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Confidence Intervals for Means

Construct and interpret confidence intervals for a population mean using the t-distribution.

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📏 Confidence Intervals for Means

When to Use t-Interval vs z-Interval

  • Use t-interval: When σ (population SD) is unknown (almost always in practice)
  • Use z-interval: Only when σ is known (rare in real applications)

One-Sample t-Interval Formula

xˉ±t∗⋅sn\bar{x} \pm t^* \cdot \frac{s}{\sqrt{n}}

where:

  • xˉ\bar{x} = sample mean
  • t∗t^* = critical t-value (depends on confidence level and degrees of freedom)
  • ss = sample standard deviation
  • nn = sample size
  • df=n−1df = n - 1

Conditions for t-Interval

  1. Random sample: Data collected randomly
  2. Independence: Sampling without replacement; use 10% rule (n ≤ 0.10N)
  3. Normality: Either population is normal OR n ≥ 30 (CLT)

Finding t∗t^* Values

  • df = n − 1
  • Look up t∗t^* in t-table using df and confidence level

Common values (df = large, approximately normal):

  • 90% CI: t∗≈1.645t^* \approx 1.645
  • 95% CI: t∗≈1.96t^* \approx 1.96
  • 99% CI: t∗≈2.576t^* \approx 2.576

For small samples:

  • df = 9, 95% CI: t∗=2.262t^* = 2.262 (larger than z)
  • df = 4, 95% CI: t∗=2.776t^* = 2.776 (even larger)

Smaller df → larger t∗t^* → wider CI.

One-Sample Example

A random sample of 16 students has mean test score xˉ=78\bar{x} = 78 with sample SD s = 8. Find a 95% CI for the population mean.

Check conditions:

  • Random sample ✓
  • n = 16 < 30, but assume population approximately normal ✓
  • Independence ✓

Calculate:

  • df=16−1=15df = 16 - 1 = 15
  • From t-table: t∗=2.131t^* = 2.131 (95% CI, df = 15)
  • SE=816=84=2SE = \frac{8}{\sqrt{16}} = \frac{8}{4} = 2
  • ME=2.131×2=4.262ME = 2.131 \times 2 = 4.262
  • CI: 78±4.262=(73.738,82.262)78 \pm 4.262 = (73.738, 82.262)

Interpretation: We are 95% confident the mean score is between 73.7 and 82.3.

Two-Sample t-Interval

Comparing two population means μ1\mu_1 and μ2\mu_2:

(xˉ1−xˉ2)±t∗⋅SE(xˉ1−xˉ2)(\bar{x}_1 - \bar{x}_2) \pm t^* \cdot SE(\bar{x}_1 - \bar{x}_2)

where:

SE(xˉ1−xˉ2)=s12n1+s22n2SE(\bar{x}_1 - \bar{x}_2) = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}

(Use technology to find df; approximately min(n1−1,n2−1)min(n_1 - 1, n_2 - 1) or more complex formula)

Common Mistakes

  1. Confusing df: Always use df = n − 1, not n
  2. Using s instead of SE: The standard error is sn\frac{s}{\sqrt{n}}, not just s
  3. Wrong critical value: Look up t∗t^*, not z, from the t-table
  4. Forgetting conditions: Particularly the normality condition; state why it's satisfied

AP Exam Tip

Free-response questions often ask you to construct a t-interval. Show all steps: state formula, check conditions, identify xˉ\bar{x}, s, n, df, and t∗t^*, calculate ME, and state the CI with interpretation. Partial credit is generous if you show correct understanding.

📚 Practice Problems

1Problem 1easy

❓ Question:

What is a confidence interval and what does the confidence level represent?

💡 Show Solution

A confidence interval is a range of plausible values for a population parameter, calculated from sample data. The confidence level (e.g., 95%) represents the long-run success rate: if we repeated our sampling procedure many times and computed a confidence interval each time, approximately 95% of those intervals would contain the true population mean μ\mu. A 95% CI does NOT mean there is a 95% probability that μ\mu is in this specific interval; rather, the interval either contains μ\mu or it does not.

2Problem 2medium

❓ Question:

Given a sample of 25 students with mean xˉ=72\bar{x} = 72, sample standard deviation s=8s = 8, construct a 95% confidence interval for the population mean.

💡 Show Solution

Conditions: Random sample, population approximately Normal or n≥30n \ge 30 (assuming met). Formula: xˉ±t∗sn\bar{x} \pm t^* \frac{s}{\sqrt{n}} with df=n−1=24df = n - 1 = 24. From t-table: t0.025,24∗≈2.064t^*_{0.025, 24} ≈ 2.064. SE=825=85=1.6SE = \frac{8}{\sqrt{25}} = \frac{8}{5} = 1.6. CI: 72±2.064(1.6)=72±3.30=(68.70,75.30)72 \pm 2.064(1.6) = 72 \pm 3.30 = (68.70, 75.30). We are 95% confident the population mean lies between 68.70 and 75.30.

3Problem 3hard

❓ Question:

Two researchers compute 90% confidence intervals for the same population mean. Researcher A uses n=64n = 64, Researcher B uses n=256n = 256. Whose interval is narrower? Explain why.

💡 Show Solution

Researcher B's interval is narrower. The margin of error is ME=t∗snME = t^* \frac{s}{\sqrt{n}}. Since nn appears in the denominator, larger nn produces smaller SESE and thus smaller MEME. For A: SE=s64=s8SE = \frac{s}{\sqrt{64}} = \frac{s}{8}. For B: SE=s256=s16SE = \frac{s}{\sqrt{256}} = \frac{s}{16}. Researcher B's standard error is half as large, so the margin of error is smaller, producing a narrower interval. This demonstrates why larger sample sizes provide more precise estimates—tighter confidence intervals.

Explain using:

⚠️ Common Mistakes: Confidence Intervals for Means

Avoid these 3 frequent errors

📌 Related Topics in Unit 7: Inference for Quantitative Data — Means

❓ Frequently Asked Questions

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Construct and interpret confidence intervals for a population mean using the t-distribution.
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Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 3 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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Confidence Intervals for Means is part of the AP Statistics course on Study Mondo, specifically in the Unit 7: Inference for Quantitative Data — Means section. You can explore the full course for more related topics and practice resources.
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Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.