Confidence Interval for a Mean

A confidence interval for a mean gives a range of plausible values for a population mean μ. When σ is unknown, you build a one-sample t-interval: x̄ ± t* · s/√n, with df = n − 1. The sample mean x̄ is the center, and the margin of error t* · s/√n sets the width.

Before this lesson: Confidence Intervals, Sampling Distribution of a Sample Mean and t-Distribution

What a t-interval does

A sample mean x̄ is your best single guess for μ. But a different sample would give a different x̄. The interval adds a cushion on each side to cover that sample-to-sample change.

The cushion is the margin of error. It is a critical value times the standard error s/√n, which measures how much x̄ typically varies. Because you use s in place of the unknown σ, the critical value comes from the t-distribution, not the normal curve.

The result is called a one-sample t-interval. It works the same way as the confidence interval for a proportion: estimate ± margin of error.

Confidence interval for a mean formula

\bar{x} \pm t^* \cdot \dfrac{s}{\sqrt{n}} \qquad df = n - 1

• x̄ is the sample mean and s is the sample standard deviation.
• n is the sample size, and s/√n is the standard error (SE).
• t* is the t critical value for your confidence level with df = n − 1. For 95%, t* = invT(0.975, df).
• The margin of error is ME = t* · SE.

Conditions

• Random: the data come from a random sample or a randomized experiment.
• 10%: when sampling without replacement, n is at most 10% of the population.
• Normal/Large Sample: the population is roughly normal, or n ≥ 30. For a small sample, graph the data and check for no strong skew and no outliers.

The n ≥ 30 rule comes from the central limit theorem. For a small sample, the t-interval is only safe when the data look roughly symmetric (see the IQR rule for outliers).

Four steps

1. State: name the parameter μ and the confidence level.
2. Plan: name the method (one-sample t-interval) and check the three conditions.
3. Do: find df, t*, SE and ME, then compute x̄ − ME and x̄ + ME.
4. Conclude: interpret the interval in context.

Calculator: on a TI-84, press STAT, choose TESTS, then 8:TInterval. Choose Stats to enter x̄, s and n, or Data to use a list. The calculator finds t* for you.

Confidence interval for a mean examples

Example 1: Build a 95% t-interval

A random sample of 25 sunflower plants from a large field has mean height x̄ = 68.2 cm and standard deviation s = 4.5 cm. A dot plot of the heights shows no strong skew and no outliers. Build a 95% confidence interval for the mean height μ.

Conditions: the sample is random, 25 plants is far less than 10% of the field, and the data show no strong skew or outliers.

df = 25 - 1 = 24

t^* = \text{invT}(0.975, 24) \approx 2.064

\sqrt{25} = 5

\text{SE} = \dfrac{4.5}{5} = 0.9

\text{ME} = 2.064 \cdot 0.9 = 1.8576

68.2 - 1.8576 = 66.3424

68.2 + 1.8576 = 70.0576

Answer: about (66.34, 70.06) cm. We are 95% confident that the mean height of all sunflowers in the field is between 66.34 cm and 70.06 cm.

Example 2: Start from raw data

Ten randomly chosen students at a large high school report how many hours they slept last night: 6.5, 7.0, 5.5, 8.0, 6.0, 7.5, 6.5, 7.0, 5.0, 6.0. The values are roughly symmetric with no outliers. Build a 90% confidence interval for the mean sleep time.

6.5 + 7.0 + 5.5 + 8.0 + 6.0 + 7.5 + 6.5 + 7.0 + 5.0 + 6.0 = 65

\bar{x} = \dfrac{65}{10} = 6.5

0 + 0.25 + 1 + 2.25 + 0.25 + 1 + 0 + 0.25 + 2.25 + 0.25 = 7.5

s^2 = \dfrac{7.5}{9} \approx 0.8333

s = \sqrt{0.8333} \approx 0.9129

df = 10 - 1 = 9

t^* = \text{invT}(0.95, 9) \approx 1.833

\sqrt{10} \approx 3.162

\text{SE} = \dfrac{0.9129}{3.162} \approx 0.2887

\text{ME} = 1.833 \cdot 0.2887 \approx 0.5292

6.5 - 0.5292 = 5.9708

6.5 + 0.5292 = 7.0292

Answer: about (5.97, 7.03) hours. We are 90% confident that the mean sleep time for all students at the school is between 5.97 and 7.03 hours. TInterval with Data gives the same result.

Example 3: Raise the confidence level

Use the sunflower data from Example 1 (x̄ = 68.2, s = 4.5, n = 25). Build a 99% interval and compare its width with the 95% interval.

t^* = \text{invT}(0.995, 24) \approx 2.797

\text{ME} = 2.797 \cdot 0.9 = 2.5173

68.2 - 2.5173 = 65.6827

68.2 + 2.5173 = 70.7173

2 \cdot 2.5173 = 5.0346

2 \cdot 1.8576 = 3.7152

Answer: about (65.68, 70.72) cm, with width 5.03 cm instead of 3.72 cm. More confidence costs a wider interval.

Example 4: Checking a battery claim (word problem)

A phone maker says its battery lasts 10 hours on average. A consumer group tests a random sample of 20 phones and finds x̄ = 9.4 hours and s = 1.1 hours. A dot plot of the 20 times is roughly symmetric with no outliers. Build a 95% interval. Is the claim plausible?

df = 20 - 1 = 19

t^* = \text{invT}(0.975, 19) \approx 2.093

\sqrt{20} \approx 4.472

\text{SE} = \dfrac{1.1}{4.472} \approx 0.2460

\text{ME} = 2.093 \cdot 0.2460 \approx 0.5149

9.4 - 0.5149 = 8.8851

9.4 + 0.5149 = 9.9149

Answer: about (8.89, 9.91) hours. The claimed mean of 10 hours is outside the interval, so it is not a plausible value. The data give good evidence that the true mean battery life is less than 10 hours.

