.9937 × 1.3781 = 2.7476 hours.
6. Interval: 3.5 ± 2.7476 = (0.75, 6.25) hours.

Check. With the conservative df = 34, t* = 2.0322, the margin of error is 2.8006 hours and the interval is (0.70, 6.30) hours: slightly wider, as expected. The point estimate 3.5 sits exactly in the middle of the interval.

What the interval tells you about the claim “North is slower” is the job of Topic 4.8.

Worked example 2: a fertiliser experiment (raw data, small samples)

Context. A fictional horticulture class grew 20 basil plants in identical pots. They randomly assigned 10 pots to a new liquid fertiliser and 10 to the standard fertiliser. After six weeks they recorded the fresh leaf mass of each plant, in grams.

• New: 48, 52, 55, 44, 58, 50, 53, 47, 56, 51
• Standard: 41, 46, 50, 39, 44, 48, 43, 52, 45, 42

Question. Construct a 90% confidence interval for the difference in mean leaf mass.

1. Procedure and parameter. Two-sample t-interval: each pot received one treatment only, so the groups are independent. Let μ_new − μ_std = the true difference (new minus standard) in the mean fresh leaf mass, in grams, of basil plants like these grown with the new fertiliser and with the standard fertiliser.

2. Conditions.

• Randomization: the fertilisers were randomly assigned to the 20 pots. ✓
• 10%: not needed, because this is a randomized experiment, not a sample drawn from a population.
• Sample data: both groups have n = 10 < 30, so look at both samples (Figure 2). New: Q1 = 48, Q3 = 55, IQR = 7, fences 37.5 and 65.5 g; all values lie between 44 and 58 g. Standard: Q1 = 42, Q3 = 48, IQR = 6, fences 33 and 57 g; all values lie between 39 and 52 g. No outliers, and neither dot plot shows strong skewness. ✓

3. Calculate.

1. Means: x̄_new = 514 ÷ 10 = 51.4 g; x̄_std = 450 ÷ 10 = 45.0 g. Point estimate = 6.4 g.
2. Standard deviations (calculator, Sx): s_new = 4.326 g; s_std = 4.082 g.
3. Variance terms (using the unrounded s values): s_new²/10 = 1.8711 and s_std²/10 = 1.6667. SE = √3.5378 = 1.8809 g.
4. Technology: df = 17.94 (between 9 and 18), t* for 90% = 1.7344.
5. Margin of error = 1.7344 × 1.8809 = 3.2622 g.
6. Interval: 6.4 ± 3.2622 = (3.14, 9.66) grams.

Check. With the conservative df = 9, t* = 1.8331 and the interval is (2.95, 9.85) g. Because this was a randomized experiment, the parameter is about the two treatments, for plants like these, not about a wider population of basil.

Worked example 3: which procedure?

Question. For each study, name the procedure for estimating a difference in means.

(a) A café records the time to prepare 30 randomly chosen orders made by hand and 30 randomly chosen orders made with a new machine. Two-sample t-interval: two independent random samples of a quantitative variable.

(b) Each of 25 runners runs 400 m once in old shoes and once in new shoes, in random order. One-sample t-interval for the mean difference: each runner gives a pair of linked times, so the data are paired. Find the 25 differences first.

(c) A survey asks random samples of 200 adults in two towns whether they cycle to work. Not a means procedure: the response is categorical (yes/no), so it is about proportions (Unit 3).

Common misconceptions

• Adding standard deviations. √(s₁²/n₁ + s₂²/n₂) is not s₁/√n₁ + s₂/√n₂. Square, divide, add, then take one square root.
• Subtracting the variance terms because the statistic is a difference. Variability adds for a difference of independent statistics, just as in Topic 4.6.
• Using z*. σ₁ and σ₂ are unknown, so the critical value comes from a t-distribution.
• Using df = n₁ + n₂ − 2 as the answer. That is only the upper bound for df. Use technology, or the conservative smaller of n₁ − 1 and n₂ − 1.
• Pooling. Do not choose the pooled option; the course does not assume equal population standard deviations.
• Treating paired data as two independent samples. Before-and-after measurements on the same people need the paired procedure.
• Checking only one sample. The 10% and sample data conditions apply to both groups. If either sample is under 30, look at both distributions.
• Checking the 10% condition in an experiment. It is not needed after random assignment.
• Defining the parameter as “the difference in the sample means”. The parameter is about populations or treatments, and it states the order of subtraction.

Where this leads

Topic 4.8 shows how to interpret the interval and its confidence level, and how to use an interval to support or reject a claim about μ₁ − μ₂. The next study guide starts from exactly the kind of interval you built here. Try the practice questions now, then use the revision notes and the checklist to consolidate. You can also look back at Topic 4.6 for where the standard error formula comes from.