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Statistical Inference Applications

Specialist Mathematics
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Statistical Inference Applications

Specialist Mathematics
01 May 2026

Applying Statistical Inference to Real Data

Key Assumptions

For the $z$-test / $z$-CI (mean):
- Population standard deviation $\sigma$ is known
- Either: population is normal, or $n$ is large enough for CLT ($n \geq 30$ typically)
- Sample is random and independent

For the $t$-test / $t$-CI (mean):
- Population is approximately normal (especially important for small $n$)
- Sample is random and independent
- Observations are independent (no repeated measures)

For proportion tests / CIs:
- $np_0 \geq 10$ and $n(1-p_0) \geq 10$ (for normality of $\hat{P}$)
- Random sample
- Population much larger than sample (finite population correction negligible)

Assumptions and Validity

If assumptions are violated, results may be misleading:

Violation Consequence
Non-random sample Bias; results may not represent population
Small $n$, non-normal data with $z$-test Incorrect p-values
$np < 10$ for proportion test Normal approximation poor
Dependent observations Underestimated standard error

Limitations of Inference

  • Statistical significance $\neq$ practical significance: A large sample can make any small effect significant.
  • Confidence level $\neq$ probability of capturing $\mu$: The CI is random; $\mu$ is fixed.
  • Cannot confirm causation: Inference from observational data shows association, not cause-and-effect.
  • Multiple testing: If 20 tests are run at $\alpha=0.05$, approximately 1 will be a Type I error by chance.

Full Worked Example with Assumptions

A nutritionist measures daily calcium intake (mg) for a random sample of $n=25$ adult women.
$\bar{x} = 820$ mg, $s = 120$ mg. Is the mean intake below the recommended 900 mg?

Assumptions: Calcium intake approximately normally distributed; random sample; $n=25$ so use $t$-test.

$H_0: \mu = 900$; $H_1: \mu < 900$. $\alpha = 0.05$.

$$t = \frac{820-900}{120/\sqrt{25}} = \frac{-80}{24} \approx -3.33$$

$t_{0.05, 24} = 1.711$ (left-tailed: compare with $-1.711$).

Since $-3.33 < -1.711$, reject $H_0$.

Conclusion: There is sufficient evidence at the 5% level that the mean daily calcium intake is below 900 mg.

95% CI: \$820 \pm 2.064 \times 24 = 820 \pm 49.5 = (770.5,\ 869.5)$ mg.

Effect Size

Effect size $d = \dfrac{|\bar{x}-\mu_0|}{s}$ measures the practical magnitude of a difference.

In the example: $d = 80/120 \approx 0.67$ — a moderate-to-large effect.

KEY TAKEAWAY: Always state and check assumptions before applying inference procedures. Report both statistical significance (p-value or CI) and practical significance (effect size or contextual meaning).

EXAM TIP: VCAA inference questions frequently ask you to identify and comment on assumptions. State them specifically — not just “the data is normal” but “the sample is a random sample from the population of interest.”

REMEMBER: Statistical inference provides a framework for decision-making under uncertainty — it cannot eliminate uncertainty or guarantee correct conclusions. Understanding limitations is as important as performing calculations.

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