Constructing a Confidence Interval for a Mean or Mean Difference: Study Guide (Statistics 4.2)

Why a new distribution?

In Topic 4.1 you used σx̄ = σ/√n. That needs the population standard deviation σ. In real studies of a quantitative variable, σ is almost never known. If you knew everything about the population, you would not need to estimate its mean.

So you replace σ with the sample standard deviation s. The estimated standard deviation of x̄ is called the standard error:

SE = s / √n

Standardising with SE instead of σx̄ gives the statistic

t = (x̄ − μ) / (s/√n)

This is not a z-score. Both x̄ and s change from sample to sample. When a sample happens to have a small s, t can be far from 0. So t varies more than z, and its distribution has heavier tails. That distribution is a t-distribution.

t-distributions

The t-distributions (often called Student’s t-distributions) are a family of curves. Each member is set by one number, the degrees of freedom (df). For one sample, or one sample of differences:

df = n − 1

Every t-distribution is:

• symmetric and bell-shaped, centred at 0;
• standardised, like the standard normal;
• heavier in the tails and lower in the peak than the standard normal.

With few degrees of freedom the difference is large. As df increases, s becomes a more reliable estimate of σ and the t-distribution moves closer to the standard normal (mean 0, standard deviation 1).

Heavier tails mean larger critical values. For 95% confidence:

df	2	4	9	19	29	99	standard normal
t*	4.303	2.776	2.262	2.093	2.045	1.984	z* = 1.960

Small samples pay a price: a larger t* and a wider interval. t-distributions are also used for test statistics about μ when σ is unknown (Topics 4.4 and 4.5).

Choosing the procedure and defining the parameter

One sample, one quantitative variable: use a one-sample t-interval for a population mean, μ.

Matched pairs: each value in one set is linked to one value in the other for a real reason: the same person measured twice, the same car tested two ways, twins. The two sets are dependent. Do not treat them as two separate samples. Instead:

1. subtract within each pair, in a fixed order, to get one sample of differences;
2. build a one-sample t-interval for the population mean difference, μd, from those differences.

Define the parameter in context. Name the mean, the response variable with units, and the population. For a mean difference, also state the order of subtraction.

• μ = the true mean response time, in minutes, of all emergency calls to the service that month.
• μd = the true mean difference in fuel economy (low-resistance tyres minus standard tyres), in km per litre, for all vans in the fleet.

The three conditions

Condition	What to check	Why it matters
Randomization	The data come from a random sample or a randomized experiment.	Makes the values independent and the estimate unbiased.
10%	When sampling without replacement, n ≤ 10% of N.	Keeps the values close enough to independent for s/√n to be the right standard error.
Sample data	You are told the population is approximately normal; or the sample size is at least 30; or, for a smaller sample, a graph of the data shows no strong skewness and no outliers.	Makes the sampling distribution of x̄ close enough to normal for t* to be accurate.

For matched pairs, apply the sample data condition to the differences: the number of differences is at least 30, or the differences show no strong skewness or outliers. The two original columns do not need to pass on their own.

When n < 30, look at the data: make a dot plot or boxplot, or use the 1.5 × IQR rule, and describe what you see. “The dot plot of the 14 times is roughly symmetric with no outliers” is a check. “n is big enough” is not.

The calculation

1. Point estimate: x̄ (or x̄d for differences).
2. Standard error: SE = s/√n (or s_d/√n).
3. Critical value: t* for the central C% of a t-distribution with df = n − 1. On a calculator, t* = invT(1 − (1 − C)/2, df). For 95%, that is invT(0.975, df).
4. Margin of error: ME = t* × SE.
5. Interval: x̄ ± t × s/√n*.

Keep unrounded values until the final step. Check with the one-sample t-interval function on your calculator.

Worked example 1: one sample from raw data

Context. A fictional county ambulance service received 2,600 emergency calls in March. An analyst selected a random sample of 14 calls and recorded each response time in minutes:

8.2, 11.5, 9.7, 13.1, 10.4, 7.6, 12.3, 9.1, 10.9, 14.0, 8.8, 11.2, 10.1, 12.7

Question. Construct a 95% confidence interval for the mean response time.

1. Procedure and parameter. One-sample t-interval for μ = the true mean response time, in minutes, of all 2,600 emergency calls in March.

2. Conditions.

• Randomization: a random sample of calls. ✓
• 10%: 14 ≤ 10% of 2,600 = 260. ✓
• Sample data: n = 14 < 30, so check the data. Ordered: 7.6, 8.2, 8.8, 9.1, 9.7, 10.1, 10.4, 10.9, 11.2, 11.5, 12.3, 12.7, 13.1, 14.0. Five-number summary: 7.6, 9.1, 10.65, 12.3, 14.0. The median is near the middle of the box, so there is no strong skew. IQR = 3.2, fences 4.3 and 17.1, so there are no outliers. ✓

3. Calculate.

1. x̄ = 149.6 ÷ 14 ≈ 10.6857 minutes; s ≈ 1.9139 minutes.
2. SE = 1.9139/√14 ≈ 0.5115 minutes.
3. df = 13, so t* = invT(0.975, 13) ≈ 2.160.
4. ME = 2.160 × 0.5115 ≈ 1.105 minutes.
5. Interval: 10.6857 ± 1.105 = (9.58, 11.79) minutes.

