Unbiased Estimator vs Point Estimate

An estimator is unbiased when the mean of its sampling distribution is exactly the parameter, $E(\hat{\theta}) = \theta$, and when that holds for every possible value of the parameter rather than for one convenient case. The sample mean qualifies, $E(\bar{x}) = \mu$ (x-bar, mu), and so does the sample proportion, $E(\hat{p}) = p$ (p-hat). Unbiasedness pins down where the sampling distribution sits and says nothing about how wide it is.

Take the four-value population 1, 2, 3, 6, which has $\mu = 3$, $\sigma^2 = 3.5$ and $\sigma = 1.8708$. Write out all 16 samples of size 2 drawn with replacement and average each statistic across them. The sample mean averages to 3, exactly $\mu$. The sample variance with the $n-1$ divisor averages to 3.5, exactly $\sigma^2$. The sample standard deviation averages to $\sqrt{2} = 1.4142$, which is 75.6 percent of $\sigma$.

So this sentence is false: "the $n-1$ divisor makes $s$ an unbiased estimator of $\sigma$." What $n-1$ buys is $E(s^2) = \sigma^2$. Taking a square root is nonlinear, and the mean of a square root is not the square root of a mean, so the correction does not survive the step. $s$ comes out too small on average, at every sample size, for any population in which it varies at all.

The shortfall shrinks with $n$ without reaching zero: in the same population, all 1,024 samples of size 5 give an average $s$ of 1.7673, or 94.5 percent of $\sigma$. Inference for means uses $s$ in place of $\sigma$ regardless, because at ordinary sample sizes the gap is small and the exact correction depends on the population's shape.

Unbiasedness is also not the whole of a good estimator. Topic 3.1, Estimators, weighs it against variability: an estimator centered perfectly but scattered widely can be worse than one that leans slightly and lands close every time.

A point estimate is the value a statistic takes in the one sample you collected, offered as the estimate of a parameter you cannot observe. Each parameter has a standard partner: $\bar{x}$ (x-bar) estimates $\mu$ (mu), $\hat{p}$ (p-hat) estimates $p$, and $s$ estimates $\sigma$ (sigma). The word point is doing work in that phrase. It signals a single value with no width attached to it.

If 63 of 200 randomly sampled voters approve, the point estimate is $\hat{p} = \frac{63}{200} = 0.315$. That number alone says nothing about how far off it might be. Attach a margin of error and the picture changes: the standard error is $\sqrt{\frac{0.315(0.685)}{200}} = 0.0328$, so with $z^* = 1.96$ the margin is 0.064 and the 95% interval runs from 0.251 to 0.379. An estimate that looked exact carries a span of nearly 13 points around it.

The misreading is a missing word. "Thirty-one and a half percent of voters approve" states a sample statistic as though it were the population parameter. It is one draw from a sampling distribution centered at $p$, and the chance it equals $p$ exactly is essentially zero. Notation carries the distinction, so writing $p$ where you computed $\hat{p}$ claims something you did not compute.

An unbiased point estimate is still allowed to be far off. Unbiasedness is a property of the procedure across all samples, not a promise about this one, so knowing that $\bar{x}$ is unbiased does not license treating $\bar{x}$ as $\mu$. For a difference, the point estimate is the difference of the two point estimates, $\hat{p}_1 - \hat{p}_2$ or $\bar{x}_1 - \bar{x}_2$, and it sits at the center of the matching interval.

Topic 3.1 is Estimators, where point estimates are judged on bias and on variability.