Example 5: Work backward from an interval

A report gives a 95% confidence interval of (42.1, 47.9) for a mean, from a sample of 16. Find x̄, the margin of error and the sample standard deviation s.

\bar{x} = \dfrac{42.1 + 47.9}{2} = \dfrac{90}{2}

= 45

\text{ME} = 47.9 - 45 = 2.9

t^* = \text{invT}(0.975, 15) \approx 2.131

\text{SE} = \dfrac{2.9}{2.131} \approx 1.361

s = 1.361 \cdot 4 = 5.444

Answer: x̄ = 45, ME = 2.9 and s ≈ 5.44. Since SE = s/√n, you undo it by multiplying by √n.

Common mistakes

Related methods

When σ is known: the z-interval

If you somehow know the population standard deviation σ, use \bar{x} \pm z^* \cdot \sigma/\sqrt{n} with z* from the normal curve. This is rare in real studies, so the t-interval is the standard method.

Paired data

For matched pairs, such as before-and-after scores for the same people, find each pair’s difference first. Then build a one-sample t-interval on those differences (see matched pairs design).

Planning the sample size

To reach a chosen margin of error, solve ME = z*σ/√n for n, using a guess for σ. The margin of error lesson shows the steps.

Practice problems

1. A sample has n = 36. Find df and t* for a 95% confidence interval.
2. A random sample of 16 has x̄ = 50 and s = 8. Build a 95% confidence interval for μ.
3. A random sample of 25 has x̄ = 12.4 and s = 3. Build a 99% confidence interval for μ.
4. A 95% confidence interval for a mean is (18.6, 23.4). Find x̄ and the margin of error.
5. A random sample of 8 has x̄ = 30 and s = 4, and the data look roughly normal. Build a 95% confidence interval.
6. A random sample of 12 home prices is strongly skewed right with one very high outlier. Is a t-interval safe? Explain.
7. A 95% confidence interval for the mean daily screen time of teens at a school is (3.1, 3.9) hours. Interpret it.
8. A 95% confidence interval for mean commute time is (24.2, 28.6) minutes. Is 30 minutes a plausible mean? Is 25 minutes?
9. A random sample of 30 has x̄ = 72 and s = 6. Build a 90% confidence interval for μ.
10. Use Example 1 (s = 4.5). If n grows from 25 to 100, find the new 95% margin of error. About how does it compare with 1.8576?

df = 36 - 1 = 35

t^* = \text{invT}(0.975, 35) \approx 2.030

t^* = \text{invT}(0.975, 15) \approx 2.131

\text{SE} = \dfrac{8}{\sqrt{16}} = \dfrac{8}{4}

= 2

\text{ME} = 2.131 \cdot 2 = 4.262

50 - 4.262 = 45.738

50 + 4.262 = 54.262

t^* = \text{invT}(0.995, 24) \approx 2.797

\text{SE} = \dfrac{3}{\sqrt{25}} = \dfrac{3}{5}

= 0.6

\text{ME} = 2.797 \cdot 0.6 = 1.6782

12.4 - 1.6782 = 10.7218

12.4 + 1.6782 = 14.0782

\bar{x} = \dfrac{18.6 + 23.4}{2} = \dfrac{42}{2}

= 21

\text{ME} = 23.4 - 21 = 2.4

t^* = \text{invT}(0.975, 7) \approx 2.365

\sqrt{8} \approx 2.828

\text{SE} = \dfrac{4}{2.828} \approx 1.414

\text{ME} = 2.365 \cdot 1.414 \approx 3.344

30 - 3.344 = 26.656

30 + 3.344 = 33.344

t^* = \text{invT}(0.95, 29) \approx 1.699

\sqrt{30} \approx 5.477

\text{SE} = \dfrac{6}{5.477} \approx 1.0955

\text{ME} = 1.699 \cdot 1.0955 \approx 1.8613

72 - 1.8613 = 70.1387

72 + 1.8613 = 73.8613

t^* = \text{invT}(0.975, 99) \approx 1.984

\text{SE} = \dfrac{4.5}{\sqrt{100}} = \dfrac{4.5}{10}

= 0.45

\text{ME} = 1.984 \cdot 0.45 = 0.8928

\dfrac{0.8928}{1.8576} \approx 0.481

Confidence interval for a mean FAQ

What is the formula for a confidence interval for a mean?

When σ is unknown, use x̄ ± t* · s/√n, where t* comes from the t-distribution with df = n − 1. The part after ± is the margin of error.

When do you use t instead of z for a confidence interval?

Use t for a mean whenever you estimate σ with the sample standard deviation s, which is almost always. Use z only if σ is truly known, and always use z for proportions.

What does a 95% confidence interval for a mean tell you?

It gives a range of plausible values for the population mean. The 95% means the method captures the true mean in about 95% of all possible samples.

What sample size do you need for a t-interval?

Any size works if the population is roughly normal. If you don’t know the shape, n ≥ 30 is the usual rule; for smaller samples, check a graph of the data for strong skew and outliers.

Summary

• A confidence interval for a mean is x̄ ± t* · s/√n, with df = n − 1.
• Check Random, 10% and Normal/Large Sample before you build it.
• Find t* with invT(left area, df) or a t-table; for 95%, the left area is 0.975.
• Higher confidence or a smaller sample gives a wider interval.
• Interpret it as a range of plausible values for the population mean μ, not for single values.

Previous lesson: t-Distribution

Next lesson: Confidence Interval for a Difference in Proportions

Back to the Statistics Index