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In statistics, the sample proportion is a critical measure used to estimate the proportion of a characteristic within a population based on a sample. To calculate this, you simply divide the number of individuals in the sample who exhibit the characteristic by the total sample size.
The sample proportion is denoted by \( \hat{p} \).
For instance, in the Google Glass exercise, none of the 500 surveyed people owned Google Glass. Thus, the sample proportion of people who do not own it is:

• Number of people without Google Glass = 500
• Total sample size = 500

The sample proportion is:\[\hat{p} = \frac{500}{500} = 1\]This signifies that, in our sample, 100% do not own Google Glass. However, this figure only represents our sample and not necessarily the wider population. Understanding this distinction is critical in statistical inference since our goal is usually to infer conclusions about the entire population.

The standard error (SE) provides insight into the precision of a sample statistic as a point estimate of a population parameter. For sample proportions, the standard error helps us gauge how much the sample proportion (\( \hat{p} \)) would differ from the true population proportion if we drew numerous samples. It's calculated using the formula for standard error of a sample proportion:\[SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\]In the given problem, with all 500 individuals sampled not owning Google Glass, our \( \hat{p} = 1 \).
Therefore, the standard error becomes:\[SE = \sqrt{\frac{1(0)}{500}} = 0\]This result indicates no variability in our sample proportion — everyone sampled fits the criterion of not owning Google Glass. However, such a perfect proportion doesn't portray realistic variability for the entire population, showing the limitations of using standard error for extreme proportions.

The Wilson score interval is a more accurate method for estimating confidence intervals, especially when dealing with extreme sample proportions (close to 0 or 1). Unlike the standard confidence interval, which might be misleading in these scenarios, Wilson's method adjusts the interval to account for the possibility of extreme values.
In the original problem discussed, using the large-sample confidence interval model led to a non-informative range due to \( SE = 0 \). The Wilson score model rectifies this by providing a smoothed approach through an adjusted sample proportion \( \tilde{p} \) and an adjusted standard error \( SE_{adj} \).\[\tilde{p} = \frac{n \hat{p} + (z^2/2)}{n + z^2}\]Here, solving gives:\[\tilde{p} \approx 0.995 \]Wilson's interval then calculates a confidence interval which reveals:\[CI_{adj} = \tilde{p} \pm z \times SE_{adj} \approx (0.993, 1)\]This interval suggests that nearly all surveyed (and by extension, possibly the broader population) do not own Google Glass. However, it leaves room to acknowledge that a tiny fraction might own it, offering a more realistic overview than the initial assessment.