Mathfolis

Confidence Interval for One Mean

Unit 7 · Inference for Quantitative Data: Means

What AP Stats asks here

Once σ\sigma is unknown — the realistic case — the t-distribution replaces the normal in the formula xˉ±t∗s/n\bar{x} \pm t^* s/\sqrt{n}. The t has heavier tails to account for using ss in place of σ\sigma; for small samples that matters a lot. AP problems test the construction, the interpretation (with units), and the use of the CI to evaluate a claim about μ0\mu_0.

One-mean CI

Estimate ± ME
xˉ±t∗⋅sn\bar{x} \pm t^* \cdot \dfrac{s}{\sqrt{n}}
Degrees of freedom
df=n−1\text{df} = n - 1

Conditions

Random sample
required\text{required}
Normality
population normal OR n≥30 OR symmetric sample, no outliers\text{population normal OR } n \ge 30 \text{ OR symmetric sample, no outliers}
10%
n<0.10⋅Nn < 0.10 \cdot N
AP Tip: Use t whenever σ is unknown — which is essentially always. The z-interval requires a known σ; in practice that is almost never the case.
Caution: A skewed sample with outliers and small n breaks the normality condition; the nominal 95% interval can capture far less than 95% of the time.
Type 1

Construct a t-CI

Use xˉ±t∗⋅s/n\bar{x} \pm t^* \cdot s/\sqrt{n} with df =n−1= n - 1. The t∗t^* depends on df and confidence level.

Example 1
An SRS of n=25n = 25 soup cans gives xˉ=12.05\bar{x} = 12.05 oz and s=0.18s = 0.18 oz; the boxplot is symmetric with no outliers. Which is the 95% CI for μ\mu? (t∗=2.064t^* = 2.064 at df = 24) (A) (11.976,12.124)(11.976, 12.124) (B) (11.954,12.146)(11.954, 12.146) (C) (12.000,12.100)(12.000, 12.100) (D) (11.988,12.112)(11.988, 12.112)

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Type 2

Interpret CI and include units

The interpretation is about the parameter (population mean), not individuals. Include units.

Example 2
A 95% CI for the mean fill weight of soup cans is (11.976,12.124)(11.976, 12.124) oz. Which interpretation is correct? (A) There is a 95% probability that the true mean is between 11.976 and 12.124 oz. (B) 95% of cans weigh between 11.976 and 12.124 oz. (C) We are 95% confident that the true mean fill weight of all cans in this lot is between 11.976 and 12.124 oz. (D) 95% of sample means will land in this interval.

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Type 3

Use CI to evaluate a claim

If μ0\mu_0 is inside the CI, the two-sided test of H0:μ=μ0H_0: \mu = \mu_0 does not reject at the matching α\alpha.

Example 3
The 95% CI for the mean fill weight is (11.976,12.124)(11.976, 12.124) oz. The label claims μ=12.00\mu = 12.00 oz. Can we conclude the label is wrong at α=0.05\alpha = 0.05? (A) Yes — the CI contains 12.00, which proves the label is wrong. (B) Yes — 12.00 is close to the lower end of the CI. (C) No — 12.00 is in the CI, so it is a plausible value and the two-sided test at α=0.05\alpha = 0.05 does not reject H0:μ=12.00H_0: \mu = 12.00. (D) Cannot tell without seeing the test statistic.

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