t-interval versus z-interval near a decision threshold
A nutrition study measures a biomarker in n=16 adults. The sample mean is 12.4 mg/L and the sample standard deviation is 3.2 mg/L. The population distribution is believed to be roughly symmetric but may have mild outliers, and the population standard deviation is unknown. Analyst A reports a 95% interval 12.4 ± 1.96(3.2/√16) = (10.83, 13.97). Analyst B reports 12.4 ± 2.131(3.2/√16) = (10.70, 14.10). A reviewer notes that the two intervals differ by about 0.14 mg/L on each side and asks which is defensible, what assumptions justify each, and how the conclusion would change if one observation were an outlier. The team must decide before reporting against a public-health threshold of 11 mg/L.
1. Which interval is appropriate? Identify the distributional assumption and the degrees of freedom.
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The t-interval is appropriate. The population standard deviation is unknown and estimated by s=3.2, and n=16 is small. Assuming the population is approximately normal, the pivot (x̄−μ)/(s/√n) has a t distribution with n−1=15 degrees of freedom. The correct 95% interval is 12.4±t₀.₉₇₅,₁₅(3.2/4)=12.4±2.131·0.8=(10.70,14.10). The z-interval is defensible only if σ were known or n were large enough for s to be a precise substitute; here it understates uncertainty.
2. Does either 95% two-sided interval establish that the population mean exceeds 11 mg/L? Explain.
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No. Both two-sided 95% intervals contain 11 mg/L: (10.83,13.97) and (10.70,14.10). A two-sided interval containing 11 means the data are compatible with μ=11 at 95% confidence. To claim μ>11, one needs a one-sided procedure or a lower confidence bound, not a two-sided interval that straddles the threshold.
3. Test H0: μ=11 versus H1: μ>11 at α=0.05 using the t-statistic, and compare with a z-test. What is the practical lesson?
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For H0: μ=11 versus H1: μ>11, t=(12.4−11)/(3.2/4)=1.4/0.8=1.75 with df=15. The one-sided 5% critical value is t₀.₉₅,₁₅=1.753, so t=1.75 is just below it; the one-sided p-value is about 0.0505, so do not reject at α=0.05. A z-test would use 1.645 and give p≈0.040, apparently rejecting. The lesson is that using z with unknown σ and n=16 can flip a decision near the threshold. Outliers would inflate s and make the t-interval even wider, so the claim is fragile.