Check. The point estimate sits in the middle: (9.58 + 11.79)/2 ≈ 10.69. Using z* = 1.96 by mistake would give a margin of error of only 1.0025 minutes: too narrow. What the interval tells you about the service is Topic 4.3’s job.

Worked example 2: matched pairs

Context. A delivery company has 180 vans. It selected 10 at random. Each van was driven on the same route once with standard tyres and once with low-resistance tyres; a coin toss decided which set came first. Fuel economy is in km per litre.

Van	1	2	3	4	5	6	7	8	9	10
Standard	9.8	10.4	8.9	11.2	10.0	9.5	10.7	9.1	10.9	9.6
Low-resistance	10.3	10.6	9.5	11.5	10.6	9.6	11.3	9.4	11.2	10.3
d = low − standard	0.5	0.2	0.6	0.3	0.6	0.1	0.6	0.3	0.3	0.7

Question. Construct a 90% confidence interval for the mean difference in fuel economy.

1. Procedure and parameter. Each van gives two linked measurements, so the data are paired. One-sample t-interval for μd = the true mean difference in fuel economy (low-resistance minus standard), in km per litre, for all 180 vans in the fleet.

2. Conditions.

• Randomization: the 10 vans were a random sample of the fleet, and the order of tyres was randomly assigned. ✓
• 10%: 10 ≤ 10% of 180 = 18. ✓
• Sample data: only 10 differences, so check them. Ordered: 0.1, 0.2, 0.3, 0.3, 0.3, 0.5, 0.6, 0.6, 0.6, 0.7. Q1 = 0.3, Q3 = 0.6, IQR = 0.3; fences −0.15 and 1.05. No outliers and no strong skew. ✓

3. Calculate.

1. x̄d = 4.2 ÷ 10 = 0.42 km/L; s_d ≈ 0.2044 km/L.
2. SE = 0.2044/√10 ≈ 0.0646 km/L.
3. df = 9, so t* = invT(0.95, 9) ≈ 1.833.
4. ME = 1.833 × 0.0646 ≈ 0.1185 km/L.
5. Interval: 0.42 ± 0.1185 = (0.30, 0.54) km/L.

Check. The two columns have standard deviations of about 0.77 and 0.76 km/L, because vans differ a lot. A two-sample method would wrongly give a standard error of about 0.3427 km/L. Pairing removes the van-to-van variation, which is why the paired interval is so much narrower.

Worked example 3: summary statistics and a large sample

Context. A fictional gym chain selected a random sample of 48 of its 9,000 members and recorded the length of each member’s most recent visit. The sample is skewed to the right with no extreme values. x̄ = 41.8 minutes and s = 15.6 minutes.

Question. Construct a 99% confidence interval for the mean visit length.

1. Procedure: one-sample t-interval for μ = the true mean length, in minutes, of the most recent visit for all 9,000 members.
2. Conditions: random sample ✓; 48 ≤ 10% of 9,000 = 900 ✓; n = 48 ≥ 30, so the skew in the sample does not stop the procedure ✓.
3. SE = 15.6/√48 ≈ 2.2517 minutes. df = 47, t* = invT(0.995, 47) ≈ 2.685. ME = 2.685 × 2.2517 ≈ 6.0447 minutes.
4. Interval: 41.8 ± 6.0447 = (35.76, 47.84) minutes.

Common misconceptions

• Using z with s.* When σ is unknown, the critical value comes from a t-distribution with n − 1 df.
• Using df = n. For one sample, or one sample of differences, df = n − 1.
• Treating paired data as two independent samples. If the values are linked in pairs, subtract first and work with one sample of differences.
• Forgetting the order of subtraction. State it in the parameter, and keep it the same throughout.
• Checking the sample data condition on the wrong values. For matched pairs, check the differences.
• Writing “n ≥ 30” when n is 14. For a small sample you must look at the data and describe its shape and any outliers.
• Citing the 10% condition for an experiment with volunteers. The 10% condition applies to sampling without replacement from a population.
• Rounding too early. Rounding x̄, s or t* early can move the endpoints in the second decimal place.

Where this leads

Topic 4.3 shows how to interpret a t-interval and its confidence level, and how to use one to judge a claim: Justifying a Claim Based on a Confidence Interval for a Population Mean or Population Mean Difference. Try the practice questions now, then use the revision notes and the checklist to consolidate. To review the sampling distribution of x̄, return to Topic 4.1.