# Point estimate of proportion

A study was conducted at the University of Waterloo on the impact characteristics of football helmets used in competitive highschool programs. There were three types of helmets considered, classified according to liner type: suspension, padded-suspension, and padded. In the study, a measurement called the Gadd Severity Index (GSI) was obtained on each helmet using a standardized impact test. A helmet was deemed to have failed if the GSI was greater than 1200. Of the 81 helmets tested 33 failed the GSI 1200 criterion. Assuming the suspension helmets tested were selected at random, calculate the point estimate of the proportion of suspension helmets that fail, and the standard error of the estimate.

Can I get some theory links that can help me solve this question? I don't know how to convert the entire population proportion of 33/81 to the particular category of suspension helmets.

Estimated probability of failure is $$\hat p = X/n = 33/91.$$ Then $$V(X) = np(1-p),$$ $$Var(\hat p) = p(1-p)/n,$$ $$SE(\hat p) =\sqrt{p(1-p)/n}.$$

Estimated standard error of $$\hat p$$ is $$\widehat{SE} = \sqrt{\hat p(1-\hat p)/n}.$$ Estimated standard errors are sometimes called simply 'standard errors when the estimation is obvious. They are often used to make confidence intervals.

For example, the (asymptotic) Wald 95% confidence interval for $$p$$ is of the form $$\hat p \pm 1.96\sqrt{\frac{\hat p(1-\hat p)}{n}}.$$

For your example: $$(0.264, 0.461).$$ [Computation in R.]

p.hat = 33/91;  p.hat + qnorm(c(.025,.975))*sqrt(p.hat*(1-p.hat)/91)
[1] 0.2638601 0.4614146


Especially for small $$n,$$ a more accurate Agresti-Coull 95% confidence interval uses point estimate $$\tilde p = (X+2)/(n+4)$$ and the CI is $$\tilde p \pm 1.96\sqrt{\frac{\tilde p(1-\tilde p)}{n+4}}.$$

For your data: $$(0.266, 0.460).$$

p.est = 35/95;  p.hat + qnorm(c(.025,.975))*sqrt(p.est*(1-p.est)/95)
[1] 0.2656372 0.4596375


Another good CI for $$p$$ is the Jeffreys interval $$(0.269, 0.464).$$ See the Wilkipedia article on binomial confidence intervals for more.

qbeta(c(.025,.975), .5+33, .5+91-33)
[1] 0.2693914 0.4644662
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• Per the question: "Can I get some theory links that can help me solve this question? I don't know how to convert the entire population proportion of 33/81 to the particular category of suspension helmets.", so this answer as it currently stands does NOT address directly the point (or its error) estimate for the subcategory of suspension helmets. All such particular helmets encompassed in the selected sample may have failed or even none! Dec 17, 2020 at 13:28
• Given my guess of the context of your question (from a few decades of teaching from elementary statistics texts), I believe my discussion and the Wikipedia link will answer your question. If you have something else in mind, please edit more particulars into your question. – Dec 18, 2020 at 20:52

In general, to address the so-called subclass estimation problem for proportions and associated error analysis, I reference this work Handbook on precision requirements and variance estimation for ESS households surveys.

Upon review of the cited source, my intuition relating to the current case given little information other than the overall proportion of 0.407 (33/81) is to assume a varying wide range of potential values. For example, select five possible values: 0.007, .107, .307, 0.407 and 0.5. For each of the latter assign class probabilities based on other sources relating to failure laws or intuition, say, 0.05, 0.10, 0.35, 0.4, 0.1. My intuition-based logic: very low subclass proportions are progressively less likely with a tilt to the mean or perhaps even higher.

Use all this as input to a Monte Carlo non-parametric bootstrap. However, per the source, to quote:

But the ordinary Monte-Carlo non-parametric bootstrap can lead to biased variance estimates when samples are drawn with unequal probabilities or without replacement...

However, bias would not be my main concern here, but gathering a rational precision estimate for a subclass failure rate and its associated sampling variability.

One can repeat the process with a different set of inputs based on other gathered information and examine the extent of variation in results.