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WAMath MethodsSyllabus dot point

How does the sample proportion behave across repeated random samples, and why is it approximately normal?

Describe the sampling distribution of the sample proportion, including its mean, standard error and approximate normality for large samples

WACE Year 12 Mathematics Methods Unit 4 the sample proportion: random sampling, the sampling distribution of p-hat, its mean p, the standard error, and approximate normality for large samples, with worked SCSA-style examples.

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  1. What this dot point is asking
  2. Random sampling and the sample proportion
  3. Mean and standard error
  4. Why approximate normality

What this dot point is asking

SCSA Unit 4 introduces statistical inference by studying how the sample proportion behaves across repeated samples. This dot point asks you to describe its sampling distribution, the mean, the standard error and the approximate normality, which underpins the confidence interval that follows. It is examined in both sections.

Random sampling and the sample proportion

When a random sample of size nn is drawn, the number of successes XX follows a binomial distribution X∼B(n,p)X\sim B(n,p), where pp is the unknown population proportion. The sample proportion is

p^=Xn.\hat{p}=\frac{X}{n}.

Different samples give different values of p^\hat{p}, so p^\hat{p} is itself a random variable with its own distribution, called the sampling distribution.

Mean and standard error

Because p^\hat{p} is 1n\dfrac{1}{n} times a binomial variable, its mean and standard deviation follow from the binomial mean npnp and variance np(1−p)np(1-p).

Since E(p^)=pE(\hat{p})=p, the sample proportion is an unbiased estimator: on average it equals the true proportion. The standard error has n\sqrt{n} in the denominator, so larger samples give less variable, more precise estimates.

Why approximate normality

For large nn the binomial distribution of XX is well approximated by a normal distribution, so p^=Xn\hat{p}=\dfrac{X}{n} is approximately normal too. A common rule of thumb is that both npnp and n(1−p)n(1-p) should be at least about 1010. This normal approximation is what allows the confidence interval to use a zz-multiplier.